Renal Microvascular Complications Of Type 2 Diabetes Essay

Renal Microvascular Complications of Type 2 Diabetes: Aetiology and Pathogenesis Abstract There have been a number of manuscripts reporting on the association of complications in type 2 diabetes with high glucose blood levels, high levels of C-Peptide, high advanced glycation end products (AGEs) and vascular cell adhesion molecule-1 (VCAM-1) and oxidative stress. In order to further investigate the aetiology and pathophysiology of renal microvascular complications in type 2 diabetes, papers were reviewed through 2000 using the NIH PubMed Literature Search System. Inclusion criteria were that manuscripts 1) be primary peer-review research article; 2) concisely explained, or investigated, the pathophysiology of renal microvascular complications in type 2 diabetes; 3) be published in English. High blood glucose levels and enhanced oxidative stress, found in both clinical and experimental diabetes, are thought to be the main cause of chronic diabetic complications. Advanced glycation end products (AGEs) resulted to be strongly correlated to renal microvascular complications via enhanced expression of adhesion molecules, including vascular cell adhesion molecule-1 (VCAM-1). Renal Microvascular Complications of Type 2 Diabetes: Aetiology and Pathogenesis Diabetes mellitus type 2 is a complex disease withShow MoreRelatedInvestigating The Aetiology And Pathophysiology Of Renal Microvascular Complications1094 Words   |  5 Pages There have been a number of manuscripts reporting on the association of complications in type 2 diabetes with high glucose blood levels, high levels of C-Peptide, high advanced glycation end products (AGEs) and vascular cell adhesion molecule-1 (VCAM-1) and oxidative stress. In order to further investigate the aetiology and pathophysiology of renal microvascular complications in type 2 diabetes, papers were reviewed through 2000 using the NIH PubMed Literature Search System. Inclusion criteriaRead MoreDiabetic Microvascular : Leading Cause Blindness762 Words   |  4 PagesDiabetic microvascular complications are the leading cause of blindness, end-stage renal diseases, and other neuropathies due to hypoxia and ischemia in the retina, the kidney, and nerves. Thickening of the capillary basement membrane result in decreased tissue perfusion. Many people with type 2 diabetes present with microvascular complications because of the long duration of asymptomatic hyperglycemia that usually precedes diagnosis (Mccaine and Huther). Diabetic Retinopathy Diabetic retinopathyRead MoreDiabetes Mellitus And The Long Term Complications1385 Words   |  6 Pagesthis paper is to give a general idea of diabetes mellitus, epidemiology, role factors and complications that arise from it, comparing and exhibiting the distinctions between type I type II diabetes, the people who are in jeopardy of developing diabetic renal diseases and hypertension due to the complications identifying the general pathogenesis of diabetes mellitus the long term complications that may transpire. Epidemiology of Diabetes Mellitus Diabetes Mellitus is one of the very prevalentRead MoreIntensive Glucose Control Of Patients With Type 1 Diabetes Mellitus Essay1606 Words   |  7 PagesTopic Intensive glucose control of patients with type 1 diabetes mellitus to maintain blood glucose levels (BGLs) near normal range Introduction Type 1 diabetes is an autoimmune disorder which can occur from insufficient insulin secretion. Insulin is a hormone which is secreted by the pancreas and regulates BGLs (Loghmani, 2005; Nussey Whitehead, 2001). Type 1 diabetes is also known as insulin dependent diabetes mellitus (IDDM). In this, the body produces no, or very little, insulin which canRead MoreType 2 Diabetes : An Overview Essay760 Words   |  4 PagesType 2 Diabetes: An Overview Diabetes mellitus is a disease that affects how an individual’s body metabolizes glucose (Blair, 2016; Franks, 2012; Scobie, McLean, Samaras, 2014; Spears, Schub, Pravikoff, 2015; WebMD, 2016c). People diagnosed with type 2 diabetes do not produce a sufficient amount of insulin to keep up with the demands of the body; which leads to hyperglycemia (Blair, 2016; Franks, 2012; Scobie et al., 2014; Spears et al., 2015; WebMD, 2016c). Glucose is the main source of energyRead MoreDiabetic Nephropathy Case Study1448 Words   |  6 PagesAim: Diabetic nephropathy (DN), classically defined by the presence of proteinuria is one of the major late microvascular complications of type 1 and type 2 diabetes mellitus and leading to a decline in renal function. In the present study, three important single nucleotide polymorphisms (SNPs) of the PPARG gene were analysed to understand the potential modifier effect of PPARG gene on the advancement of chronic kidney disease in DN. Methods: A total of 187 diabetic nephropathy patients (101 maleRead MoreThe Use of GLP 1 Analogues1608 Words   |  6 Pagesanalogues Patients with type 2 diabetes are at risk of macrovascular and microvascular complications. Evidence has shown that an improvement in blood glucose control, as shown by a reduction in HbA1c levels, is associated with a reduction in microvascular complications (2,3,4). Evidence from ACCORD (2), UKPDS(3) and ADVANCE(4) suggests that intensive blood glucose control reduced composite endpoints for microvascular outcomes, but the impact on patient oriented outcomes such as renal failure or dialysisRead MoreEssay On The Road To Health With Prediabetes1205 Words   |  5 Pagesbut not high enough to be diagnosed as type 2 diabetes. Despite advances in medical technology, treatments, and diagnoses, uncontrolled diabetes continues to rise in the United States (US) (American Diabetes Association [ADA], 2016). Between 2012-2014, 33.9 % of the US population were diagnosed with prediabetes (Center for Disease Control and Prevention [CDC], 2016). According to the ADA (2016) in 2010 18.8 million of the population was diagnosed with diabetes, 7 million were undiagnosed, comparedRead MoreUsing Sglt 2 Inhibitors With Insulin Essay729 Words   |  3 PagesUSING SGLT-2 INHIBITORS WITH INSULIN IN TYPE 2 DIABETES: CLINICAL IMPLICATIONS Mathew John, Deepa Gopinath Introduction Diabetes and its complications account for significant healthcare costs in developed countries. The control of hyperglycemia forms the cornerstone in managing diabetes. United Kingdom Prospective Diabetes Study (UKPDS) has shown that improving blood sugars can result in reduction in onset of retinopathy, microalbuminuria and neuropathy. Follow up of the subjects in UKPDS afterRead MoreHealth Disease Choice : Diabetes802 Words   |  4 PagesHealth Disease Choice - Diabetes Type 2 diabetes is one of the most common chronic conditions encountered in the primary care setting. It is a lifetime condition that requires careful management to prevent debilitating complications, such as peripheral neuropathy, amputations, diabetic nephropathy, kidney failure, diabetic retinopathy, and blindness. It is also a cardiovascular risk factor to heart attack and stroke. Basing on my experience with my patient encounters as a nurse, most diabetics do

High Noon Analysis - 938 Words

Honor, abstract in definition, is elucidated by communities and its individuals. It is a changing force, that can be approached from many different perspectives and has been interpreted countless ways. Honor used to be interpreted as a code that members of a community would follow in order to be recognized with esteem and respect. Over time, honor has been determined by the individual, regardless of other interpretations. An individual would build their honor manifesto on the values they deem necessary to live an honorable life, honorable according to their definition. In The Crucible, by Arthur Miller, John Proctor prioritizes love. Will Kane, from the film â€Å"High Noon†, directed by Fred Zinnemann, values responsibility. John Proctor and†¦show more content†¦He was the one that originally arrested and agitated Miller, therefore, he felt it was his responsibility to address the consequences. After he had killed Miller and the additional accomplices, thus completin g his personal mission, he was able to throw away the tin star and leave everything behind. That tin star symbolizes his job and the town, and by throwing it away, he demonstrates that he no longer cares or feels the need to stay. This display shows that the town, nor its inhabitants, were his source of motivation. He may have acted selflessly, but he did so for his own personal need of fulfilling his duty. Both characters developed a personal definition of honor, and both characters were rejected by their communities as result of living in accordance with their manifesto. The communities of Salem and Hadleyville, each community belonging to a unique time period and environment, has their own code of honor. However, both communities overlap in their need for self-preservation and greediness. In Salem’s case, when the witch trials erupted, many people, supposedly Mr. Putnam being one of them, and Abigail another took this as a chance to gain power and prey on those they hated. In fact, the overall community took it as a chance to spread hate and distrust. It was at this time, the community lackedShow MoreRelatedCritical Analysis of Film785 Words   |  4 PagesGive a critical analysis of the Western Holly Wood film High Noon. This essay will focus on the current representation of women and men in the classical Holly Wood western film High Noon, focusing on the gender roles of each character and the stereotypical roles that are given. High Noon  is a 1952 Western film  directed by  Fred Zinnemann, one which broke genre rules of masculine ideals and popular themes of cowboys and indians (Johans;1994). The male protagonist Marshal  Kane (Gary Cooper  ) startsRead MoreJ. Galsworthy. the Broken Boot1005 Words   |  5 PagesA Sample of Complex Stylistic Analysis J. Galsworthy. The Broken Boot (E.M. Zeltin et. Al. English Graduation Course, 1972, pp.88-89: finishing with the words .. .walked side by side.) Text Interpretation The passage under analysis is taken from John Galsworthys story The Broken Boot. It is about an actor whose name is Gilbert Caister. For six months he had been without a job and a proper meal. He ran into a man whom he had come to know in a convalescent camp, a man who thought aRead MoreJ. Galsworthy. the Broken Boot1021 Words   |  5 PagesA Sample of Complex Stylistic Analysis J. Galsworthy. The Broken Boot (E.M. Zeltin et. Al. English Graduation Course, 1972, pp.88-89: finishing with the words .. .walked side by side.) Text Interpretation The passage under analysis is taken from John Galsworthys story The Broken Boot. It is about an actor whose name is Gilbert Caister. For six months he had been without a job and a proper meal. He ran into a man whom he had come to know in a convalescent camp, a man who thought a lot ofRead MoreIndividual Report (50%)806 Words   |  4 Pagesmain features of the marketing plan. The new market that you recommend is your choice but it must not be one where the Company is already present. The report should be 3000 words max and the hand in date is 13th March 2013 via Blackboard before 12 noon. Your recommendation needs to contain the following: 1 Reasons for selection of new market for entry by the Company 2. Alternative markets which were candidates and reasons not selected 3 Proposed market entry strategy, including rationale Read MoreA Brief Note On Industrial And Organizational Psychology1819 Words   |  8 Pagesinsights on increasing productivity in the workplace and related matters such as physical and mental wellbeing of employees (Youssef Noon, 2012, sec.1.1). Industrial organizational psychology is known to be the most important asset in the organization, in which the factors are many; influence, performance, creation, strategies, innovation, and superiority (Youssef Noon, 2012, sec.1.1). These levels perform a wide variety of tasks in industrial/organization psychology involving studying worker attitudesRead MoreEssay On Eta1356 Words   |  6 Pagestake a wider Determination of dry and wet edges in the LST– NDVI scatter is necessary, to estimate pixel by pixel ET and EF using Eqs. (4) and (5). In arid and semi-arid areas, it should be noted that, for given vegetation cover, spatial pixels with high surface temperature and low EF are detectable by satellite remote sensors. On the other hand, the saturated soil water which evaporates potentially pixels is rarely and hardly existed in these conditions (see red lined triangle inside Fig. 2). ObtainingRead MoreJob Description Essay755 Words   |  4 PagesDescription RFR No. HRMT 70002 Job Position (Specialist Type): On-line/Cloud Project Coordinator Quantity: 1 Estimated Start Date: Immediately Employment Type: Full Time Reports to: Raja Ramanathan Issue Date: November 14 2017 Closing Deadline: 12:00 noon (local Toronto time) November 30th, 2017 1.0 Summary Coordinate activities related Technical Business SMEs in the development and implementation of SharePoint Office365 with MS Azure in the Cloud Project: - Assist in a Pilot to determine feasibilityRead MoreEmpirical Relationship between the Price of Gold and Three Other Variables1221 Words   |  5 Pagescharacteristics. It’s popularly used to decorate luxurious jewelries. With the improving standard of living and increasing income, people consume demand more and more gold. Because its unique physical properties of gold, it’s also widely used in modern high-tech industries such as electronics, telecommunications, aerospace, chemical and so forth. In the history, gold has been used as measure of value, means of circulation, means of payment and even world currency where major currencies were tied toRead MoreEssay about Analysis of German Film Run Lola Run906 Words   |  4 PagesAnalysis of German Film Run Lola Run Run Lola Run, is a German film about a twenty-something woman (Lola) who has 20 minutes to find $100,000 or her love (Manni) will be killed. The search for the money is played through once with a fatal ending and one would think the movie was over but then it is shown again as if it had happened ten seconds later and changed everything. It is then played out one last time. After the first and second sequence, there is a red hued, narrative bridgeRead MoreEquity and Discrimination at Big Red Kangaroo Airlines1389 Words   |  5 Pagescomplex and dynamic industry, characterized by numerous elements. For instance, the airline industry is now, unlike its inception period, populated by privately owned companies, rather than state owned enterprises. Then, the airline sector is marked by high levels of financial resource consumption, employment concerns or pollution complaints. As a result of the 9/11 terrorist attacks, the airline industry was faced with a decreasing demand, leading to numerous bankruptcies in the sector. Additionally

