Engineering Economy ISYE 6225
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Stock Price Dynamics Homework Solutions 1 a i T1 050 72 7020 UA 12rl 72 015 110 1 n1 rg71 00954 ii T1 050 72 7033 UA 12rl 72 0083 110 x1n1rg7 1 0 b Usually the geometric mean rate of return is the most appropriate for measurement of investment performance 2 a Using equations 38 and 41 of the notes we rst calculate 0392p 71 025020 005 402p17 p 006 When n 10 the mean is 10005 05 and the variance is 10006 06 When n 100 the mean is 100005 5 and the variance is 100006 6 When n is 1 million the mean is 507 000 and the variance is 607 000 standard deviation is 245 b We seek 1 000 000 7 507 000 497 000 7 507 000 P S gt 49 000 P gt 1900900 1 60 000 i 60 000 49000 7 507 000 z 17 ltlgtlt 607 000 since 71 1000000 is so large the random variable 51900900 7 50000 607 000 essentially has the standard normal distribution The answer is 0999978 3 We have u 0712 02 02 712 016 012 a Eln 51 012 Stdev In 51 040 b E51 5021 122 Stdev 51 E512 7 E512 52th2 22001 7 122 1e2quot2 72171222 05088 c We have 5025 140 lt lt 7 P50 25140 P1n 50 70150 S 025 V i Pln 45W 7 012025 lt ln 7 012 025 040025 040 025 7 lt1n140 7 012025gt 0400025 7 ltIgt15326 093727 1 Stock Price Dynamics and Related Topics Introduction These notes are intended for students in Financial Engineering or Advanced Engineering Econ omy as a supplement to the lectures They are not intended as a substitute for an in depth study of the theory of stochastic processes applied to nance 11 Overview Here is a road map of what is presented in these notes Section 2 introduces stochastic processes Properties of stationary and independent in crements are de ned Section 3 introduces Brownian Motion a fundamental building block of nancial engi neering and mathematics Basic properties are covered as well as a few notable results While Brownian Motion itself cannot be used to model the stochastic evolution of stock prices functions of Brownian Motion can Section 4 introduces the Random Walk a simple model of a stochastic process It is shown that the Random Walk converges in a certain sense to Brownian Motion It turns out that event probabilities associated with Brownian Motion can be determined by analyzing a limiting Random Walk and vice versa Section 5 introduces Geometiic Brownian Motion which is the most ubiquitous model of stochastic evolution of stock prices Basic properties are established Section 6 revisits the Binomial Lattice and shows that the choices for the parameters we have been using will approximate Geometric Brownian Motion when the number of periods is suf ciently large and the period length tends to zero Section 7 provides a number of numerical examples of the results obtained up to this point Section 8 derives the famous Black Scholes Call Option formula Section 9 illustrates how to use the Black Scholes formula with a number of examples Section 10 introduces onedimensional lto Processes stochastic processes that encompass Geometric Brownian Motion processes Essentially lto Processes may be represented as a stochastic integral involving functions of Brownian Motion We will heuristically derive the famous lto Doeblin s Formula otherwise known as lto Doeblin s Lemma Examples of the use of lto Doeblin s Formula are provided Section 11 revisits the Black Scholes formula by rst deriving the famous Black Scholes equation using lto Doeblin s Lemma The Black Scholes formula solves a partial differ ential equation which involve the sensitivities or the Greeks of the call option value to the stock price and the time to maturity Section 12 closes with an application of continuous time stochastic optimal control to a real options valuation problem 12 Notation In these notes the notation X N Na02 means the random variable X has the normal distribution with mean a and variance 02 Its density function is given by 1 yiu My 7 6 202 1 7 27m and its cumulative distribution function is given by 12 M PltX z may lt2 The abbreviation iid stands for independent and identically distributed as in X are iid random variables77 2 Stochastic Processes Stochastic processes are used to model a variety of physical phenomena for example a stock price over time or a project value over time Each outcome to of a continuous time stochastic process Xt t 2 0 may be identi ed with a function Xw de ned on 0 00 whose graph tXwt t 2 0 is called a sample path The index t is commonly identi ed with time To ease notational burdens the symbol to is often suppressed and one writes Xt or X for the random value of the process at time t It is important to always keep in mind that Xt is a random variable Stochastic processes differ in the properties they possess Often properties are imposed for natural reasons or simply for analytical convenience Two such convenient properties are stationary and independent increments De nition 1 A stochastic process Xt t 2 0 possesses independent increments if the random variables Xt1 7 Xt0 Xt2 7 Xt1 Xtn 7 Xtn1 are independent random variables for any 0 g to lt t1 lt lt tn lt 00 and stationary increments if the distribution of Xt s 7 Xt depends only on difference t s 7 t 5 Example 1 A stochastic process that possesses stationary and independent increments is the counting process Nt associated with a homogeneous Poisson Process A counting process Nt records the number of arrivals that have entered a system by time t Each sample path is a nondecreasing step function that always starts at zero ie NO O and its discontinuities de ne the arrival times of the events At most one arrival may enter the system at any point in time and there are always a nite number of arrivals in any nite interval of time Thus each sample path does not have too many jumps discontinuity points For a homogeneous Poisson counting Process the discrete random variable Nt s 7 Nt is Poisson with parameter As that is its probability mass function is given by 57As 8 k PNt s 7 Nt k 3 for each non negative integer k The mean and variance of Nt s 7 Nt equal As The parameter A de nes the expected rate of arrivals per unit time The times between arrivals are iid exponential random variables One can simulate a39sample path by sampling N exponential random variables X1 X2 XN and de ning Tl 221 Xi 3 Brownian Motion Another example of a stochastic process that possesses stationary and independent increments is Brownian Motion one of the most famous and important stochastic processes of probability and nance 31 De nition De nition 2 Standard Brownian Motion is a stochastic process Bt t 2 0 with the following properties i it always starts at O ie B0 0 ii it possesses stationary and independent increments iii each sample path Bw is continuous almost surely and iv Bt s 7 Bt N NO s A stochastic process Xt t 2 0 is called a a Brownian Motion if it may be represented as M M it amt lt4 where B is standard Brownian Motion and X0 is independent of B By property iv Xts 7Xt IasaBs Nas025 5 The parameter a is called the drift and the parameter 02 is called the variance of the process X The expression Brownian Motion will be used to describe all such processes X the symbol B will be reserved for standard Brownian Motion in which a 0 and 02 1 32 Discussion It was the French mathematician Bachelier who used Brownian Motion in his doctoral disser tation 1900 to describe the price movements of stocks and commodities Brownian Motion is an example of a continuoustime continuous state space Markov process Each increment Xt s 7 Xt is independent of the history of the process up to time t Alternatively if we know Xt m then any knowledge of the values of Xr for r lt t will have no effect on the probability law governing Xt s 7 Xt Two serious problems with using Brownian Motion to model stock price movements are 0 Since the price of the stock has a normal distribution it could theoretically become neg ative which can never happen in practice due to limited liability The price difference over an interval of xed length has the same distribution regardless of the price at the beginning of the period For example the model assumes the probability the stock price will increase by 10 over the next month will be the same whether the initial price is 100 or 20 In the rst case the increase is 10 whereas in the second case it is 50 It is more reasonable to assume the probability of a xed percentage increase is the same From the nancial modelling perspective one may ask Why study Brownian Motion It turns out that modelling the percentage change in the stock price over an interval as a normal random variable underlies the most ubiquitous basic model of stock price dynamics which we shall cover shortly This particular model will be seen as a simple transform of Brownian Motion and so its properties are fundamentally tied to the properties of Brownian Motion Remark 1 It is a famous theorem of probability that the Brownian Motion process exists In a subsequent section we shall heuristically show that Brownian Motion may be seen as an appropriate limit of a particular type of discretetime stochastic process called a Random Walk Remark 2 Another important theorem states that if X is a stochastic process with station ary and independent increments whose sample paths are continuous almost surely then X must be Brownian Motion This means that Brownian Motion could be de ned by stationary and independent increments and sample path continuity alone with the normal distribution property of each increment being a consequence of these assumptions Note that the Poisson Process which has stationary and independent increments is not Brownian Motion since its sample paths are discontinuous 33 A few useful results The structure of Brownian Motion permits closed form expressions for a variety of results though some require advanced techniques to establish We note a few of these results now 331 Covariance The covariance of the random variables BHS and 13 is determined as follows COVBt9Bt EBtth7EBtSEBt Bts Bt BtBtl Bts BtBtl ElBtZl Bts BtlElBtl l t 6 where the second line follows since the means are zero the third line applies a simple but very useful identity the fourth line follows by the linearity of the expectation operator E the fth line follows by the independent increments of Brownian Motion and the last line follows since each increment has zero mean and VarBt In other words for any times 739 and 7quot COX13 BTI EBB min739 H 7 Formula 7 and the algebraic identity are used frequently when analyzing properties of Brow nian Motion in particular stochastic integrals 332 Differentiability Fix a time t and consider the random ratio Bt5 B E lts mean Bt5 B E El 0 9 since BHE 7 13 N N0 6 Its variance VWlBtJre BtlEZ 157 10 which goes to in nity as e 7 0 Since the limit of this ratio as e 7 0 would ideally represent the derivative of B at t it would appear at least heuristically that the limit does not exist A fundamental theorem of Brownian Motion is that each of its sample paths is nowhere di erentiable almost surely Essentially each sample path wiggles too much 333 Variation Variation is a measure of how much a function wiggles With respect to one measure of variation the sample paths of Brownian Motion wiggle far too much with respect to a different measure of variation the sample paths of Brownian Motion are just ne First some de nitions Let Pt ti0 denote a nite collection oftime points for which 0 to lt t1 lt lt tn t Given a function f 0 oo 7 R for each R de ne map 7 lm 7 mm lt11 i1 and de ne the total variation of 1 over 0t as 121 suppt vfPt 12 The function is said to be of bounded variation if 121 lt 00 for all t gt 0 Each nondecreas ing function f has bounded variation since 121 ft 7 f0 A