Most Significant Events of Each Decade Free Essays

Most Significant Events Final Project K A University of Phoenix: Axia College Jonathan Tietz November 28, 2010 Most Significant Events of the 50`s, 60`s 70`s 80`s and 90`s World War II lasted from 1939 to 1945. This was a war that involved most of the world’s nations and all of the world’s greatest powers. When the war ended the events that followed over the next five decades had a great effect on the American people. We will write a custom essay sample on Most Significant Events of Each Decade or any similar topic only for you Order Now There are some events that had more of an effect than others. This paper will discuss five major events over five decades that has had a powerful effect on the American people.Beginning in 1950 and ending in 1990, this paper will discuss the most significant events from each decade that either positively or negatively changed the American way of living. The communist scare took place in the 1950`s. McCarthyism, named after a man named Joseph McCarthy was a republican U. S. senator. In the 1950`s anticommunism created fear among the American people (McCarthyism, 2006). McCarthy played on Americans fears in an effort to better his political campaign. He instilled this fear by convicting anyone who was a part of the communist party or had anything to do with it.During court hearing Americans remained silent so that they wouldn`t be accused of communism. He accused some of the United States federal government of being communist and soviet spies. McCarthy was unable to prove his claims and was therefore censored by the American government (McCarthyism, 2006). Americans lost their jobs if they were accused of communism, library books were burned to hide evidence of communist acts. Americans suffered greatly during this time as they had to walk on around on egg shells (McCarthyism, 2006). IfAmericans did not agree with anticommunism, they were considered to be communist and were punished for not supporting the American way. People feared McCarthy, but it all ended when he made a public mockery of senate procedures (McCarthyism, 2006). He ended his career to be known as reckless and dishonest man. Americans want peace took place in the 1960`s. The Vietnam War was well overdue considering some events that took place after World War II. How did America get involved in the Vietnam War? It all started with the Atlantic Charter. Franklin D. Roosevelt and Winston S.Church hill created this charter in hopes of a better world. Russia and China were not part of this charter and ultimately lead the U. S. to believe that Russia and China were involved in colonialism. Communism increased in South East Asia, mostly in Korea, Vietnam, China, and Cuba. The U. S. took action and attacked and conflict arose with Korea. The U. S. gets involved in Southeast Asia`s politics. North Korea invades South Korea. Nothing is accomplished and North Korea remains the same and South Korea remains the same. Next the U. S. inhabits South Vietnam.The U. S. helps rebuild South Vietnams economy. The U. S. tried to gain control over South Vietnams politics to avoid Vietnam turning communist. In 1965 North Vietnam attacked South Vietnam and American bases. North Vietnam lost the Vietnam War because they were no match for the American troops. The war had a dramatic effect on the American way of living. Americans rallied for peace and fell into the hippie era during the Vietnam War. During this time sex and drugs were on the rise. New drugs were introduced to the American people as the drug population grew.People began to have more sex during this time, which may have something to do with the increased drug use. The hippie era was a way of Americans expressing themselves of how they felt about the war. Many people wrote songs that told stories of things that were going on during that time. Mainly people wanted the war to end. Americans wanted peace, and sex, drugs and music was their way of getting away from it all. Detente took place in the 1970`s. Nixon`s visit to china in 1972 was an important step to build a relationship between America and China. This was the first time an American president had visited China.President Nixon visited China from February 21-28, 1972. Nixon reaffirmed interest for a peaceful settlement, and the U. S. continued to have relations with the Republic of China. Nixon was a strong advocate against communism so it surprised everyone when he went to visit china. He felt that if all nations cooperated with each other, then they could have reduced revenue and prevented a third world war. Nixon`s engagement with the communist was called Detente. Nixon held diplomatic conversations with China to establish a relationship and use that relationship against the Soviet Union.Nixon began the Nuclear Arms Reduction Treaty. Soon after came the end of the cold war. All About the Benjamin’s, took place in the 1980`s the decade of corporate greed. During this time Ronald Regan was the president and he was looking for a way to improve the American economy. His economic policy focused on focused on four points. First Regan wanted to reduce government spending, reduce income, reduce government regulation and control the money supply (Reagonomics 2010). His approach involved reducing tax cuts for wealthy Americans, and cutting funding for lower class Americans ( Reagonomics 2010).Regan felt if he cut funds for people on welfare that those Americans would try to get jobs. He introduced the earned income credit. He felt that this tax credit would be a way to encourage the unemployed to get out and work. Regan economic policies had both positive and negative effects on the American way of living (Reganomics 2010). The wealthy became richer and the poor became poorer. The unemployment rate declined but there was an increase in homeless and hungry Americans. Regan thought that he could create wealth for the U. S. by allowing business owners and free market corporations to compete for wealth.Reagan lowered the oil windfall profit tax. He lifted the petroleum price, deregulated airlines, and most of the airlines went bankrupt. He thought he could fix things that he could not fix. His motives were in the right place but the actions he took were not completely in the Americas best interest. Many Americans suffered during this time, but still Reagan was elected for a second term. He was liked by many. Some believe that Reganomics benefited America. Some of the policies introduced by Reagan are still being used today. Some of Reagan’s policies served as a blueprint on different ways to better the conomy. The Never Ending War took place in the 1990`s. In the books it is written that the gulf war began in August 1990 and ended in February 1991, but a war in Iraq still goes on today (1990`s, 2010). The gulf war was a war waged by the U. N. led by the United States and The United Nations against Iraq. Iraq troops invaded Kuwait in 1990 and brought immediate economic sanctions against Iraq. United States president George H. W. Bush sent American soldiers to Saudi Arabia six months later (1990`s, 2010). Many nations joined the coalition with America but America had the majority in military forces.Following The United States was Saudi Arabia, The United Kingdom, and Egypt (Gulf War, 2010). There had already been friction with Iraq and the United States that date back to the Cold War. Iraq was an ally of the Soviet Union (Gulf War, 2010). The United States had a concern involving Iraq`s position on Israel and Palestinian politics because Iraq disapproved of peace between Israel and Egypt. On August 12, 1990 Saddam Hussein wanted to compromise (Gulf War, 2010). He requested an immediate freeze of all boycott and siege decisions and wanted normalization of relations with Iraq.The United States expressed that there would be no negotiations until Iraq came out of Kuwait. Many resolutions were passed regarding the invasion made by Iraq. The most important was Resolution 678, passed in November of 1990, which set a deadline for Iraq to withdrawal from Kuwait (Gulf War, 2010). The deadline was for January 15, 1991. Ultimately if Iraq did not withdrawal but that time, they would have to be forced out. This brings us back to the coalition. Some countries did not join the coalition but sent money in support of it. Some countries did not want to increase U. S. nfluence in the Middle East (1990`s, 2010). In the end many nations were persuaded. Nations were promised economic aid, debt forgiveness or threats to withhold aid. Today the war in Iraq continues. There are still American troops overseas. How long will this war continue? It is unknown, but what is known is that this war has been indirectly been going on for decades, and all other wars and significant events before this war ultimately led to this one. So what is going to happen to America for the years to come? How will our current president deal with current issue that are results of past presidents decisions?America still has the same economic problems that Regan faced. We are still at war with Iraq even after the first President Bush left office. In the future I see the war continuing. After the attacks of 9/11 I don’t think that The United States government would feel comfortable if they pulled all of the soldiers out of Iraq. I see the United States falling into a deeper depression. After all we are dealing with years of economic recovery. The unemployment rate over the past years has dropped. It is extremely hard for some people to get jobs. What is so hard to understand is how the United States began as one of the strongest countries economically, socially and politically and now we are falling so far behind. The United States is falling behind in education and this used to be the leading country in education. The United States has fallen behind in production. How could a country that is not social developed, compete with other countries that are so far ahead? It is unknown where the United States will be a decade from now but based on the past compared to now, we might be dealing with same old issues.It may be worst. There may be more homeless, hungry, unemployed Americans a decade from now. The way the deficit is Funding for welfare programs may get cut again. Things are still the same. The rich are getting richer and the poor are getting poorer. If things do not soon change, there will be no hope for America’s future.Works CitedMcCarthyism. (2006, 8 23). Retrieved 11 23, 2010, from pbs. org: http://www. pbs. org/wnet/americanmasters/episodes/arthur-miller/mccarthyism/484/ 1990`s. (2010).Retrieved 11 28, 2010, from Wikipedia: http://en. wikipedia. org/wiki/1990s Gulf War. (2010).Retrieved 11 27, 2010, from Wikipedia: http://en. wikipedia. org/wiki/Gulf_War Vietnam War. (2010).Retrieved 11 23, 2010, from wikipedia: http://en. wikipedia.org/wiki/Vietnam_War How to cite Most Significant Events of Each Decade, Papers

Things Not Seen free essay sample

Things Not Seen Author: Andrew Clements Genre: Fiction Favorite Character amp; Why: My favorite character in this story was Robert Phelps also known as â€Å"Bobby†, because Bobby feels invisible to his parents, even before he actually is invisible. He feels like they put work before him, and don’t notice what he wants or who he is, which I can relate to. Reasons I liked this Book: The reasons I liked this book is because it shows how this teenager has to deal with being invisible and not knowing if he’ll ever go back to normal. Also because this book makes you think about how you would feel and deal with the fact knowing that you’re invisible and no one can see you. Summary: Fifteen year old Bobby Phillips wakes up one morning and discovers he’s become invisible. After he proves to his parents that je has turn invisible he is forced to stay at home and not to tell anyone that he has become invisible. We will write a custom essay sample on Things Not Seen or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Then Bobby’s parents get into a car accident that causes them to be hospitalized. So Bobby then meets a blind girl named Alicia in a library. And later on learns that about Bobby’s â€Å"condition† that he has. Later on the law gets involved and investigates why Bobby has not been seen, which could end up with his parents in jail. On this difficult journey that Bobby has to face with his parents and his new friend Alicia by his side to help him find out why he has become invisible. 1st Quote: I cant see myself! You cant see me. I cant be seen like, Im invisible! † (3) This quote is important because it shows how scared Bobby is and how he reacts to his parents about him not being able to see himself. This can give you an idea on how he will deal with him being invisible later on. 2nd Quote: â€Å"But then come the tears. Mom slumps down in her chair and starts crying, and I cant take that. I can never take that. †   (6) This shows how Bobby’ s Condition is begging to affect his mother.

10 Signs It’s Time to Quit Your Job

10 Signs It’s Time to Quit Your Job Let’s be honest: many of us aren’t in our dream jobs, for whatever reasons. And even if you are working in your ideal field, there’s a good chance that the experience isn’t what you daydreamed it would be. It’s important to know when it might be time to cut bait and start over in a new role. 1. You dread going to work in the morning.If you hit your snooze button 15 times or roll out of bed every morning frowning about what your day holds, this is a problem. Even the most chipper coworker in your office has less enthusiastic days, but if it becomes an everyday dread, this could impact your overall happiness and health.2. You can’t hide your disdain at work.If other people are noticing that you are cranky or unhappy, it’s not good. It could impact your relationship with your boss and be noted as a performance issue.3. You dislike your team.If you have issues working together with your immediate group on projects or their everyday habits are like nails on a chalkboard to you, the problem might not be them. It could be that you would fit in better somewhere else.4. You dislike your team leader.There are plenty of terrible bosses out there: mean, arrogant, and just plain incompetent. It’s also possible that he or she is a great person, but you just can’t flourish under their style of management. If you find yourself rolling your eyes every time you get an email from this person, it could be time to leave.5. Your personal life is affected by your job.Feeling overworked and unhappy can drag down other parts of your life. This can show itself in a variety of ways: feeling short-tempered with family or friends, having issues with sleep, or feeling anxiety over things that may not seem work-related. If you find that your general feeling of well-being is lower because of your work activities, it’s better to err on the side of self-interest.6. Your health is affected by your job.Stress-related illnesses are very much a thing. Working too hard or experiencing consistent stress can make you more susceptible to colds, flu, or any number of illnesses just waiting for a gap in your immune system caused by poor self-care. Anxiety disorders and depression are also conditions that can be made worse by staying in a job that causes you consistent stress.  Few jobs are forever, but your health is always going to be with you.7. Company morale is low.Shared misery can be a rallying point for coworkers when things are rough, but it could be that everyone is unhappy because something is seriously wrong at the upper management level. Even if you feel loyal to your company, it’s important to keep an eye on how things are going in general and to decide whether any issues are likely to be resolved in the short term. If not, you are not obligated to stick around if there are other opportunities.8. You don’t see how this job will advance your career.If you’ve moved up as far as y ou can in your current role without major personnel or company changes, consider whether there would be better chances for advancement somewhere else. Be proactive, instead of waiting patiently for someone else to retire or quit.9. You’re consistently bored at work.If you find yourself slacking or filling long hours between projects, it may be because you’re not being challenged enough by your job. Every job will have its boring moments, but overall it should be a role where the tasks make you feel engaged and productive. If that’s not happening, this job likely isn’t the right fit for you.10. You feel undervalued.We all have moments where we grumble that we aren’t being paid enough to deal with this *bleep*, but if you feel like your job responsibilities have seriously outpaced your paycheck, it’s time to re-evaluate. Do some research into salary ranges and compensation for your role at other companies. If you still feel like you are being under-compensated (and there’s no chance to negotiate more for yourself at your current place), it’s probably time to move on.