fundamental property is that a continuous function has bounded variation if and only if it may be represented as the dif ference of two nondecreasing functions This fundamental property makes it possible to de ne the Riemann Stieltjes Integral f hdf as the limit in a certain sense of Z htf 7 ft1 If we replace the function with Bw it turns out that almost all sample paths of Brownian Motion have in nite variation Heuristically at least this is not too surprising since the sample paths are in nitely jagged De ne the quadratic uariation of 1 over 0t as 610 supp qfPt 13 where V L qfR Zltfltti ftiil2 14 i1 A fundamental property of Brownian Motion is that almost all of its sample paths have the same quadratic variation given by azt This property contains the essence of the famous lto Doeblin s formula which is the key tool for analysis of stochastic processes related to Brownian Motion Remark 3 A heuristic explanation begins by noting that EqX 02tu2 2ti iti712 When Pt is chosen so that t int then EqX a azt as n a 00 A bit more work shows that VarqXPt a 0 as n a 00 334 Hitting time distribution Let X represent Brownian Motion such that X0 O For each t let M denote the maximum ualue ofX on the interual 0t lts distribution is given by PM lt y af x e shy02m lt15 For y gt 0 let Ty denote the rst time t at which X y It is called a one sided hitting or passage time It should be clear that PTy gt t PMt lt y 16 from which the onesided hitting time distribution may be calculated via 15 The results when X0 z are obtained by considering the process Xt 7 m See Section 7 for an example 335 Ruin probabilities Let X represent Brownian Motion such that the drift a 7 0 and X0 m Imagine an individual who purchases a unit of a good whose price is governed by the X process The individual wishes to hold the good until its price reaches 1 gt z To limit his losses the individual will sell the good should its price decline to a lt m Let Tab denote the rst time the process X reaches either a or b It records the time when the individual will sell the good and either reap the gain I 7 x if XTab b or take a loss of z 7 a if XTab a The chance the individual will go home a winner is e72nm02 7 e72Ma72 PXTable0mWv 17 and the chance the individual will go home ruined is of course 17PXTab b l X0 The formula 17 is valid so long as a lt z lt I even if these numbers are negative Formula 17 is not valid when u 0 however when u z 0 we may substitute 51 z 1 z in 17 to obtain that PXTab b XO z if when M 0 18 See Section 7 for an example 336 Versions of standard Brownian Motion It is possible for a function of standard Brownian Motion to be another version of standard Brownian Motion Examples include o B105 cBt02 for xed 0 gt O o B205 tB1t ift gt 0 and 0 otherwise 0 B305 Bt h 7 1301 for xed h gt O For each of these examples it should be clear that every increment Bit s 7 1305 i 1 2 3 is normally distributed with zero mean and that the increments over disjoint time intervals determine independent random variables To complete the veri cation one must show the sample paths are continuous and that variance of each EBt s 7 Bit2 is indeed s The veri cation of the variance is left as an exercise 4 Random Walks A cornerstone of the theory of stochastic processes is the random walk It provides a simple model of a wealth process of an individual who repeatedly tosses a possibly biased coin and places bets The most common description of the price of a non dividend paying stock will turn out to be a limit of a suitable simple transformation of a random walk 41 De nition We imagine an individual who sequentially tosses a fair coin and who will receive one dollar if the coin comes up heads and will lose one dollar ifthe coin comes up tails Let X1 1 g i lt oo denote an in nite sequence of iid random variables with common distribution given by PXi 1 PXi 71 12 19 The value n 5 Z X 20 i1 is the individual s wealth after n tosses of the coin The process Sn 0 g n lt 00 is called a simple random walk The expected winnings on each toss is of course zero and so ESn 0 21 Moreover Varn n 22 since the X are independent and the variance of each X is one The second moment is always one and the mean is zero Next we allow the coin to be biased so that the probability the coin will come up heads is given by the parameter p E 0 1 In this setting the X have the same distribution given by PXi1p7 PXi 11ip 23 Of course the expected winnings after n tosses is no longer zero and its variance changes too In particular ElSnl nElXil 71 11P1l 2102971 24 Varn nVarlXi n17 2p 712 4np17 p 25 Finally we not only allow the coin to be biased but we now scale the amount won or loss by the parameter a gt 0 In this setting the X now have common distribution given by PXi039P7 PXi01197 26 and ElSnl nElXil 71W 1 P0 l 710219 1 27 Var5n nVarlXi n02 7 02p 712 4n02p17p 28 De nition 3 A random walk shall refer to the positions Sn generated by the general process de ned by 26 A biased random walk shall refer to the special case when a 1 and a simple random walk shall refer to the special case when a 1 and p 12 4 2 Standardization We now incorporate a time dimension as follows Suppose each At gt 0 units of time the individual tosses the coin The value Sn records the individual s wealth after Tn nAt units of time If we instead set a time T gt 0 and insist that the number of tosses n and the time between each toss At are such that the game is always over by time T then n and At determine one another via the equation T nAt 29 It is understood that At is always chosen so that n is an integer For example if T 1 year we could toss the coin quarterly monthly daily hourly or even by the minute which would results in 4 12 365 8760 and 525600 tosses respectively There are two parameters p and a required to specify a random walk If we choose p and a in 26 to be independent of n then for a biased coin 27 and 28 show that o ESn goes to plus or minus in nity depending on whether p gt 12 or p lt 12 as n tends to in nity and o The coef cient of variation WEE is proportional to 1N5 which converges to zero as n tends to in nity Consequently the value of wealth at time T will be unbounded 7 as At a 0 Now keeping in mind identity 29 suppose instead we de ne pi 1 gm 30 an mE 31 and set p pn and a an into 26 Then Em 7 imam 71 7 namngm 7 VT 32 Varn 7 471333917 1 7 4n02At 17 7At 7 UZT 33 then the limiting value of wealth at time T ST lim 5 34 L700 will exist and have a non trivial distribution In particular its mean is always VT and its variance is UZT The parameter 1 in 30 is an arbitrary nite number that does not depend on n and it is understood that At is suf ciently small or n is suf ciently large to ensure that 19W 6 01 43 Limiting distribution Let Xi i 1 2 describe a random walk with parameters pn and an de ned in 30 and 31 respectively Consider the standardized sum 21Xi7 VT rm1 simw The numerator of 35 is a sum of random variables minus its mean and the denominator is the standard deviation of that sum What happens to the standardized sum as it tends to in nity A general1 Central Limit Theorem due to Lindeberg Feller ensures that for every xed real value x 35 3330 Pz lt z a m 36 As a direct consequence to any desired degree of accuracy for each xed real number s we may pick it sufficiently large to ensure that PSn lt s PnX lt s 37 i1 22L Xi 7 VT 5 7 VT la vi lt m lt38 57 VT z ltIgt U 39 To all intents and purposes the the distribution ofSn conuerges to NVT UZT the distribution of the random variable we call ST Remark 4 The rewardloss scale parameter 0 directly determines the variance of the lim iting random variable ST Furthermore given the rewardloss scale if the form of pn follows 30 then each choice of llO39 not only determines the likelihood of success on each coin toss but it determines the mean of the limiting random variable ST The argument above can be extended to all points in time This fundamental extrapolation is known as Donsher s theorem which states that to the naked eye the random walk SM looks like a 1039 Brownian Motion stochastic process The implication of this result can not be understated the chances of a reasonable euent occurring for a Brownian Motion process may be estimated by computing the chance the euent would occur for a random walk with su iciently many steps 1Keep in mind the distribution of Xi depends on ri 44 Functions of a random walk We rst consider a biased random walk ie general 197 a 1 Let the positive integer L represent the maximum amount of money the individual is prepared or can lose7 and let the integer M represent the amount of money the individual wishes to win We suppose the individual will continue to toss the coin until either he reaches his goal of M or is forced to quit by losing L Let 739M7 L denote the rst index integer for which STML M or 7 L If we identify each toss with a unit of time7 then 739M7 L would record the time at which the game ends Note that by de nition of 739M7 L7 STML must equal either M or 7L A fundamental probability of interest is the so called rum probability given by 7139M7 L PSTML reaches M before L 40 Let q 1 7p It can be shown that L71 7rML W 41 E7ML L 7ML7rML 42 q i P q i P The formulas do not apply for the simple random walk since p q However7 by setting p 12 e and 17p 12 7 6 into 41 and 42 and letting e a 07 it may be shown that for the simple random walk 7rM7L MiL 43 ETML ML 44 Example 2 For a simple random walk there would be a 23 chance of winning 100 before losing 200 On the other hand7 when p 0497 051049200 71 M L 00183 45 7 051049300 71 a less than 2 chance of being successful The cost of an even a small bias is surprisingly high Table 44 shows the probability of winning 100 before losing 100 as the probability of winning varies Remark 5 We analyzed ruin probabilities in our discussion about Brownian Motion We have argued that the random walk converges in a certain sense to Brownian Motion It can be shown 12 Table 1 Sample benchmarks l Prob p l 05000 l 04950 l 04900 l 04800 l 04700 l Prob of winning 05000 01191 00179 00003 0000006 l Expected duration of game l 10000 l 7616 l 4820 l 2498 l 1667 l that the probabilities shown in the Table can be estimated using those earlier formulas It will not be exact as the probabilities will be equal only in the limit It turns out that event probabilities concerning continuous functions of Brownian Motion may be found by taking the limit of the corresponding event probabilities of the corresponding random walk7 and vice versa The choice as to which approach to take depends on which analytical techniques will be more suited for the task at hand This fundamental result is knowns as Donsker s Invariance Principle 5 Geometric Brownian Motion 51 De nition A stochastic process Yt 0 g t lt 00 is a Geometric Brownian Motion process if it may be represented in the form Yt Y05Xlttgt 46 where X is a 17 a Brownian Motion Process and Y0 is a positive constant independent of X Note that each sample path of the Y process is strictly positive By taking the natural logarithm of both sides of 467 3t 2 l 7 N N t t 47 n m u a gt lt gt A random variable whose natural logarithm has a normal distribution is called a lognormal random variable and its distribution is called a lognormal distribution Thus7 each ln W3 is lognormally distributed 52 Calculation of probabilities and moments Probabilities of events on Y are easily calculated via the normal distribution For 0 lt a lt b7 PaltYtltb PaY0ltYtY0ltbY0 ln aYoizt lt ln YtYoizt lt ln bYoilt P N N N mln 1323 ut7 mln 113 125 48 Recall the moment generating function for a normal random variable X N Na7 b2 is E59X esa92b2239 49 Using this fact and 467 the mean of Y is given by