STUDIES ON USE OF AQUATIC MACROPHYTES FOR removal of pollut essays

STUDIES ON USE OF AQUATIC MACROPHYTES FOR removal of pollut essays Since the last decade Effluent Treatment Plant has become an integrated part of each and every industry that disposes off wastewater. Not only just industries but Domestic and Municipal waste also have to be treated before disposing as all these effluents contain pathogenic microorganisms, toxic organic For removal of these chemicals various conventional methods are used. But these methods were found to be uneconomical and in absence of regular maintenance, the processes are delicate and prone to failure. Besides, these methods cannot be applied where volume of effluent is low like the municipal waste in rural areas. Lower the quantity of wastewater, higher is its cost of treatment per litre. As a result, in most of its cases it is left untreated and that leads to outbreak of water borne diseases. To all these problems Biotechnology has brought one simplest, easiest 5. It can be applied in varied conditions and environment. It is found that the macrophytes perform several different functions: 1. Take up minerals thereby lowering toxicity of the effluent. 2. Transporting from atmosphere to root zone facilitating aeration of the effluent. 3. Provide substrate for microbial activity. 4. Remove Nitrates, Phosphates, sulphates, Ca, Mg, K 5. Remove heavy metals like Cu, Pb, Cd, etc that cannot be completely removed by conventional methods. Since the introduction of this technique a decade ago, it was mainly used for secondary and tertiary treatment of sewage. However, later it was found to be equally effect ...

Hedge accounting under IAS 39 and IFRS 9 - A critical comparison Dissertation

Hedge accounting under IAS 39 and IFRS 9 - A critical comparison - Dissertation Example IFRS 9 is still in its development phase and such studies help in changing accounting methods from a rules-focused approach to more of a principle- based view; this is done to encourage more personalities and institutions to apply hedging as it has benefits associated with good pricing and effective financial reporting. All the selected 32 hedge fund managers participated in the survey and the researcher’s contribution helped in bringing them up to speed with the scope of the study and ensured these participants were previously using IAS 39 for relevance purposes. The researcher also queried the achievements and financial performance across all accounting spectrum since these UK based hedge fund managers started applying IFRS 9. ... 41 Results 41 Demographics 41 Research Questions and Hypotheses 41 Quantitative aspects 42 Qulaitative analysis 45 CHAPTER 5: SUMMARY OF FINDINGS, CONCLUSIONS AND RECOMMENDATIONS 49 Summary of findings 49 Data collected revealed that a majority of the respondents were familiar with the provisions of IFRS 9 having encountered the instrument in their line of work. The proposed changes, therefore, had a direct effect on how the respondents and their organizations would operate upon full adoption. From the analysis of participant opinions, majority had in-depth knowledge of IFRS 9 and IAS 39 judging by the precision of their responses. This was attributed to both their routine responsibilities within the various Hedge Funds and level of education that inclined towards advance financial management for a majority of the sample population. It was therefore fair to form a preliminary conclusion that the respondents were authorities in their own individual rights within the region of hedge ac counting. 49 With the majority of respondents willing and ready to fully move from IAS 39 to IFRS 9, opening conclusions were drawn that the new system was found acceptable by the bulk of industry players making the instrument much better that IAS 39 on the strength of its acceptability. This was further supported by the present rate of the instrument’s application where it was revealed through the collected data, that 97% were actually enjoying larger provisions of the system. Revelations of IFRS 9 eliminating 80-125 effectiveness testing rule and hedging net options stood out as significant positive changes that the new instrument would bring upon adoption. 49 Conclusions 52 Recommendations 53 Bibliography 55 Abstract This study will seek to provide a

Training Module Essay Example | Topics and Well Written Essays - 500 words

Training Module - Essay Example It will be part of their work duties. The training will teach the workers how to maintain a healthy lifestyle outside of the workplace to foster a better performance inside the workplace. The purpose of conducting this training is partially to reduce drama in the workplace. Many people come into the workplace with lots of drama issues surrounding their personal lives, which is not acceptable in the workplace. Relationships are important, but details about private relationships should be kept just that—private. People who are working with other workers do not want to hear their co-workers talk about how much someone had to drink the night before, their problems with their significant other or others, and who is sleeping with whom. It’s quite unprofessional and does not reflect the kind of worker that the company wishes to employ. Workers who do not abide by the work/life balance training will be considered for termination or face some kind of penalties. This work/life balance training module has several objectives. One of the objectives is getting employees to â€Å"prioritize† tasks in their lives so that their lives will be more efficient, and thus make them better workers.1 People who prioritize their tasks appropriately are more able to complete their work tasks with more efficiency. Another objective will be to make sure that employees complete their tasks by the stated deadlines. Two other objectives (which now total four) include having employees develop respect for themselves as well as others by being polite and extending various social courtesies and graces. While many people believe this could be a sort of morality training, it is basically training to help balance life at work and at home. The main issue that will be dealt with in meeting the goals or objectives is to try to stop workers from bringing home life drama into the workplace and to realize that there is a difference between

Text and tradition Essay Example | Topics and Well Written Essays - 1000 words

Text and tradition - Essay Example Socrates objects, pointing to existence of various Gods and, as a result, different ways of pleasing them; there is a logical contradiction when one the same action can be represented at the same time pious and improper: Later Euthyphro modifies definition by saying that pious is something that causes unanimously positive reaction in all Gods without exception (and vice versa). Socrates in reply formulates "Euthyphro`s dilemma": the act is pious because it is dear to Gods or it is dear to Gods because it is pious in nature? â€Å"And a thing is not seen because it is visible, but conversely, visible because it is seen; nor is a thing led because it is in the state of being led, or carried because it is in the state of being carried, but the converse of this. And now I think, Euthyphro, that my meaning will be intelligible; and my meaning is, that any state of action or passion implies previous action or passion" Socrates offers his own variant of the definition combining piety with justice. However, reflections on this concept, make the interlocutors reject such definition because all pious is fair, thus not all fair is pious. Attempt to specify the concept of justice leads Socrates and Euthyphro to the thought that justice involves interaction with Gods in this or that sense. Euthyphro formulates the definition according to which piety is intervened with sacrificing and praying. However, both interlocutors agree that sacrificing is reasonable in case when someone needs gifts while Gods cannot require anything, and, therefore, the acts of praying and sacrificing do not make sense and cannot solely characterize piety. On the contrary these acts remind trade. As a result Socrates and Euthyphro return to the thought that piety is a gratification to God. The general sense of "Euthyphro" is clear. It is correct that piety is a gratification to Gods but it is not exact. It is also correct that piety is an aspiration to justice but it is not precise as well.