Yoevt02t2 Y08zxc722t7 50 and its variance is given by mm ElYfl 7 Elm Eli0W 7 Elm Y02l62 2t 7 WWW lt51 6 Binomial Lattice Revisited 61 Convergence Consider an N period binomial lattice in which Sk uSkA with probability p and Sk dSkA with probability 1 7 197 with u gt 1 and d lu By taking the natural logarithm of both sides of the following identity7 Sn Sn Sn7 1 1 So Snil Sniz 5707 52 it follows immediately that Sn ln amp iXi 53 SO i1 where X ln Sil 54 for each 1 g i g N For the binomial lattice the X are iid random variables with common distribution PXilnup7 PXi7lnu17p 55 Now compare 55 to 26 It should be clear that the Sn de ne a random walk in which ln u plays the role of 0 Accordingly for a given T gt O we know if we set u d and p such that ln u av At ltgt u eam 56 ln d faxAt ltgt d gram 57 1 p 7 1 5W lt58 2 a then as At a 0 i the distribution of ln it conver es to a normal distribution with mean 1t and variance so g 2 a t and ii the distribution of ln 549 will be normal with mean 15 and variance 025 Thus the natural logarithm of the normalized price process normalized by SO is a Brownian Motion and so the price process St 0 g t lt 00 is a Geometric Brownian Motion process 62 Interpretation of the parameter V The parameter 1 determines the mean of Zn StSO By de nition the random continuously compounded rate of return over the interval 0 t call it a satis es the equation 5 See It is random since it depends on the value of St Since a ln StSolt wt 59 its expectation Ea 1 60 Consequently the parameter 1 is the expected continuously compounded rate of return on the stock 63 Interpretation of the parameter a From 50 Em Emqu sown2 505 61 By de nition the random growth rate or proportional rate of return of the stock over a short interval 0 At is given by SmSO 71At 62 whose expected value from 61 is 5W4 1iAt717 At At 7 63 when At is close to 0 Recall that cm x 1 z when x z 0 Consequently the parameter a 1 1 022 64 is the expected growth rate or proportional rate of return of the stock 64 Relationship between V and a For the stochastic price process described here the expected continuously compounded rate of return and the expected proportional rate of return are linked via the identity 64 and are most de nitely not equal If the price process 5 were deterministic with St Seem then clearly the continuously compounded rate of return would be a and a would equal 1 In the stochastic setting an adjustment of 022 must be made when moving from the logarithm of normalized prices to the normalized prices themselves Often the parameter a is given as well as a from which one calculates 1 The next section gives some example calculations 7 Examples We follow common convention and measure time in years so that t 1 corresponds to 1 year from now A blue chip stock like IBM would have its annual standard deviation 0 of say 28 and a stock like Microsoft would have its 0 around 40 Keep in mind the variance of the natural logarithm of the price process is proportional to time and so the standard deviation is proportional to the square root of time Suppose the expected proportional rate of return on IBM or Microsoft stock is a 12 per year or 1 per month For t 112 1 month ln StSO X N N012 7 2822 112 008082 0 A positive 30 event corresponds to ln StSo 000673 30808 or St 128350 0 A negative 30 event corresponds to ln StSO 000673 7 30808 or St 079050 When t 312 3 months ln StSo X N N012 7 2822 312 0142 o A positive 30 event corresponds to ln StSo 00202 314 or St 155350 0 A negative 30 event corresponds to ln StSO 00202 7 314 or St 067050 For Microsoft the corresponding numbers are S 141950 and St 071050 for t 112 and St 1847SO and St 055650 for t 312 Suppose the current price for Microsoft Stock is So 100 and assume a 012 and a 040 Here 1 a7 022 012 7 04022 004 Let t 112 which corresponds to the end of this month We shall use the fact that the stochastic process X ln StSo is 110 Brownian motion with 1 004 o What is the probability that the stock price will exceed 120 by time t We seek ln StSO 7 1t gt ln12 7 00412 a 01112 17 lt1gt155 00657 PStgt120 P 0 Suppose we hold one share of Microsoft stock and want to know the chance that over the next month the stock price will never exceed 120 The event that S g 120 for each 739 E 0t is equivalent to the event that X ln STSO y for each 739 E 0t where y ln 12 As a result we can directly apply 15 with the a parameter there set to 004 to obtain that POr17agxtST 120 POr17agxt ln 57100 12 PMt S y 7 20 o4ln12 7 7 ltln 12 04112 7 SW ln 12 04112 040112 040112 ltIgt155 7 10954ltIgt7161 09343 7 1095400537 08806 and so there is an 88 chance the stock price will never exceed 120 over the next month In addition we know the chance that it will take at least one month for the stock price to exceed 120 is also 88 0 Suppose we buy one share of Microsoft stock and decide to hold it until it either rises to 110 our cash out point or it falls to 80 our stoploss point What is the probability we will cash out with a pro t of 10 We can directly apply 17 with b ln 12 x 0 and a ln 080 to obtain that PSTab120150100 PXTabln12lX00 17 1 7 5720040402a 5720040402b 7 5720040402a 70118034 05753 8 Derivation of the BlackScholes Call Option Pricing Formula We calculate the value of a call option on a non dividend paying stock by modeling the price process via an N step binomial lattice and then computing the discounted expected value of the options payoff at time T using the risk neutral probability For the binomial lattice we set the parameter p to be the risk neutral probability denoted by q The risk neutral probability is the unique value for q that ensures the discounted expectation of next period s stock prices using the risk free rate of return rAt always equals the current price namely qUSk71 1 qd5k71 Si k1 1rAt 65 The unique value for q is 1 rAtSk1 7 dSk1 1 rAt 7 d 66 q