Chemistry of Essential Oils

Chemistry of Essential Oils Rhea Hughes Table of Contents 1. Introduction 2. Basic Chemical Structure 2.1 Hydrocarbons Monoterpenes Sesquiterpenes 2.2 Oxygenated Compounds 3. Extraction Methods 4. How essential oils are analysed 5. Discussion 6. Conclusion 7. Bibliography 1. Introduction Essential oils are used as alternative medical treatments, fragrances for perfumes and also as flavours in food and beverages (Djilani Dicko, 2012). They are made up of fragranced mixtures that are found in different parts of plants such as the seeds, stems or flowers. (doTERRA, 2014). In this essay the following topics of essential oils will be discussed; the chemical structure, how they are extracted from plants and how to determine their components through analysis. 2. Basic Chemical Structure Essential oils are extracted from plants as the compounds that make up essential oils are created naturally by chemical reactions that occur in different plants. Essential oils are released when oils are chemically extracted from the herb or when the herb is compressed (EBSCO, 2014).There are two groups of chemicals that are made from nature; they are classified into primary and secondary metabolites. Primary metabolites can be divided further into carbohydrates, lipids, nucleic acids and proteins. Secondary metabolites are divided into alkaloids, polyketides, shikimates and terponiods (Baser Buchbauer, 2010). The secondary metabolism of a plant is responsible for their scent and also plays a role in the defence system of the plant protecting it from pests Essential oils are made up of hydrocarbons and their derivatives oxygentated compounds that are created from plants secondary metabolism (Chamorro et al., 2012). As it can be seen in Figure 1 in the biosysthesis of secondary metabolites, water and carbon dioxide are converted to glucose through photosynthesis. Phosphoenolpyrutave is a main element in the shikimate group of natural products. Decarboxylation of phosphoenolpyrutave produces acetate which is esterified with coenzyme-A to produce acetyl CoA. Acetyl CoA is a starting point for Mevalonic Acid, which is the starting compound for terpenoids. Figure 1 shows the chemical structure of the biosynthesis of secondary metabolites. (Baser Buchbauer, 2010). Figure 1: Biosythesis of secondary metabolites (Baser Buchbauer, 2010). Essential oils molecules are prepared mainly from carbon, hydrogen, and oxygen (essentials, 2014). Essential oils can be divided into two groups hydrocarbons and their derived oxygenated compounds (Martin, 2014). The hydrocarbon group is divided into monoterpenes and sesquiterpenes. Hydrocarbon chains are kept together by carbon atoms that are linked together. At different points in the chains atoms are attached to make up alternative oils (essentials, 2014). The oxygenated compounds can be broken down into smaller compounds such as Phenols, Alcohols, Aldehydes, Ketones, Esters, Lactones, Coumarins, Ethers and Oxides (EsotericOils, 2014). 2.1 Hydrocarbons Terpenes are made up of isoprene units. Isoprenes are five-carbon molecules. Isoprenes are assembled in different formations to make up terpenes (Cyberlipid, 2014). Terpenes are a group of molecules that is based on a number of isoprene units in a head to tail fashion. Figure 2 : Isopene Unit (Cyberlipid, 2014) Modified terpenes (terpenoids) are where methyl groups have been moved or removed or oxygen atoms added to the structure. Terpenes are easily decomposable under different effects such as air light and moisture which can cause spoilage of the oil. Terpenes have a narrow boiling range, making them difficult to purify. (Parry, 1922). Monoterpenes and sesquiterpenes are the main components of essential oils as they are adequately volatile (Cyberlipid, 2014), (Baser Buchbauer, 2010). Monoterpenes Monoterpenes have 2 isoprenes units linked together. Monoterpenes are very volatile due to the molecular structure (Chamorro et al., 2012). Monoterpenes increase the therapeutic functions of other components in the oil and stop the collection of toxins (Martin, 2014). An example of an essential oil that is a monoterpene is menthol which is a cyclic monoterpene. Menthol has cooling properties and has a unique scent of the oil residue that is extracted from it (Kamatoua et al., 2013). There is a difference in the content of methone and menthol detected, as the plant mature there is a higher content of menthol and menthyl acetate content (Baser Buchbauer, 2010). Figure 3: Steoreoisomers of menthol (-) –menthol is the most commonly used synthetic and natural form. Menthol has microbial properties but is not a principal compound in essential oils as it is only a component in of a restricted number of aromatic plants (Kamatoua et al., 2013). Sesquiterpenes Sesquiterpenes have 3 isoprene units linked together. They are common in essential oils and are less volatile than monterpenes so they blend well with volatile oils (Martin, 2014). An example of a sestquiterpene is Zingiberene that is used in the oil. It is acquired by fractional distillation under reduced pressure (Parry, 1922). Figure 4: Structure of Zingiberene (Chemistry, 2014). 2.2 Oxygenated Compounds Oxgenated compounds or terpenoids are derivates of terpenes. They have a stronger aroma and are normally more stable as they do not oxidise as easily under different conditions (Fresholi, 2014). Some examples of oxygentated compounds are alcohol, ketones and esters. Alcohols can further be divided into monoterpene alcohol and sesquiterpene alcohols (EsotericOils, 2014). Alcohols are found in in their free state and in the form of esters in essential oils. Methyl alcohols which are soluble in water which are main components in essential oils are removed during the distillation process so they are not actually found in the essential oil but can be found in the distilled water. But when in ester form methyl alcohols can be found in essential oils such as winter green. (Parry, 1922). Alcohols are beneficial components that have antimicrobial and antiseptic properties. Esters are naturally occurring in plants. Esters such as terpineol attribute to the aroma in essential oils (Pharmacognosy, 2012). They also have calming and sedative properties (Oils, 2007). Linalyl acetate is a naturally occurring ester that is found herbs and flowers. It is used to make Lavender oil (Hermitageoils, 2014).Ketones have skin healing properties and also helps break down fats (Oils, 2007). Verbenone is an example of a natural ketone that is extracted from plants. It is an ingredient in the oil Rosemary Other examples of oxygenated compounds are aldehydes that have anti-inflammatory and anti-infectious agents such as geraniol found in Rose Geranium. Oxides that have anti-inflammatory properties such as eucalyptole are found in Eucalyptus. Phenols have anti-pathogenic properties (Oils, 2007). Thymol is an example of a phenol, it is a compound of thyme and ajowan seed oil (Parry, 1922) and ethers help regulate hormones and the central nervous system such as chavicol found in basil (Oils, 2007). 3. Extraction Methods There are different extraction methods that are used for the extraction of components for essential oils. There are two main types of extraction methods these are distillation and expression. Solvent extraction and CO2 extracts are also other techniques that can be used for the release of essential oils from plants. Different distillation methods that are used are steam distillation, water distillation and steam and water distillation. For distillation process the material from which the material is being extracted is placed on a grid in the still, the steam or/and water depending on which method is being used breaks through the plant material and removes the volatile compounds in it. The volatile compounds rise up into the condenser which cools the vapour into liquid form. This oil liquid will be form a separate layer with water and can be drawn off separately from the water (NAHA, 2014). Expression of essential oils is done through a technique known as ecuelle a piquer. This technique involves placing the rind of a fruit in a container with spikes that puncture the peel while it is being rotated. This technique allows the essential oils that are contained in the fruit to be released when it is punctured. Centrifugal force can then be used to separate the fruit juice from the essential oils (NAHA, 2014). Solvent extraction is used when the plants are too fragile to go through the distillation process. In solvent extraction the odoriferous lipophilic is extracted from the plant along with other tissues in the plant. This causes a thick solution to be extracted that contains waxes, fats and other odoriferous material. This solution is then mixed with alcohol which extracts the aromatic compounds (NAHA, 2014). Hypercritical carbon dioxide (CO2) extraction involves putting C02 under pressure to turn it into a liquid from a gas. This liquid is then used as an inert liquid solvent which can extract aromatic compounds from the plant by diffusing through it. C02 can contain some elements that are not found in the corresponding essential oils. During extraction methods some main considerations must be addressed such as the cost, pesticide residue on the plants and also the safety and therapeutic benefits of the essential oil being produced. These considerations help decide which extraction method is most beneficial (NAHA, 2014). 4. How essential oils are analysed Terpenes are structurally varied therefore the methods used for analyses have to account for a great number of molecular compounds (Baser Buchbauer, 2010). With essential oils there are cheaper versions available on the market. Sensory tests are preformed first to help determine if the oil is poor quality. The clarity, colour, odour and viscosity are looked at to determine the quality. When these sensory tests are completed, physical parameters are measured through refractive index, optical rotation and their specific gravity (Lyth, 2014). Chromatography techniques are then used in the separation and identification of compounds (Baser Buchbauer, 2010).Gas chromatography/Mass Spectrometry is widely used in the separation and identification of compounds in essential oils. In gas chromatography the identification and quantification of the different chemical compounds in essential oils can be detected. Each individual compound can be identified by the retention time of the peaks. The data collected can then be compared against standards to determine the purity. (Lyth, 2014).One of the main objectives in separation is the resolution of the compound with a short retention time. To achieve this, the appropriate parameters must be provided. Chiral stationary phases allow for the separation of compounds with optical isomerism. When identifying essential oils in gas chromatography the temperature must be changed as essential oils contain votaile compounds and less volatile compounds. The temperature must start off low and then be raised every minute until 200C to obtain elution of heavy terpenoids. This allows for shorter elution times, separate and narrow peaks (Baser Buchbauer, 2010) 5. Discussion The chemistry of essential oils is important as they have such diverse effects. The different structures of the compounds is what causes the effects of the essential oil (EBSCO, 2014).It also depends on what part of the plant is being used and also the development stage of the plant. There are also changes in the amount of oil yield and the different compounds that are observed between the process of the bud of a flower forming and full flowering (Baser Buchbauer, 2010). But a plant’s compounds can have different effects depending on the country of origin, the type of soil it was grown in and also the climate conditions of where it was grown. These factors can all effect natural variations in the essential oils that are extracted from the plant (Lyth, 2014). Even though essential oils are known for their healing properties it is difficult to prove that they work because of the unique aroma of essential oils it is also difficult to have a control group. A lot of published studies on essential oil fail to reach accurate and reliable scientific standards (EBSCO, 2014) 6. Conclusion Essential oils are used in a variety of industries, but it is important to know the chemistry behind the process of producing them. This helps produce high quality and pure products that will produce better results. The extraction of them from the plant and also where the plant was growing and the conditions that it grew in are important factors when analysing the product for quality and purity. 7. Bibliography Anon., 2014. Organic facts. [Online] Available at: https://www.organicfacts.net/organic-oils/natural-essential-oils/list-of-essential-oils.html [Accessed 05 November 2014]. Baser, K.H.C. Buchbauer, G., 2010. Handbook of essential oils : science, technology, and applications. Boca Raton: Taylor and Francis Group. Chamorro, E.R. et al., 2012. INTECH. [Online] Available at: http://www.intechopen.com/books/gas-chromatography-in-plant-science-wine-technology-toxicology-and-some-specific-applications/study-of-the-chemical-composition-of-essential-oils-by-gas-chromatography [Accessed 16 October 2014]. Chamorro, E.R. et al., 2012. Study of the Chemical Composition of Essential Oils by Gas Chromatography. Gas Chromatography in Plant Science,Wine Technology, Toxicology and Some Specific Applications, pp.307-25. Chemistry, R.S.o., 2014. ChemSpider. [Online] Available at: http://www.chemspider.com/Chemical-Structure.83751.html [Accessed 28 October 2014]. Cyberlipid, 2014. Cyberlipid centre. [Online] Available at: http://www.cyberlipid.org/simple/simp0004.htm [Accessed 08 November 2014]. doTERRA, I., 2014. doTERRA. [Online] Available at: http://www.doterra.com/sgen/essentialDefinition.php [Accessed 05 November 2014]. EBSCO, C.R.B., 2014. NYU Langone Medical Centre. [Online] Available at: http://www.med.nyu.edu/content?ChunkIID=37427 [Accessed 05 November 2014]. EsotericOils, 2014. Esoteric Oils. [Online] Available at: http://www.essentialoils.co.za/components.htm [Accessed 08 November 2014]. essentials, T.r., 2014. The Real Essentials. [Online] Available at: http://www.therealessentials.com/chemistry.html [Accessed 05 November 2014]. Fresholi, 2014. Fresholi. [Online] Available at: http://www.fresholi.com/index.php?option=com_contentview=articleid=653:essential-oil-componentscatid=88:aromatherapy [Accessed 28 October 2014]. Hermitageoils, 2014. Hermitageoils. [Online] Available at: http://www.hermitageoils.com/essential-synthetics/linalyl-acetate [Accessed 06 November 2014]. Inc, M.W.P., 2013. International WellnessDirectory. [Online] Available at: http://www.mnwelldir.org/docs/therapies/essentia02.htm [Accessed 30 October 2014]. Kamatoua, G.P.P., Vermaaka, I. Viljoen, A.M., 2013. Phytochemistry. Menthol: A simple monoterpene with remarkable biological properties [Online], 96, pp.15-25. Available DOI: 10.1016/j.phytochem.2013.08.005 [Accessed 14 November 2014]. Martin, N., 2014. Experience-Essential-Oils.com. [Online] Available at: http://www.experience-essential-oils.com/chemistry-of-essential-oil.html [Accessed 08 November 2014]. MintNews, 2014. MintNews. [Online] Available at: http://www.mintnews.in/Product-Directories.php?id=cpbThD2wlbs- [Accessed 06 November 2014]. NAHA, 2014. National Association for Holistic Aromatherapy. [Online] Available at: http://www.naha.org/explore-aromatherapy/about-aromatherapy/how-are-essential-oils-extracted [Accessed 05 November 2014]. NobelMediaAB, 2014. Norbal Prize.org. [Online] Available at: http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1910/press.html [Accessed 29 October 2014]. Oils, N.O., 2007. NHR Organic Oils. [Online] Available at: http://www.nhrorganicoils.com/frame.php?page=info_21 [Accessed 05 November 2014]. OMICS, 2013. Open access Scientific Reports. [Online] Available at: http://omicsonline.org/scientific-reports/srep129.php [Accessed 25 October 2014]. Parry, E.J., 1922. The chemistry of Essential oils and Artifical perfumes. 4th ed. London: D. Van Nostrand Company. Pharmacognosy, 2012. Pharmacognosy. [Online] Available at: http://www.epharmacognosy.com/2012/05/ester-volatile-oils-esters-of-aliphatic.html [Accessed 03 November 2014]. Wikipedia, 2014. Wikipedia. [Online] Available at: http://en.wikipedia.org/wiki/Verbenone [Accessed 06 November 2014]. Wikipedia, 2014. Wikipedia. [Online] Available at: http://en.wikipedia.org/wiki/Thymol [Accessed 06 November 2014]. http://www.biospiritual-energy-healing.com/essential-oil-chemistry.html http://en.wikipedia.org/wiki/Essential_oil http://www.zenitech.com/documents/Toxicity_of_essential_oils_p1.pdf http://depa.fquim.unam.mx/amyd/archivero/LipidosPaulaYurkanis_11315.pdf Page 1 of 10

Jems Journal: Chapter Summary :: essays research papers

Jem's Journal: Chapter Summary Dan Latham Chapter 4 - I think at times my sister, Scout can be disgusting. I came home from a long day at school. I found Scout on the porch chewing a wad of gum. I knew it was gum because she had it in her mouth for a long time and plus I could see it in her mouth. So just like any normal brother would do, I told her not to eat things you find and she said that she didn't find it on the ground but rather in a tree. I put an expression on my face that clearly communicated to her that I didn't think she was funny. I also growled at her. She told my like an innocent girl that it was sticking in a tree on the way home from school. I really didn't care about where she got it from I just wanted that disgusting piece of trash out of her mouth immediately before she caught some germs. I told Scout to spit it out immediately. She was actually pretty obedient and spit the gum out. She told me that she had been chewing it all afternoon and that if she wasn't dead and didn't feel sick. She was obviously mad at me for ruining her chewing enjoyment but I didn't want her getting sick because knowing Atticus, I'd be the one who would have to take care of her and that wouldn't go over to well with me since I know I could have stopped her from getting sick. I think I yelled at her and said that isn't she supposed to know that she isn't allowed even near those trees. We all know about that weird guy Boo Radley and we know that property is off limits. I told her that she would probably get killed if she were caught. She said in defense that I touched the house once. This was a clear reference to Boo Radley's house but I ignored her comment and told her that it was different. I also ordered her to go and use some mouth wash to get rid of the germs that she collected from that stick of gum she just spit out. She wasn't pleased that I just ordered her to wash her mouth out and told me in defense that washing her mouth out will dissolve the taste in her mouth but I still didn't care. I wanted her to wash her mouth out so I told her that I would tell Calpurnia about how she found some gum in a tree near Boo Jems Journal: Chapter Summary :: essays research papers Jem's Journal: Chapter Summary Dan Latham Chapter 4 - I think at times my sister, Scout can be disgusting. I came home from a long day at school. I found Scout on the porch chewing a wad of gum. I knew it was gum because she had it in her mouth for a long time and plus I could see it in her mouth. So just like any normal brother would do, I told her not to eat things you find and she said that she didn't find it on the ground but rather in a tree. I put an expression on my face that clearly communicated to her that I didn't think she was funny. I also growled at her. She told my like an innocent girl that it was sticking in a tree on the way home from school. I really didn't care about where she got it from I just wanted that disgusting piece of trash out of her mouth immediately before she caught some germs. I told Scout to spit it out immediately. She was actually pretty obedient and spit the gum out. She told me that she had been chewing it all afternoon and that if she wasn't dead and didn't feel sick. She was obviously mad at me for ruining her chewing enjoyment but I didn't want her getting sick because knowing Atticus, I'd be the one who would have to take care of her and that wouldn't go over to well with me since I know I could have stopped her from getting sick. I think I yelled at her and said that isn't she supposed to know that she isn't allowed even near those trees. We all know about that weird guy Boo Radley and we know that property is off limits. I told her that she would probably get killed if she were caught. She said in defense that I touched the house once. This was a clear reference to Boo Radley's house but I ignored her comment and told her that it was different. I also ordered her to go and use some mouth wash to get rid of the germs that she collected from that stick of gum she just spit out. She wasn't pleased that I just ordered her to wash her mouth out and told me in defense that washing her mouth out will dissolve the taste in her mouth but I still didn't care. I wanted her to wash her mouth out so I told her that I would tell Calpurnia about how she found some gum in a tree near Boo

Stochastic Calculus Solution Manual

Stochastic Calculus for Finance, Volume I and II by Yan Zeng Last updated: August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you have any comments or ? nd any typos/errors, please email me at [email  protected] edu. The current version omits the following problems. Volume I: 1. 5, 3. 3, 3. 4, 5. 7; Volume II: 3. 9, 7. 1, 7. 2, 7. 5–7. 9, 10. 8, 10. 9, 10. 10. Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. 1. 1. Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model Proof. If we get the up sate, then X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ); if we get the down state, then X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a positive probability of being strictly positive, then we must either have X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, then ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) > 0. Plug in X0 = 0, we get u? 0 > (1 + r)? 0 . By condition d < 1 + r < u, we conclude ? 0 > 0.In this case, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 [d ? (1 + r)] < 0. (ii) If X1 (T ) > 0, then we can similarly deduce ? 0 < 0 and hence X1 (H) < 0. So we cannot have X1 strictly positive with positive probability unless X1 is strictly negative with positive probability as well, regardless the choice of the number ? 0 . Remark: Here the condition X0 = 0 is not essential, as far as a property de? nition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can de? ne arbitrage as a trading strategy such that P (X1 ?X0 (1 + r)) = 1 and P (X1 > X0 (1 + r)) > 0. First, this is a generalization of the case X0 = 0; second, it is â€Å"proper† because it is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X0 as borrowed from money market account. Then at time 1, we have to pay back X0 (1 + r) to the money market account. In summary, arbitrage is a trading strategy that beats â€Å"safe† investment. Accordingly, we revise the proof of Exercise 1. 1. as follows.If X1 has a positive probability of being strictly larger than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) > 0, i. e. ?0 > 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) < (1 + r)X0 . The second case can be similarly analyzed. Hence we cannot have X1 strictly greater than X0 (1 + r) with positive probability unless X1 is strictly smaller than X0 (1 + r) with positive probability as well. Finally, we comment that the above formulation of arbitrage is equivalent to the one in the textbook. For details, see Shreve [7], Exercise 5. . 1. 2. 1 5 Proof. X1 (u) = ? 0 ? 8 + ? 0 ? 3 ? 5 (4? 0 + 1. 20? 0 ) = 3? 0 + 1. 5? 0 , a nd X1 (d) = ? 0 ? 2 ? 4 (4? 0 + 1. 20? 0 ) = 4 ? 3? 0 ? 1. 5? 0 . That is, X1 (u) = ? X1 (d). So if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. Remark: Note the above relation X1 (u) = ? X1 (d) is not a coincidence. In general, let V1 denote the ? ? payo? of the derivative security at time 1. Suppose X0 and ? 0 are chosen in such a way that V1 can be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . Using the notation of the problem, suppose an agent begins ? replicated: (1 + r)(X with 0 wealth and at time zero buys ? 0 shares of stock and ? 0 options. He then puts his cash position ? 0 S0 ? ?0 X0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the expression of V1 and sort out terms, we have ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X0 , then there will no arbitrage. 1. 3. S0 1 Proof. V0 = 1+r 1+r? d S1 (H) + u? ? r S1 (T ) = 1+r 1+r? d u + u? 1? r d = S0 . This is not surprising, since u? d u? d u? d u? d this is exactly the cost of replicating S1 . Remark: This illustrates an important point. The â€Å"fair price† of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1 (H) and S1 (T ) are given, we could have two current prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are determined by S0 and S0 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, 1+r? d u? d S1 (H) S0 = + u? 1? r u? d S1 (T ) 1+r S0 = 1+r? d u ? d S1 (H) + u ? 1? r u ? d S1 (T ) 1+r . Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicating component cannot determine its own â€Å"fair† price via the risk-n eutral pricing formula. 1. 4. Proof. Xn+1 (T ) = = ? n dSn + (1 + r)(Xn ? ?n Sn ) ?n Sn (d ? 1 ? r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ? ? Vn+1 (H) ? Vn+1 (T ) (d ? 1 ? r) + (1 + r) = u? d 1+r = p(Vn+1 (T ) ? Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ? ? ? = pVn+1 (T ) + q Vn+1 (T ) ? ? = Vn+1 (T ). 1. 6. 2 Proof. The bank’s trader should set up a replicating portfolio whose payo? s the opposite of the option’s payo?. More precisely, we solve the equation (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 Then X0 = ? 1. 20 and ? 0 = ? 2 . This means the trader should sell short 0. 5 share of stock, put the income 2 into a money market account, and then transfer 1. 20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0. 8(1 + r) in money market account will cancel out with the option’s payo?. Therefore we end up with 1. 20(1 + r) in the separate money market account. Remark: This problem illustrates why we are in terested in hedging a long position.In case the stock price goes down at time one, the option will expire without any payo?. The initial money 1. 20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payo? at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1. 7. Proof. The idea is the same as Problem 1. 6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1. 2. 4. More precisely, he should short sell 0. 1733 share of stock, invest the income 0. 933 into money market account, and transfer 1. 376 into a separate money market account. The portfolio consisting a short position in stock and 0. 6933-1. 376 in money market account will replicate the opposite of the option’s payo?. After they cancel out, we end up with 1. 376(1 + r)3 in the separate money market ac count. 1. 8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1. 696. (iii) Proof. ?n (s, y) = vn+1 (us, y + us) ? vn+1 (ds, y + ds) . (u ? d)s 1. 9. (i) Proof. Similar to Theorem 1. 2. 2, but replace r, u and d everywhere with rn , un and dn .More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . Then un ? dn Vn = pn Vn+1 (H) + qn Vn+1 (T ) . 1 + rn (ii) Proof. ?n = (iii) 3 Vn+1 (H)? Vn+1 (T ) Sn+1 (H)? Sn+1 (T ) = Vn+1 (H)? Vn+1 (T ) . (un ? dn )Sn 10 10 Proof. un = Sn+1 (H) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at time n are pn = u1? dnn = 1 and qn = 1 . Risk-neutral pricing implies the price of this call at time zero is ? ? 2 2 n ? d 9. 375. 2. Probability Theory on Coin Toss Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it su? ces to work on the case N = 2.When A1 and A2 are disjoint, P (A1 ? A2 ) = A1 ? A2 P (? ) = A1 P (? ) + A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, using the result when they are disjoint, we have P (A1 ? A2 ) = P ((A1 ? A2 ) ? A2 ) = P (A1 ? A2 ) + P (A2 ) ? P (A1 ) + P (A2 ). 2. 2. (i) 1 3 1 Proof. P (S3 = 32) = p3 = 8 , P (S3 = 8) = 3p2 q = 3 , P (S3 = 2) = 3pq 2 = 8 , and P (S3 = 0. 5) = q 3 = 8 . 8 Ac P (? ) + A P (? ) = P (? ) = 1. (ii) Proof. E[S1 ] = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, E[S2 ] = 16p2 + 4  · 2pq + 1  · q 2 = 6. 25, and 3 1 E[S3 ] = 32  · 1 + 8  · 8 + 2  · 3 + 0.  · 8 = 7. 8125. So the average rates of growth of the stock price under P 8 8 5 are, respectively: r0 = 4 ? 1 = 0. 25, r1 = 6. 25 ? 1 = 0. 25 and r2 = 7. 8125 ? 1 = 0. 25. 5 6. 25 (iii) 8 1 Proof. P (S3 = 32) = ( 2 )3 = 27 , P (S3 = 8) = 3  · ( 2 )2  · 1 = 4 , P (S3 = 2) = 2  · 1 = 2 , and P (S3 = 0. 5) = 27 . 3 3 3 9 9 9 Accordingly, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13. 5. So the average rates of growth of the stock price 9 6 under P are, respectively: r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Apply conditional Jensen’s inequality. 2. 4. (i) Proof.En [Mn+1 ] = Mn + En [Xn+1 ] = Mn + E[Xn+1 ] = Mn . (ii) 2 n+1 Proof. En [ SSn ] = En [e? Xn+1 e? +e ] = 2 ? Xn+1 ] e? +e E[e = 1. 2. 5. (i) 2 2 Proof. 2In = 2 j=0 Mj (Mj+1 ? Mj ) = 2 j=0 Mj Mj+1 ? j=1 Mj ? j=1 Mj = 2 j=0 Mj Mj+1 + n? 1 n? 1 n? 1 n? 1 2 2 2 2 2 2 2 2 Mn ? j=0 Mj+1 ? j=0 Mj = Mn ? j=0 (Mj+1 ? Mj ) = Mn ? j=0 Xj+1 = Mn ? n. n? 1 n? 1 n? 1 n? 1 n? 1 (ii) Proof. En [f (In+1 )] = En [f (In + Mn (Mn+1 ? Mn ))] = En [f (In + Mn Xn+1 )] = 1 [f (In + Mn ) + f (In ? Mn )] = 2 v v v g(In ), where g(x) = 1 [f (x + 2x + n) + f (x ? 2x + n)], since 2In + n = |Mn |. 2 2. 6. 4 Proof. En [In+1 ?In ] = En [? n (Mn+1 ? Mn )] = ? n En [Mn+1 ? Mn ] = 0. 2. 7. Proof. We denote by Xn the result of n-th coin toss, where Head is represented by X = 1 and Tail is 1 represented by X = ? 1. We also suppose P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 ,  ·  ·  · , Xn )Xn+1 , where bn ( ·) is a bounded function on {? 1, 1} , to be determined later on. Clearly (Sn )n? 1 is an adapted stochastic process, and we can show it is a martingale. Indeed, En [Sn+1 ? Sn ] = bn (X1 ,  ·  ·  · , Xn )En [Xn+1 ] = 0. For any arbitrary function f , En [f (Sn+1 )] = 1 [f (Sn + bn (X1 ,  ·  ·  · , Xn )) + f (Sn ? n (X1 ,  ·  ·  · , Xn ))]. Then 2 intuitively, En [f (Sn+1 ] cannot be solely dependent upon Sn when bn ’s are properly chosen. Therefore in general, (Sn )n? 1 cannot be a Markov process. Remark: If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealth at time n. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2. 8. (i) Proof. Note Mn = En [MN ] and Mn = En [MN ]. (ii) Proof. In the proof of Theorem 1. 2. 2, we proved by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In other words, the sequence (Vn )0? n?N can be realized as the value process of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a martingale under P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale under P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), then use (i). 2. 9. (i) (H) S1 (H) 1 = 2, d0 = S1S0 = 2 , S0 (T and d1 (T ) = S21 (TT)) = 1. S 1 1 0 ? d So p0 = 1+r? d0 0 = 2 , q0 = 2 , p1 (H) u0 5 q1 (T ) = 6 . Therefore P (HH) = p0 p1 (H) = 1 , 4 5 q0 q1 (T ) = 12 . Vn (1+r)n = En VN (1+r)N , so V0 , V1 1+r ,  ·Ã‚ ·Ã‚ ·, VN ? 1 , VN (1+r)N ? 1 (1+r)N is a martingale under P . Proof. u0 = u1 (H) = =S2 (HH) S1 (H) = 1. 5, d1 (H) = S2 (HT ) S1 (H) = 1, u1 (T ) = S2 (T H) S1 (T ) =4 1+r1 (H)? d1 (H) u1 (H)? d1 (H) 1 = 1 , q1 (H) = 2 , p1 (T ) = 2 1 4, 1+r1 (T )? d1 (T ) u1 (T )? d1 (T ) 1 12 1 = 6 , and P (HT ) = p0 q1 (H) = P (T H) = q0 p1 (T ) = and P (T T ) = The proofs of Theorem 2. 4. 4, Theorem 2. 4. 5 and Theorem 2. 4. 7 still work for the random interest rate m odel, with proper modi? cations (i. e. P would be constructed according to conditional probabilities P (? n+1 = H|? 1 ,  ·  ·  · , ? n ) := pn and P (? n+1 = T |? 1 ,  ·  ·  · , ? n ) := qn . Cf. notes on page 39. ). So the time-zero value of an option that pays o?V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . )(1+r (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2. 4, V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9, and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ? 1. 5 (iii) Proof. ?0 = (iv) Proof. ?1 (H) = 2. 10. (i) Xn+1 Proof. En [ (1+r)n+1 ] = En [ ? n Yn+1 Sn + (1+r)n+1 (1+r)(Xn n Sn ) ] (1+r)n+1 Xn (1+r)n . V2 (HH)? V2 (HT ) S2 (HH)? S2 (HT ) V1 (H)? V1 (T ) S1 (H)? S1 (T ) = 1 2. 4? 9 8? 2 = 0. 4 ? 1 54 ? 0. 3815. = 5? 1 12? 8 = 1. = ?n Sn (1+r)n+1 En [Yn+1 ] + Xn Sn (1+r)n = ?n Sn (1+r)n+1 (up + dq) + Xn n Sn (1+r)n = ?n Sn +Xn n Sn (1+r)n = (ii) Proof . From (2. 8. 2), we have ? n uSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (H) ? n dSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (T ). So ? n = Xn+1 (H)? Xn+1 (T ) uSn ? dSn and Xn = En [ Xn+1 ]. To make the portfolio replicate the payo? at time N , we 1+r VN X must have XN = VN . So Xn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ]. Since (Xn )0? n? N is the value process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations), the no-arbitrage price of VN at time n is Vn = Xn = En [ (1+r)N ? ]. (iii) Proof. En [ Sn+1 ] (1 + r)n+1 = = < = 1 En [(1 ? An+1 )Yn+1 Sn ] (1 + r)n+1 Sn [p(1 ? An+1 (H))u + q(1 ? An+1 (T ))d] (1 + r)n+1 Sn [pu + qd] (1 + r)n+1 Sn . (1 + r)n Sn (1+r)n+1 (1? a)(pu+qd) Sn+1 If An+1 is a constant a, then En [ (1+r)n+1 ] = Sn (1+r)n (1? a)n . = Sn (1+r)n (1? a). Sn+1 So En [ (1+r)n+1 (1? a)n+1 ] = 2. 