ask1 7 15k1 u 7 d Since u 50m 1 547 and At z 0 we may use the approximation 1 z z22 for em in 66 to obtain 1 rAt 7 17 axAt UZAtQ 1 axAt UZAtQ 7 17 axAt UZAtQ r 7 022At axAt 20 At g 117 KxAt 67 2 a where Vr7022 68 In light of previous developments it should be clear that when there are many steps to the binomial lattice the distribution of STS0 is lognormal with parameters 1 r 7 022 and 0 Since the expected growth rate of the stock is 1 1 022 we see that under the risk neutral probability the expected growth mte of the stock is the risk free mte In the limit as the number of steps of the binomial lattice goes to in nity the value of a European call option on a non dividend paying stock is given by Ee TTMaxST 7 K 0 69 18 when the underlying distribution is ST lnm Nr 7 022T UZT 70 It is possible to compute this expectation and derive a closed form analytical solution for a call option value known as the famous Black Scholes formula to which we now turn To simplify the derivation to follow we express ST S0eX where X N Na b2 Upon substitution 69 may be expressed via the following equalities 00 1 1 aka 2 77T m 7 7 77 M S K 0 d e Leo ax 05 mbe 2 b x 00 1 1 aka 2 e TT Semi 57 b dm ln KSo 0 x27rb 00 1 1 aka 2 00 1 1 aka 2 S e TT em b dmiKe TT 557 b dz 71 0 ln KSo x27rb ln KSO x27rb By completing the square in the exponent of the exponential the rst integral on the right hand side of 71 is equivalent to w 2 505T5a122 57lt bb2dz 72gt ln KSo 19 which may be further simpli ed via the following chain of equalities 00 1 SOE TTeaJFl Zl 7 2 7 HKSob ab m ln KSO 7 a b2 1 57y2dy 7 soarTeWVZ 1 7 lt1gt ln SoK 7022T axT 3 50 d1 73 after substituting the de nitions for a r 7 022T and b2 UZT 50 The second integral on the right hand side of 71 is equivalent to K TT 00 71 lyZd 74 e 72752 a lt gt b which may be further simpli ed via the following chain of equalities Ke TT 17 ltIgt ln KS 7 a 19 ln SOK a Ke lT W l Ke rT fTfi Kerrdeg 75 after substituting once again for the de nitions for a and b We conclude with the famous Black Scholes European call option formula is given by CK T SltIgtd1 7 Ke TTd2 76 where recall that denotes the cumulative standard normal density7 and ln SK r 022T d 77 1 mT ln r 7 022T d d 7 xT 78 2 mT 1 U Notice that the parameter la is nowhere to be found in this formula Remark 6 Note that d2 d1 7 U 79 Remark 7 Let CS denote the value of the call option as a percentage of the current stock price7 let m SKe TT 80 a2 axT 81 Note that C 1 1 1 H lt61 62 n quot91 quot92 7 7 7 7 lt1gt 77 82 S l 92 2 91 l 92 2 l7 which shows that one only needs the values for 1 and 2 to compute the value of the call option 9 Call Option Pricing Examples Consider a call option on IBM stock Suppose current value of its stock is So 100 The annual standard deviation of log volatility is 28 and the annual risk free rate will be set to 5 As a function of the K and T we have that ln 100K 00892T 028T d2K T d1K T 7 02sxT CK T 100ltIgtd1K T 7 Ke O39O5TltIgtd2K T 011K7 T Now consider the following options 0 Cost of a 1 month call option with strike price 100 Here 11 009207 12 001117 ltIgtd1 053677 ltIgtd2 05044 and C100711210005367799584205044 344 0 Cost of a 3 month call option with strike price 100 Here 11 015937 12 001937 ltIgtd1 056327 ltIgtd2 05076 and C1007 312 10005632 7 987577805076 619 0 Cost ofa 1 month call option with strike price 105 Here 11 7051177 12 7059257 ltIgtd1 030437 ltIgtd2 02768 and C1057112 10003043 7 104563402768 149 0 Cost of a 3 month call option with strike price 105 Here 11 7018927 12 7032927 ltIgtd1 042507 ltIgtd2 03710 and C1057312 10004250 7 103695703710 403 Now consider what happens when a 40 Cost of a 1 month call option with strike price 100 Here 11 009387 12 7002177 ltIgtd1 053757 ltIgtd2 04916 and C100711210005375799584204916 479 Cost of a 3 month call option with strike price 100 Here 11 016257 12 7003757 ltIgtd1 056467 ltIgtd2 04850 and C1007 312 10005646 7 987577804850 856 Cost of a 1 month call option with strike price 105 Here 11 7032877 12 7044417 ltIgtd1 037127 ltIgtd2 03284 and C1057112 10003712 7 104563403284 278 Cost of a 3 month call option with strike price 105 Here 11 7008157 12 7028157 ltIgtd1 046257 ltIgtd2 03891 and C1057312 10004625 7 103695703891 590 Notice that as the time to expiration increases the call option value increases7 and that as the strike price increases the call option value decreases Both observations are true in general 21 10 OneDimensional Ito Processes A stochastic process that encompasses Brownian Motion and Geometric Brownian Motion is the Ito Process also known as a stochastic integral We shall describe a subset of Ito processes that possess the Markov property 101 De nition and interpretation In differential form X is a Markov Ito process if it may be represented as dXt aXt tdt 0Xt tdBt 83 for appropriate choices for the functions a and 02 Equations of type 83 are known as stochastic di erential equations One loosely interprets 83 to mean that for sufficiently small At Xt At 7 Xt z aXttAt 0Xt tBt At 7 Bt 84 Keep in mind that the both sides of 84 are random variables and that BaAaimaNmAa as One may use 84 to simulate the X process as follows Fix a small period length At gt 0 and de ne t iAt for integer i 2 1 For each i let Z be iid random variables with common distribution given by N0 At Beginning with the known constant XO simulate Z1 Z2 and sequentially set Xm4XMMWmNdMWmA X052 3 XtltMXt17t1At0Xt17t1Z2 Xtk1 3 Xtk MXtktkAt 039XtktkZk1 Example 3 The simplest example is when aXt t a and 0Xt t a gt O in which case dXt idt adBt 86 Although we have not formally discussed what it means to integrate both sides of 86 any reasonable de nition of integration would suggest that XtiX0LLt039Bt which is precisely Brownian Motion 2General Ito processes permit the functions a and a to depend on the whole history of the X process up