11. (i) Proof. FN + PN = SN ? K + (K ? SN )+ = (SN ? K)+ = CN . (ii) CN FN PN Proof. Cn = En [ (1+r)N ? n ] = En [ (1+ r)N ? n ] + En [ (1+r)N ? n ] = Fn + Pn . (iii) FN Proof. F0 = E[ (1+r)N ] = 1 (1+r)N E[SN ? K] = S0 ? K (1+r)N . (iv) 6 Proof.At time zero, the trader has F0 = S0 in money market account and one share of stock. At time N , the trader has a wealth of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Since F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En [ (1+r)N ?n ] = Sn ? So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ? 1. (1+r)N S0 (1+r)N ? n (1+r)N S0 (1+r)N = 0, C0 = P0 . = Sn ? S0 (1 + r)n . 2. 12. Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P ), where C=E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payo? of max(C, P ) at time m. Therefore, its current no-arbitrage price should be E[ max(C,P ) ]. (1+r)m K K By the put-call parity, C = Sm ? (1+r)N ? m + P . So max(C, P ) = P + (Sm ? (1+r)N ? m )+ . Therefore, the time-zero price of a chooser option is E K (Sm ? (1+r)N ? m )+ P +E (1 + r)m (1 + r)m =E K (Sm ? (1+r)N ? m )+ (K ? SN )+ . +E (1 + r)N (1 + r)mThe ? rst term stands for the time-zero price of a put, expiring at time N and having strike price K, and the K second term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)N ? m . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ max(C,P ) ], (1+r)m due to the economical argument involved (like â€Å"the chooser option is equivalent to receiving a payo? of max(C, P ) at time m†), then we have the following mathematically rigorous argument. First, we can construct a portfolio ? 0 ,  ·  ·  · , ? m? 1 , whose payo? at time m is max(C, P ).Fix ? , if C(? ) > P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (SN ? K)+ ; if C(? ) < P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) > P (? ) if C(? ) < P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In Xm particular, V0 = X0 = E[ (1+r)m ] = E[ max(C,P ) ]. (1+r)m 2. 13. (i) Proof.Note under both actual probability P and risk-neutral probability P , coin tosses ? n ’s are i. i. d.. So n+1 without loss of generality, we work on P . For any function g, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a function of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov un der P . (ii) 7 Sn+1 Sn Sn )] Proof. Set vN (s, y) = f ( Ny ). Then vN (SN , YN ) = f ( +1 Vn = where En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r ,Yn+1 ) ] N n=0 Sn N +1 ) = VN . Suppose vn+1 is given, then = 1 1+r [pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since coin tosses ? n ’s are i. i. d. under P , (Sn , Yn )0? n? M is Markov under P . More precisely, for any function h, En [h(Sn+1 )] = ph(uSn ) + h(dSn ), for n = 0, 1,  ·  ·  · , M ? 1. For any function g of two variables, we have EM [g(SM +1 , YM +1 )] = EM [g(SM +1 , SM +1 )] = pg(uSM , uSM )+ n+1 n+1 qg(dSM , dSM ). And for n ? M +1, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + SSn Sn )] = pg(uSn , Yn +uSn )+ qg(dSn , Yn + dSn ), so (Sn , Yn )0? n? N is Markov under P . (ii) y Proof. Set vN (s, y) = f ( N ? M ).Then vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . Suppose vn+1 is already given. a) If n > M , then En [vn+1 (Sn+1 , Yn+1 )] = pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ). So vn (s, y) = pvn+1 (us, y + us) + qvn+1 (ds, y + ds). b) If n = M , then EM [vM +1 (SM +1 , YM +1 )] = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (us, us) + qvM +1 (ds, ds). c) If n < M , then En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. State Prices 3. 1. Proof. Note Z(? ) := P (? ) P (? ) = 1 Z(? ) . Apply Theorem 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of properties (i)-(iii) of Theorem 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. E[Y ] = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = A Z(? )P (? ) = 0, by P (Z > 0) = 1, we conclude P (? ) = 0 for any ? ? A. So P (A) = A P (? ) = 0. (v) Proof. P (A) = 1 P (Ac ) = 0 P (Ac ) = 0 P (A) = 1. (vi) A P (? ) = Z(? )P (? ) = E[Z] = 1. Y (? )P (? ) = Y (? )Z(? )P (? ) = E[Y Z]. Z(? )P (? ). Since P (A) = 0, P (? ) = 0 for any ? ? A. So P (A) = 0. 8 Proof. Pick ? 0 such that P (? 0 ) > 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 Then P (Z ? 0) = 1 and E[Z] = if ? = ? 0 .  · P (? 0 ) = 1. =? 0 Clearly P (? {? 0 }) = E[Z1? {? 0 } ] = Z(? )P (? ) = 0. But P (? {? 0 }) = 1 ? P (? 0 ) > 0 if P (? 0 ) < 1. Hence in the case 0 < P (? 0 ) < 1, P and P are not equivalent. If P (? 0 ) = 1, then E[Z] = 1 if and only if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P have to be equivalent. In summary, if we can ? nd ? 0 such that 0 < P (? 0 ) < 1, then Z as constructed above would induce a probability P that is not equivalent to P . 3. 5. (i) Proof. Z(HH) = (ii) Proof. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )P (? 2 = T |? 1 = H) = 3 E1 [Z2 ](T ) = Z2 (T H)P (? 2 = H|? = T ) + Z2 (T T )P (? 2 = T |? 1 = T ) = 2 . (iii) Proof. V1 (H) = [Z2 (HH)V2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )V2 (HT )P (? 2 = T |? 1 = T )] = 2. 4, Z1 (H)(1 + r1 (H)) [Z2 (T H)V2 (T H)P (? 2 = H|? 1 = T ) + Z2 (T T )V2 (T T )P (? 2 = T |? 1 = T )] 1 = , Z1 (T )(1 + r1 (T )) 9 3 4. 9 16 , Z(HT ) = 9 , Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ? 1. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3. 6. Proof. U (x) = have XN = 1 x, (1+r)N ? Z so I(x) = = 1 Z] 1 x. Z (3. 3. 26) gives E[ (1+r)N 1 X0 (1 + r)n Zn En [Z  ·X0 N Z (1 + r) . 0 = Xn , where ? Hence Xn = (1+r)N ? Z X En [ (1+r)N ? n ] N ] = X0 . So ? = = En [ X0 (1+r) Z n 1 X0 . By (3. 3. 25), we 1 ] = X0 (1 + r)n En [ Z ] = the second to last â€Å"=† comes from Lemma 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26), we have E[ (1+r)N ( (1+r)N ) p? 1 ] = X0 . Solve it for ? , we get ? ?p? 1 1 1 ? ? =? ? X0 p E 1 Z p? 1 Np ? ? ? = p? 1 X0 (1 + r)N p (E[Z p? 1 ])p? 1 1 p . (1+r) p? 1 ? Z So by (3. 3. 25), XN = ( (1+r)N ) p? 1 = 1 1 Np ? p? 1 Z p? 1 N (1+r) p? 1 = X0 (1+r) p? 1 E[Z p p? 1 Z p? 1 N (1+r) p? 1 = (1+r)N X0 Z p? 1 E[Z p p? 1 1 . ] ] 3. 8. (i) 9 d d Proof. x (U (x) ? yx) = U (x) ? y. So x = I(y) is an extreme point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a maximum point. Therefore U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Following the hint of the problem, we have E[U (XN )] ? E[XN ? Z ? Z ? Z ? Z ] ? E[U (I( ))] ? E[ I( )], N N N (1 + r) (1 + r) (1 + r) (1 + r)N ? ? ? ? ? i. e. E[U (XN )] ? ?X0 ? E[U (XN )] ? E[ (1+r)N XN ] = E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. 3. 9. (i) X Proof. Xn = En [ (1+r)N ? n ]. So if XN ? 0, then Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x < ? and 0 < y ? ? , then U (x) ? yx = ? yx ? and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 b) If 0 ? x < ? and y > ? , then U (x) ? yx = ? yx ? 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 c) If x ? ? and 0 < y ? ? , then U (x) ? yx = 1 ? yx and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 1 ? yx. So U (x) ? yx ? U (I(y)) ? yI(y). 1 d) If x ? ? and y > ? , then U (x) ? yx = 1 ? yx < 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). (iii) XN ? Z Proof. Using (ii) and set x = XN , y = (1+r)N , where XN is a random variable satisfying E[ (1+r)N ] = X0 , we have ?Z ? Z ? E[U (XN )] ? E[ XN ] ? E[U (XN )] ? E[ X ? ]. (1 + r)N (1 + r)N N ? ? That is, E[U (XN )] ? ?X0 ? E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. (iv) Proof. Plug pm and ? m into (3. 6. 4), we have 2N 2N X0 = m=1 pm ? m I( m ) = m=1 1 pm ? m ? 1{ m ? ? } . So X0 ? X0 ? {m : = we are looking for positive solution ? > 0). Conversely, suppose there exists some K so that ? K < ? K+1 and K X0 1 m=1 ? m pm = ? . Then we can ? nd ? > 0, such that ? K < < ? K+1 . For such ? , we have Z ? Z 1 E[ I( )] = pm ? m 1{ m ? ? } ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 Hence (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1{ m ? ? } . Suppose there is a solution ? to (3. 6. 4), note ? > 0, we then can conclude 1 1 1 m ? ? } = ?. Let K = max{m : m ? ? }, then K ? ? < K+1 . So ? K < ? K+1 and K N m=1 pm ? m (Note, however, that K could be 2 . In this case, ? K+1 is interpreted as ?. Also, note = (v) ? 1 Proof. XN (? m ) = I( m ) = ? 1{ m ? ? } = ?, if m ? K . 0, if m ? K + 1 4. American Derivative Securities Before proceeding to the exercise problems, we ? rst give a brief summary of pricing American derivative securities as presented in the textbook. We shall use the notation of the book.From the buyer’s perspective: At time n, if the derivative security has not been exercised, then the buyer can choose a policy ? with ? ? Sn . The valuation formula for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative security exercised according to ? : N Vn (? ) = k=n En 1{? =k} 1 1 Gk = En 1{? ?N } G? . (1 + r)k? n (1 + r)? ?n The buyer wants to consider all the possible ? ’s, so that he c an ? nd the least upper bound of security value, which will be the maximum price of the derivative security acceptable to him. This is the price given by 1 De? nition 4. 4. 1: Vn = max? ?Sn En [1{? ?N } (1+r)? n G? ]. From the seller’s perspective: A price process (Vn )0? n? N is acceptable to him if and only if at time n, he can construct a portfolio at cost Vn so that (i) Vn ? Gn and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can ? nd (? n )0? n? N and (Cn )0? n? N so that Cn ? 0 and Sn Vn+1 = ? n Sn+1 + (1 + r)(Vn ? Cn ? ?n Sn ). Since ( (1+r)n )0? n? N is a martingale under the risk-neutral measure P , we conclude En Cn Vn+1 Vn =? ? 0, ? n+1 n (1 + r) (1 + r) (1 + r)n Vn i. e. ( (1+r)n )0? n? N is a supermartingale. This inspired us to check if the converse is also true.This is exactly the content of Theorem 4. 4. 4. So (Vn )0? n? N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the requirement 0? n? N Vn ? Gn ). In summary, a price process (Vn )0? n? N is acceptable to the seller if and only if (i) Vn ? Gn ; (ii) is a supermartingale under P . 0? n? N Theorem 4. 4. 2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable to both. Theorem 4. 4. 3 gives a speci? c algorithm for calculating the price, Theorem 4. 4. establishes the one-to-one correspondence between super-replication and supermartingale property, and ? nally, Theorem 4. 4. 5 shows how to decide on the optimal exercise policy. 4. 1. (i) Proof. V2P (HH) = 0, V2P (HT ) = V2P (T H) = 0. 8, V2P (T T ) = 3, V1P (H) = 0. 32, V1P (T ) = 2, V0P = 9. 28. (ii) Proof. V0C = 5. (iii) Proof. gS (s) = |4 ? s|. We apply Theorem 4. 4. 3 and have V2S (HH) = 12. 8, V2S (HT ) = V2S (T H) = 2. 4, V2S (T T ) = 3, V1S (H) = 6. 08, V1S (T ) = 2. 16 and V0S = 3. 296. (iv) 11 Proof. First, we note the simple inequality max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). >† holds if and only if b1 > a1 , b2 < a2 or b1 < a1 , b2 > a2 . By induction, we can show S Vn = max gS (Sn ), S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ? max gP (Sn ) + gC (Sn ), ? max gP (Sn ), P C = Vn + Vn . P P pVn+1 + Vn+1 1+r + max gC (Sn ), S P C As to when â€Å" C C pVn+1 +qVn+1 1+r or gP (Sn ) > P P pVn+1 +qVn+1 1+r and gC (Sn ) < C C pVn+1 +qVn+1 }. 1+r 4. 2. Proof. For this problem, we need Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Then ? 1 (H) = and ? 0 = V2 (HH) ? V2 (HT ) 1 V2 (T H) ? V2 (T T ) = ? , ? 1 (T ) = = ? 1, S2 (HH) ? S2 (HT ) 12 S2 (T H) ?S2 (T T ) V1 (H) ? V1 (T ) ? ?0. 433. S1 (H) ? S1 (T ) The optimal exercise time is ? = inf{n : Vn = Gn }. So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the agent borrows 1. 36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borr ow again and buy 0. 433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio 1 is X1 (T ) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (T ) = (1 + 4 )(? 1. 36 ? 0. 433 ? 4) + 0. 433 ? 2 = ? 