to time t Example 4 Here let uXt t uXt and 0Xt t TX gt O in which case dXt LLXtdt O39XtdBt Since Xt appears on both sides of 88 we cannot directly integrate both sides to solve for Xt By dividing both sides by Xt dXt Xt Since we interpret dXtXt to mean Xt At 7 XtXt for small At and since udt 0013 89 Xt At 7 Xt N Xt At Xt n X05 lnXtAt7lnXt it would appear that 89 is equivalent to d In Xt udt adBt 90 Since Xt no longer appears on the right hand side of 90 we can integrate both sides as before to obtain ln Xt7ln X0 pt039Bt 91 The conclusion is that here X is a Geometric Brownian Motion process Actually this heuristic derivation is correct up to a point but the calculus performed is not correct The drift parameter does not equal In as suggested in 91 but must be adjusted to u 7 022 This is not too much of a surprise given our discussion relating to 64 Example 5 For the mean reversion model uXtt 11 7 Xt 0Xtt aXt Here the parameter 1 denotes the long run average value of the process X and the parameter a gt 0 determines the speed with which the X process adjusts towards 1 Note that when X lt b the instantaneous drift is positive and there will be a tendency for X to increase when X gt b the instantaneous drift is negative and there will be a tendency for X to decrease Mean reversion is often used to model prices of natural resources As the price increases above the long term average producers increase supply to take advantage of the higher prices which then has a natural tendency to bring down the prices The opposite would hold true when the price decreases below the long term average 102 Functions of Ito processes and ItoDoeblin7s lemma Let X be an lto process as in 83 Let gt x be a twice continuously differentiable function de ned on 0 00 x B What can be said about the process Y gt Xt 23 The heuristic development proceeds as follows Using a second order Taylor series expansion about the point t Xt for At sufficiently small N 89 89 829 829 2 829 2 Yt At 7 Yt 5 At a AXt 8583 AtAXt 12 At 12AXt 92 where we let AXt Xt At 7 Xt 93 The idea from here on is to eliminate those terms on the right hand side of 92 that involve powers of At greater than 1 As At a 0 such terms will become increasingly inconsequential The term involving At must obviously be kept and so must the term involving AXt since its standard deviation goes as the xE which becomes large relative to At when At is small Obviously the At2 may be eliminated The term involving AtAXt has conditional mean of zero and variance of At3 and so this random variable converges to zero fast enough as At a O and so it may be eliminated too We are left with the AXt2 For notational convenience let AB C Bltt At 7 Bt th C LLXtt 039 C 039Xt t Using 84 Am HEW 2mAtAXt a ABt 95 By similar reasoning as before the rst and second terms on the right hand side of 95 may be eliminated but the third term must be kept In particular it converges to afAtfl Putting it all together as At a 0 dyt YtAt7Yt At 31 ltAX12ltAXtgt2 g N g mAt atABt12UAt 3 m 12a gt At at A3 g g t 1227ZUEgt dt 79 at dBt 96 Equation 96 is known as Ito Doeblin s formula proved by Ito and Doeblin in what is now known as Ito Doeblin s Lemma 3The type of convergence is where the quadratic Variation of Brownian motion plays a central role 24 103 Examples of Ito Calculus lto Doeblin s formula is a powerful tool for analyzing functions of Brownian Motion We illus trate with a number of examples Example 6 Consider the lto process described in 88 Let Y gt7 Xt ln Xt Then dYt Xitht 12 72UZX3gt dt XitaXtgt dB a 7 022dt adBt 97 We may now integrate both sides of 97 to obtain that In Xt ln X0 zt a3 where 1 a 7 722 98 Equivalently7 Xt Xert with Z being 17 a Brownian Motion Now since Xt fZt XOeZ 99 another application of lto Doeblin s Lemma yields dXt f ZtdZt12f ZtdZt2 XtzdtadBt 12Xt02dt u022Xtdt aXtdBt aXtdt aXtdBt 100 which is identical to 887 as it should We may use lto Doeblin s Formula to evaluate stochastic integrals7 as the following example illustrates Example 7 The expression t Bsst 101 0 is a stochastic integral Roughly speaking7 it is the limit of a sum of random variables t V L 0 139st gigozBalMBalH 7 132 102 13970 where 0 to lt t1 lt t2 lt lt tn t represent a suitable partition of the interval 07 t We can use the lto calculus to evaluate this stochastic integral From classical calculus one would 25 suspect the term 12 B would crop up somewhere So7 let gtBt 12 Bf Here7 X is simply Bt and M 0 and at is identically one We have a a a a dYt ltglpt1287Za gt dtlt7g gtdBt 8t 8m 8m 7 0 Bt0121gt dt Bt1dBt 103 which when integrated gives 12 Bf12tOt Bsst 104 In other words7 0t 39st 12 13372 105 We can use 105 to calculate the mean and variance of the stochastic integral E Ot Bsstl Var Ot Bsstl Eli2 133725 0 since 1933 t 2 BsdBS since the mean is zero7 0 14EBt4 7 2th 252 143t2 7 2t2 t2 2522 since EB 1 3252 Remark 8 The Ito Isometry can be used to directly calculate the second moment as EOtBsdBS2 EOtB ds OtEB52ds Otsds t22 106 11 BlackScholes Revisited 111 BlackScholes differential equation Let CSt7 t denote the value of a European call option on a non dividend paying stock7 whose stock price 5 follows the stochastic differential equation 88 Recall that under the risk neutral probability the growth rate of the stock is the risk free rate r and so u r By lto Doeblin s Formula7 dC Ct was 12 7252055 dt 7505ch 107 where for convenience we let 0 QCQt7 Cs 8085 and 055 82 0852 and we suppressed the if from St Let H ihS C denote a portfolio of 7h units of S and 1 unit of C The instantaneous change in the portfolio H is given by 111 ihdS 10 7hrSdt anBt d0 108 In light of 107 if we set h 05 then the stochastic term involving 13 vanishes In particular for this choice of h we have 111 0 12 0252055dt 109 Since the portfolio is instantaneously risk free it must earn the risk free rate namely