3. The agent should exercise the put at time one and get 3 to pay o? is debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The agent should borrow to buy 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.The agent should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We need Figure 1. 2. 2 for this problem, and calculate the intrinsic value process and price process of the put as follows. 2 For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the other outcomes of G is negative. 12 2 5 For the price process, V0 = 0. 4, V1 (T ) = 1, V1 (T H) = 3 , V1 (T T ) = 3 , V3 (T HT ) = 1, V3 (T T H) = 1. 75, V3 (T T T ) = 2. 125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge t he insider more than 1. 36. 4. 5. Proof. The stopping times in S0 are (1) ? ? 0; (2) ? ? 1; (3) ? (HT ) = ? (HH) = 1, ? (T H), ? (T T ) ? {2, ? } (4 di? erent ones); (4) ? (HT ), ? (HH) ? {2, ? }, ? (T H) = ? (T T ) = 1 (4 di? rent ones); (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? {2, ? } (16 di? erent ones). When the option is out of money, the following stopping times do not exercise (i) ? ? 0; (ii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H), ? (T T ) ? {2, ? } (8 di? erent ones); (iii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H) = ? (T T ) = 1 (2 di? erent ones). ? 4 For (i), E[1{? ?2} ( 4 )? G? ] = G0 = 1. For (ii), E[1{? ?2} ( 5 )? G? ] ? E[1{? ? ? 2} ( 4 )? G? ? ], where ? ? (HT ) = 5 5 1 4 4 ? 2, ? ? (HH) = ? , ? ? (T H) = ? ? (T T ) = 2. So E[1{? ? ? 2} ( 5 )? G? ? ] = 4 [( 4 )2  · 1 + ( 5 )2 (1 + 4)] = 0. 96. For 5 (iii), E[1{? ?2} ( 4 )? G? has the biggest value when ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This value is 1. 36. 4. 6. (i) Proof. The value of the put at time N , if it is not exercised at previous times, is K ? SN . Hence VN ? 1 = VN K max{K ? SN ? 1 , EN ? 1 [ 1+r ]} = max{K ? SN ? 1 , 1+r ? SN ? 1 } = K ? SN ? 1 . The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K ?S0 . Remark: We cheated a little bit by using American algorithm and Theorem 4. 4. 5, since they are developed for the case where ? is allowed to be ?. But intuitively, results in this chapter should still hold for the case ? ? N , provided we replace â€Å"max{Gn , 0}† with â€Å"Gn †. (ii) Proof. This is because at time N , if we have to exercise the put and K ? SN < 0, we can exercise the European call to set o? the negative payo?. In e? ect, throughout the portfolio’s lifetime, the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N . So, we must have V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. Let V0EP denote the time-zero value of a European put with strike K and expiration time N . Then V0AP ? V0EP = V0EC ? E[ K SN ? K ] = V0EC ? S0 + . (1 + r)N (1 + r)N 4. 7. VN K K Proof. VN = SN ? K, VN ? 1 = max{SN ? 1 ? K, EN ? 1 [ 1+r ]} = max{SN ? 1 ? K, SN ? 1 ? 1+r } = SN ? 1 ? 1+r . K By induction, we can prove Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn > Gn for 0 ? n ? N ? 1. So the K time-zero value is S0 ? (1+r)N and the optimal exercise time is N . 5. Random Walk 5. 1. (i) Proof. E[ 2 ] = E[? (? 2 1 )+? 1 ] = E[? (? 2 1 ) ]E[ 1 ] = E[ 1 ]2 . (ii) Proof. If we de? ne Mn = Mn+? ? M? m (m = 1, 2,  ·  ·  · ), then (M · )m as random functions are i. i. d. with (m) distributions the same as that of M . So ? m+1 ? ?m = inf{n : Mn = 1} are i. i. d. with distributions the same as that of ? 1 . Therefore E [ m ] = E[? (? m m? 1 )+(? m? 1 m? 2 )+ ·Ã‚ ·Ã‚ ·+? 1 ] = E[ 1 ]m . (m) (m) (iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe , so f (? ) > 0 if and only if ? > f (? ) > f (0) = 1 for all ? > 0. (ii) 1 1 1 n+1 Proof. En [ SSn ] = En [e? Xn+1 f (? ) ] = pe? f (? ) + qe f (? ) = 1. 1 2 (ln q ? ln p). Since 1 2 (ln q ln p) < 0, (iii) 1 Proof. By optional stopping theorem, E[Sn 1 ] = E[S0 ] = 1. Note Sn 1 = e? Mn 1 ( f (? ) )n 1 ? e?  ·1 , by bounded convergence theorem, E[1{? 1 1 for all ? > ? 0 . v (ii) 1 1 Proof. As in Exercise 5. 2, Sn = e? Mn ( f (? ) )n is a martingale, and 1 = E[S0 ] = E[Sn 1 ] = E[e? Mn 1 ( f (? ) )? 1 ? n ]. Suppose ? > ? 0 , then by bounded convergence theorem, 1 = E[ lim e? Mn 1 ( n>? 1 n 1 1 ? 1 ) ] = E[1{? 1 K} ] = P (ST > K). Moreover, by Girsanov’s Theorem, Wt = Wt + in Theorem 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t ( )du 0 = Wt ? ?t is a P -Brownian motion (set ? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v > ? d+ (T, x) T = N (d+ (T, x)). P (ST > K) = P (xe? WT +(r+ 2 ? )T > K) = P 46 5. 4. First, a few typos. In the SDE for S, â€Å"? (t)dW (t)† > â€Å"? (t)S(t)dW (t)†. In the ? rst equation for c(0, S(0)), E > E. In the second equation for c(0, S(0)), the variable for BSM should be ? ? 1 T 2 1 T r(t)dt, ? (t)dt? . BSM ? T, S(0); K, T 0 T 0 (i) Proof. d ln St = X = ? is a Gaussian with X ? N ( (ii) Proof. For the standard BSM model with constant volatility ? and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R? 1 ? 2 )T +? WT ? N ((R? 1 ? )T, ? 2 T ), and E[(S0 eY ? K)+ ] = 2 2 eRT BSM (T, S0 ; K, R, ? ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St ? 2St d S t = rt dt + ? t dWt ? 2 ? t dt. So ST = S0 exp{ 0 (rt ? 2 ? t )dt + 0 T 1 2 2 ? t )dt + 0 ? t dWt . The ? rst term in the expression of X is a number and the T 2 random variable N (0, 0 ? t dt), since both r and ? ar deterministic. Th erefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt }. Let second term ST = S0 eX , 1 T (E[Y ] + 1 V ar(Y )) and ? = 2 T, S0 ; K, 1 T 1 T V ar(Y ), we can get 1 V ar(Y ) . T E[(S0 eY ? K)+ ] = eE[Y ]+ 2 V ar(Y ) BSM So for the model in this problem, c(0, S0 ) = = e? ? T 0 1 E[Y ] + V ar(Y ) , 2 rt dt E[(S0 eX ? K)+ ] e BSM T, S0 ; K, 1 T T 0 T 0 1 rt dt E[X]+ 2 V ar(X) 1 T ? 1 E[X] + V ar(X) , 2 1 V ar(X) T ? = 1 BSM ? T, S0 ; K, T 0 T rt dt, 2 ? t dt? . 5. 5. (i) 1 1 Proof. Let f (x) = x , then f (x) = ? x2 and f (x) = 2 x3 . Note dZt = ? Zt ? t dWt , so d 1 Zt 1 1 1 2 2 2 ? t ? 2 t = f (Zt )dZt + f (Zt )dZt dZt = ? 2 (? Zt )? t dWt + 3 Zt ? t dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5. 2. 2. , for s, t ? 0 with s < t, Ms = E[Mt |Fs ] = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs |Fs . That is, E[Zt Mt |Fs ] = 47 Proof. dMt = d Mt  · 1 Zt = 1 1 1 ? M t ? t M t ? 2 ? t ? t t dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt (iv) Proof. In part (iii), we have dMt = Let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to show Wi (t) is an Ft -martingale under P and [Wi , Wj ](t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt E[Wi (t)|Fs ] = E |Fs . Zs By It? ’s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(? Zt )? (t)  · dWt + Zt (dWi (t) + ? i (t)dt) + (? Zt ? t  · dWt )(dWi (t) + ? i (t)dt) d t M t ? t M t ? 2 ? t ? t ? t M t ? t t dWt + dWt + dt + dt = (dWt + ? t dt) + (dWt + ? t dt). Zt Zt Zt Zt Zt Zt then dMt = ? t dWt . This proves Corollary 5. 3. 2. ?t +Mt ? t , Zt = Wi (t)(? Zt ) j=1 d ?j (t)dWj (t) + Zt (dWi (t) + ? i (t)dt) ? Zt ? i (t)dt = Wi (t)(? Zt ) j=1 ?j (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover,  ·  · [Wi , Wj ](t) = Wi + 0 ?i (s)ds, Wj + 0 ?j (s)ds (t) = [Wi , Wj ](t) = t? ij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5. 7. (i) Proof. Let a be any strictly positive number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . Then P X2 (T ) ? X2 (0) D(T ) = P (a + X1 (T ) ? a) = P (X1 (T ) ? 0) = 1, and P X2 (T ) > X2 (0) = P (X1 (T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Then X1 (0) = 0, P (X1 (T ) ? 0) = P X2 (T ) ? X2 (0) D(T ) = 1, P (X1 (T ) > 0) = P X2 (T ) > X2 (0) D(T ) > 0. 5. 8.The basic idea is that for any positive P -martingale M , dMt = Mt  · sentation Theorem, dMt = ? t dWt for some adapted process ? t . So martingale must be the exponential of an integral w. r. t. Brownian motion. Taking into account d iscounting factor and apply It? ’s product rule, we can show every strictly positive asset is a generalized geometric o Brownian motion. (i) Proof. Vt Dt = E[e? 0 Ru du VT |Ft ] = E[DT VT |Ft ]. So (Dt Vt )t? 0 is a P -martingale. By Martingale Represent tation Theorem, there exists an adapted process ? t , 0 ? t ? T , such that Dt Vt = 0 ? s dWs , or equivalently, ? 1 t ? 1 t ? 1 Vt = Dt 0 ? dWs . Di? erentiate both sides of the equation, we get dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i. e. X > 0 a. s. ) de? ned on the probability space (? , G, P ). Let F be a sub ? -algebra of G, then Y = E[X|F] > 0 a. s. Proof. By the property of conditional expectation Yt ? 0 a. s. Let A = {Y = 0}, we shall show P (A) = 0. In? 1 1 deed, note A ? F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A? {X? 1} ] + n=1 E[X1A? { n >X? n+1 } ] ? 1 1 1 1 1 P (A? {X ? 1})+ n=1 n+1 P (A? n > X ? n+1 }). So P (A? {X ? 1}) = 0 and P (A? { n > X ? n+1 }) = 0, ? 1 1 ? n ? 1. This in turn implies P (A) = P (A ? {X > 0}) = P (A ? {X ? 1}) + n=1 P (A ? { n > X ? n+1 }) = 0. ? ? t Dt dWt . T 1 Mt dMt . By Martingale Repre? dMt = Mt ( Mtt )dWt , i. e. any positive By the above lemma, it is clear that for each t ? [0, T ], Vt = E[e? t Ru du VT |Ft ] > 0 a. s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3. 4)), we have the following stronger result: for a. s. ?, Vt (? ) > 0 for any t ? [0, T ]. (iii) 1 1 Proof. By (ii), V > 0 a. s. so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + ? t Dt dWt ? t = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T ?t Vt dWt , where ? t = 5. 9. ?t Vt Dt . This shows V follows a generalized geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d ± = then f (y) = ? yf (y), cK (0, T, x, K) = xf (d+ ) 1 v ? T x (ln K + (r  ± 1 ? 2 )T ). Let f (y) = 2 y v1 e? 2 2? 2 , ?d+ ? d? ? e? rT N (d? ) ? Ke? rT f (d? ) ? y ? y ? 1 1 = xf (d+ ) v ? e? rT N (d? ) + e? rT f (d? ) v , ? TK ? T 49 and cKK (0, T, x, K) x ? d? e? rT 1 ? d+ d? ? v ? e? rT f (d? ) + v (? d? )f (d? ) xf (d+ ) v f (d+ )(? d+ ) 2 ? y ? y ? y ? TK ? TK ?T x xd+ ? 1 ? 1 e? rT d? ?1 v v ? e? rT f (d? ) v ? v f (d? ) v f (d+ ) + v f (d+ ) ? T K2 ? TK K? T K? T ? T K? T x d+ e? rT f (d? ) d? v [1 ? v ] + v f (d+ ) [1 + v ] 2? T K ? T K? T ? T e? rT x f (d? )d+ ? 2 2 f (d+ )d? . K? 2 T K ? T = = = = 5. 10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = max{C(t0 ), P (t0 )} = max{C(t0 ), C(t0 ) ? F (t0 )} = C(t0 ) + max{0, ? F (t0 )} = C(t0 ) + (e? r(T ? t0 ) K ? S(t0 ))+ . (ii) Proof. By the risk-neutral pricing formula, V (0) = E[e? rt0 V (t0 )] = E[e? rt0 C(t0 )+(e? rT K ? e? rt0 S(t0 )+ ] = C(0) + E[e? rt0 (e? r(T ? t0 ) K ? S(t0 ))+ ]. The ? st term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e? r(T ? t0 ) K. 5. 11. Proof. We ? rst make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the backward SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Multiply Dt on both sides of the ? rst equation and apply It? ’s product rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Integrate from 0 to T , we have DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the terminal T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash ? ow, provided we can ? nd a trading strategy ? that solves the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . Take the proper change of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the new measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ?t d(Dt St ) = D0 X0 + 0 ?t (Dt St )? t dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Representation Theorem and get a formula for ? by comparison of the integrands. 50 (Formal proof) Let MT = Xt = ?1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = E[MT |Ft ]. Then by Martingale Representation Theot 0 rem, we can ? nd an adapted process ? t , so that Mt = M0 + + t 0 ?t dWt . If we set ? t = T 0 ?u d(Du Su ) ? t 0 ?t Dt St ? t , we can check Cu Du du), with X0 = M0 = E[ Ct Dt dt] solves the SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Indeed, it is easy to see that X satis? es the ? rst equation.To check the terminal condition, we note T T T XT DT = D0 X0 + 0 ? t Dt St ? t dWt ? 0 Ct Dt dt = M0 + 0 ? t dWt ? MT = 0. So XT = 0. Thus, we have found a trading strategy ? , so that the corresponding portfolio X replicates the cash ? ow and has zero T terminal value. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage price of the cash ? ow at time zero. Remark: As shown in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Integrate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take conditional expectation w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E[ t Cu Du du|Ft ]. So Xt = Dt E[ t Cu Du du|Ft ].This is the no-arbitrage price of the cash ? ow at time t, and we have justi? ed formula (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Since dBi (t)dBi (t) = P. (ii) Proof. dSi (t) = = = R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + (? i (t) ? R(t))Si (t)dt ? ?i (t)Si (t)? i (t)dt d d ? ij (t) ? ij (t) d d j=1 ? i (t) ? j (t)dt = j=1 ? i (t) dWj (t) + ? ij (t)2 d e j=1 ? i (t)2 dt = dt, by L? vy’s Theorem, Bi ? ij (t) d j=1 ? i (t) dWj (t). So Bi is a is a Brownian motion under R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + j=1 ?ij (t)? j (t)Si (t)dt ? Si (t) j=1 ?ij (t)? j (t)dt R(t)Si (t)dt + ? (t)Si (t)dBi (t). (iii) Proof. dBi (t)dBk (t) = (dBi (t) + ? i (t)dt)(dBj (t) + ? j (t)dt) = dBi (t)dBj (t) = ? ik (t)dt. (iv) Proof. By It? ’s product rule and martingale property, o t t t E[Bi (t)Bk (t)] = E[ 0 t Bi (s)dBk (s)] + E[ 0 t Bk (s)dBi (s)] + E[ 0 dBi (s)dBk (s)] = E[ 0 ?ik (s)ds] = 0 ?ik (s)ds. t 0 Similarly, by part (iii), we can show E[Bi (t)Bk (t)] = (v) ?ik (s)ds. 51 Proof. By It? ’s product formula, o t t E[B1 (t)B2 (t)] = E[ 0 sign(W1 (u))du] = 0 [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0. Meanwhile, t E[B1 (t)B2 (t)] = E[ 0 t sign(W1 (u))du [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0 t = 0 t [P (W1 (u) ? ) ? P (W1 (u) < u)]du 2 0 = < 0, 1 ? P (W1 (u) < u) du 2 for any t > 0. So E[B1 (t)B2 (t)] = E[B1 (t)B2 (t)] for all t > 0. 5. 13. (i) Proof. E[W1 (t)] = E[W1 (t)] = 0 and E[W2 (t)] = E[W2 (t) ? (ii) Proof. Cov[W1 (T ), W2 (T )] = E[W1 (T )W2 (T )] T T t 0 W1 (u)du] = 0, for all t ? [0, T ]. = E 0 T W1 (t)dW2 (t) + 0 W2 ( t)dW1 (t) T = E 0 W1 (t)(dW2 (t) ? W1 (t)dt) + E 0 T W2 (t)dW1 (t) = ? E 0 T W1 (t)2 dt tdt = ? 0 1 = ? T 2. 2 5. 14. Equation (5. 9. 6) can be transformed into d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt [dSt ? rSt dt ? adt]. So, to make the discounted portfolio value e? t Xt a martingale, we are motivated to change the measure t in such a way that St ? r 0 Su du? at is a martingale under the new measure. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . Hence dSt ? rSt dt? adt = [(? t ? r)St ? a]dt+? St dWt = ? St Set ? t = (? t ? r)St ? a ? St (? t ? r)St ? a dt ? St + dWt . and Wt = t ? ds 0 s + Wt , we can ? nd an equivalent probability measure P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the rational for formula (5. 9. 7). This is a good place to pause and think about the meaning of â€Å"martingale measure. † What is to be a martingale?The new measure P should be such that the discounted value pro cess of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payo? at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ? [0, T ]. The di? culty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: ? 1 ? 1 Vt = Xt = Dt E[DT XT |Ft ] = Dt E[DT VT |Ft ]. You can think of martingale measure as a calculational convenience.That is all about martingale measure! Risk neutral is a just perception, referring to the actual e? ect of constructing a hedging portfolio! Second, we note when the portfolio is self-? nancing, the discounted price process of the underlying is a martingale under P , as in the classical Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case the relation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no longer holds, as shown in formula (5. 9. 6). The portfolio is no longer self-? nancing, but self-? nancing with consumption. What we still want to retain is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the above paragraph. Let VT be a payo? at time T , then for the martingale Mt = E[e? rT VT |Ft ], by Martingale Representation rt t Theorem, we can ? nd an adapted process ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , then the ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by setting X0 = M0 = E[e? T VT ], we must have e? rt Xt = Mt , for all t ? [0, T ]. In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justi? es the risk-neutral pricing of Europea n-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is di? erent from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the backward SDE dXt = ? t dSt ? a? t dt + r(Xt ? ?t St )dt XT = VT for two unknowns, X and ?. To do so, we ? nd a probability measure P , under which e? rt Xt is a martingale, t then e? rt Xt = E[e? T VT |Ft ] := Mt . Martingale Representation Theorem gives Mt = M0 + 0 ? u dWu for some adapted process ?. This would give us a theoretical representation of ? by comparison of integrands, hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5. 9. 7) under P , then d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , where X is given by (5. 9. 6), is a P -martingale. (ii) 1 1 Proof. By It? ’s formula, dYt = Yt [? dWt + (r ? 2 ? 2 )dt] + 2 Yt ? 2 dt = Yt (? dWt + rdt). So d(e? rt Yt ) = o t a ? e? rt Yt dWt and e? rt Yt is a P -martingale.Moreover, if St = S0 Yt + Yt 0 Ys ds, then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (? dWt + rdt) + adt = St (? dWt + rdt) + adt. Ys This shows S satis? es (5. 9. 7). Remark: To obtain this formula for S, we ? rst set Ut = e? rt St to remove the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Just like solving linear ODE, to remove U in the dWt term, we consider Vt = Ut e Wt . It? ’s product formula yields o dVt = = e Wt dUt + Ut e Wt 1 ( )dWt + ? 2 dt + dUt  · e Wt 2 1 ( )dWt + ? 2 dt 2 1 e Wt ae? rt dt ? ? 2 Vt dt. 2 53 Note V appears only in the dt term, so multiply the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt Wt + 2 ? t dt. Set Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys E[ST |Ft ] = S0 E[YT |Ft ] + E YT 0 t a ds + YT Ys T t T a ds|Ft Ys E YT |Ft ds Ys E[YT ? s ]ds t = S0 E[YT |Ft ] + 0 a dsE[YT |Ft ] + a Ys t t T = S0 Yt E[YT ? t ] + 0 t a dsYt E[YT ? t ] + a Ys T t = = S0 + 0 t a ds Yt er(T ? t) + a Ys ads Ys er(T ? s) ds S0 + 0 a Yt er(T ? t) ? (1 ? er(T ? t) ). r In particular, E[ST ] = S0 erT ? a (1 ? erT ). r (iv) Proof. t dE[ST |Ft ] = aer(T ? t) dt + S0 + 0 t ads Ys a (er(T ? ) dYt ? rYt er(T ? t) dt) + er(T ? t) (? r)dt r = S0 + 0 ads Ys er(T ? t) ? Yt dWt . So E[ST |Ft ] is a P -martingale. As we have argued at the beginning of the solution, risk-neutral pricing is valid even in the presence of cost of carry. So by an argument similar to that of  §5. 6. 2, the process E[ST |Ft ] is the futures price process for the commodity. (v) Proof. We solve the equation E[e? r(T ? t) (ST ? K)|Ft ] = 0 for K, and get K = E[ST |Ft ]. So F orS (t, T ) = F utS (t, T ). (vi) Proof. We follow the hint. First, we solve the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our analysis in part (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Integrate fr om 0 to t on both sides, we get Xt = St ? S0 ert + a (1 ? ert ) = St ? S0 ert ? a (ert ? 1). In particular, XT = ST ? S0 erT ? a (erT ? 1). r r r Meanwhile, F orS (t, T ) = F uts (t, T ) = E[ST |Ft ] = S0 + t ads 0 Ys Yt er(T ? t) ? a (1? er(T ? t) ). So F orS (0, T ) = r S0 erT ? a (1 ? erT ) and hence XT = ST ? F orS (0, T ). After the agent delivers the commodity, whose value r is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious.Note the form of Z is similar to that of a geometric Brownian motion. So by It? ’s o formula, it is easy to obtain dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), then Xt = Yt Zt = x  · 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au ? ?u ? u ? u du + dWu Zu Zu [Yu bu Zu + (au ? ?u ? u ) + ? u ? u ]du + (? u Zu Yu + ? u )dWu Yu (bu Zu du + ? u Zu dWu ) + Zu (bu Xu + au )du + (? u Xu + ? u )dWu . + ? u Z u ? u du Zu Remark: To see how to ? nd the above solution, we manipulate the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we multiply on both sides of (6. 2. 4) the integrating factor e? bv dv . Then d(Xu e? ? Let Xu = e? u t u t bv dv ) = e? u t bv dv (au du + (? u + ? u Xu )dWu ). u t bv dv Xu , au = e? ? u t bv dv au and ? u = e? ? bv dv ? ? u , then X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To deal with the term ? u Xu dWu , we consider Xu = Xu e? ? dXu = e? u t u t ?v dWv . Then ?v dWv ?v dWv ? ? [(? u du + ? u dWu ) + ? u Xu dWu ] + Xu e? a ? u t u t 1 ( u )dWu + e? 2 u t ?v dWv 2 ? u du ? +(? u + ? u Xu )( u )e? ? ?v dWv du 1 ? 2 ? ? ? = au du + ? u dWu + ? u Xu dWu ? ?u Xu dWu + Xu ? u du ? ?u (? u + ? u Xu )du ? ? ? 1 ? 2 = (? u ? ?u ? u ? Xu ? u )du + ? u dWu , a ? ? 2 where au = au e? ? ? ? 1 d Xu e 2 u t ?v dWv 2 ? v dv and ? u = ? u e? ? ? = e2 1 u t 2 ? v dv u t ?v dWv . Finally, use the integrating factor e u t 2 ? v dv u 1 2 ? dv t 2 v , we have u t 1 ? ? 1 2 (dXu + Xu  · ? u du) = e 2 2 [(? u ? ?u ? u )du + ? u dWu ]. a ? ? Write everything back into the original X, a and ? , we get d Xu e? i. e. d u t bv dv? u t 1 ? v dWv + 2 u t 2 ? v dv = e2 1 u t 2 ? v dv? u t ?v dWv ? u t bv dv [(au ? ?u ? u )du + ? u dWu ], Xu Zu = 1 [(au ? ?u ? u )du + ? u dWu ] = dYu . Zu This inspired us to try Xu = Yu Zu . 6. 2. (i) 55 Proof.The portfolio is self-? nancing, so for any t ? T1 , we have dXt = ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) + Rt (Xt ? ?1 (t)f (t, Rt , T1 ) ? ?2 (t)f (t, Rt , T2 ))dt, and d(Dt Xt ) = ? Rt Dt Xt dt + Dt dXt = Dt [? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = Dt [? 1 (t) ft (t, Rt , T1 )dt + fr (t, Rt , T1 )dRt + frr (t, Rt , T1 )? 2 (t, Rt )dt 2 1 +? 2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )? 2 (t, Rt )dt 2 ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = ? 1 (t)Dt [? Rt f (t, Rt , T1 ) + ft (t, Rt , T1 ) + ? t, Rt )fr (t, Rt , T1 ) + ? 2 (t, Rt )frr (t, Rt , T1 )]dt 2 1 +? 2 (t)Dt [? Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + ? (t, Rt )fr (t, Rt , T2 ) + ? 2 (t, Rt )frr (t, Rt , T2 )]dt 2 +Dt ? (t, Rt )[Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]]dWt = ? 1 (t)Dt [? (t, Rt ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )dt + ? 2 (t)Dt [? (t, Rt ) ? ?(t, Rt , T2 )]fr (t, Rt , T2 )dt +Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]dWt . (ii) Proof. Let ? 1 (t) = St fr (t, Rt , T2 ) and ? 2 (t) = ? St fr (t, Rt , T1 ), then d(Dt Xt ) = Dt St [? (t, Rt , T2 ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt |[? t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. Integrate from 0 to T on both sides of the above equation, we get T DT XT ? D0 X0 = 0 Dt |[? (t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. If ? (t, Rt , T1 ) = ? (t, Rt , T 2 ) for some t ? [0, T ], under the assumption that fr (t, r, T ) = 0 for all values of r and 0 ? t ? T , DT XT ? D0 X0 > 0. To avoid arbitrage (see, for example, Exercise 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? [0, T ]. This implies ? (t, r, T ) does not depend on T . (iii) Proof. In (6. 9. 4), let ? 1 (t) = ? (t), T1 = T and ? (t) = 0, we get d(Dt Xt ) = 1 ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + ? (t, Rt )fr (t, Rt , T ) + ? 2 (t, Rt )frr (t, Rt , T ) dt 2 +Dt ? (t, Rt )? (t)fr (t, Rt , T )dWt . This is formula (6. 9. 5). 1 If fr (t, r, T ) = 0, then d(Dt Xt ) = ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? 2 (t, Rt )frr (t, Rt , T ) dt. We 1 2 choose ? (t) = sign ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? (t, Rt )frr (t, Rt , T ) . To avoid arbitrage in this case, we must have ft (t, Rt , T ) + 1 ? 2 (t, Rt )frr (t, Rt , T ) = Rt f (t, Rt , T ), or equivalently, for any r in the 2 range of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We note d ? e ds s 0 bv dv C(s, T ) = e? s 0 bv dv [C(s, T )(? bs ) + bs C(s, T ) ? 1] = ? e? s 0 bv dv . So integrate on both sides of the equation from t to T, we obtain e? T 0 bv dv C(T, T ) ? e? t 0 t 0 T bv dv C(t, T ) = ? t s 0 e? T t s 0 bv dv ds. Since C(T, T ) = 0, we have C(t, T ) = e 1 ? a(s)C(s, T ) + 2 ? 2 (s)C 2 (s, T ), we get A(T, T ) ? A(t, T ) = ? bv dv T t e? bv dv ds = T e t s bv dv ds. Finally, by A (s, T ) = T a(s)C(s, T )ds + t 1 2 ? 2 (s)C 2 (s, T )ds. t