dH ert M7055 Cdt 110 Putting 109 and 110 together it follows that the Call Option value must satisfy the partial differential equation given by To 0 was 12 17252055 111 Equation 111 is known as the Black Scholes di erential equation One may verify that the previously given closed form expression for the Black Scholes formula does indeed satisfy 111 1 12 The Greeks The instantaneous units of stock to hold h is called the hedge ratio It is so named since h 8085 ACAS It is the limit of the hedge ratio obtained in the replicating portfolio obtained in our Binomial Lattice models lnstantaneously H acts like a riskless bond and the replicating portfolio is given by hS H Let c A Cs denote the sensitivity of the call option value to the price S o 9 0 denote the sensitivity of the call option value to the remaining time t and o P 055 denote the sensitivity of A to S From the Black Scholes differential equation we see that To e rSA 12 172521 112 It is not too difficult to show that A ltIgtd1 113 from which it follows that Md p 71 114 Sz T 27 where recall that denotes the density function of a standard normal random variable We know that C SltIgtd1 7 rKe TT d2 Consequently after a little algebra the value of 9 is determined as 775 gtD107 54 of NT K TltIgtd2 115 12 Application to Real Options Valuation Let the value of a project be governed by the stochastic differential equation th thdt thdBt 116 Let p denote the appropriate discount rate cost of capital for projects of this type We assume p gt u gt 0 Let the xed constant I denote the required investment when the project is undertaken The only question is one of timing namely when should the project be undertaken Clearly the optimal timing decision should be based on the current value of the project Vt Since the problem parameters are independent of time this is the only piece of information required ie it is the single state variable Let denote the value of the project with this embedded delay option At the current point in time only two actions are possible invest now in the project or ii continue to wait At the instant the investment takes place the value of the project is V 7 I If on the other hand it is not optimal at this time to invest then the current value must equal the discounted expectation of the future value For the next At units of time this discounted expectation is given by 1 7E 1 pAt t where the expectation is taken with respect to the information known at time 15 By Bellman s Principle of Optimality lFl1iAtlv 117 Fm max Vt 7 I FVtm 118 1 7E 1 pAt t Suppose the value of V is such that we are in the continuation region Then Fm m mm 119 Multiply both sides of 119 by 1 pAt and then subtract to obtain pmmm mwman am It is okay to put the expression inside the expectation since it is a known constant at time 15 Using lto Doeblin s formula dF 1101th anBt 12 FVVanzdt 121 MVFV 12 UZVZFvvdt aVFVdBt 122 As before the subscripts on F indicate partial derivatives It then follows that EtFVtAt 7 FVt EtdF MVFV 12 UZVZFVVAt 123 Keep in mind that EtABt 0 We conclude then that in the continuation region the value function F satis es the following differential equation 0 uVFV 12 UZVZFVV 7 pF 124 It may be readily checked that FV AV is a solution to 124 Substituting into 124 we have 0 26 12 026 71 p cm 125 Let quadratic form q is convex its value at 0 is negative its value at 1 is negative too since p gt u by assumption and it is unbounded as B or 73 tend to in nity Consequently q has two real roots 31 lt 0 and 32 gt 1 The function A1V 1 A2V 2 126 solves 124 Clearly if V 0 then FV 0 too which rules out the negative root Thus FV sz Z 127 It remains to determine the value of A2 and to determine the optimal control policy The optimal control policy is a threshold policy when V reaches a critical threshold V1 then it is time to implement At implementation it must be the case that Hmmik am the left hand side can never be less than the right hand side and if it were greater then implementation should have commenced a few moments ago Equation 128 is one of two necessary equations to pin down the two values V1 and A2 It turns out that at implementation d d WFW 1W WW 1 1 129 29 Consumption Investment and the Fisher Separation Principle 1 Consumption with a Perfect Capital Market Consider a simple two period world in which a single consumer must decide between consump tion 00 today in period 0 and consumption 01 tomorrow in period 1 The consumer is endowed with money me today and m1 tomorrow Consistent with his endowment the con sumer has the opportunity to borrow or lend be today at interest rate r Borrowing corresponds to a positive 0 whereas lending corresponds to a negative 0 11 Budget constraint The equations governing the consumer s feasible actions today and tomorrow are as follows Consumption in period 0 equals the original endowment adjusted for the borrowing or lending CO mg b0 We assume the consumer prefers more consumption to less and so we write this constraint as an equality Since consumption cannot be negative the consumer cannot lend more than me ie 0 Z 77710 Consumption in period 1 equals the original endowment adjusted for the repayment or receipt of the principal and interest associated with the borrowing in period 0 ie Cl m1 7 1 Tbo 2 Once again since the consumer is assumed to prefer more consumption to less we write this constraint as an equality Since consumption cannot be negative there is a limit as to how much the consumer can borrow in period 0 namely b0 After substituting co 7 mg for be in 2 the consumer s budget constraint is cl 7 m1 1 r 7 m0 1 r The symbol W0 represents the consumer s present value of wealth 30 2 W0 Example 1 Suppose m0 100 m1 990 and r 10 The present value of wealth is W0 100 99011 1000 The consumption plan 0001 500550 satis es the budget constraint To achieve this consumption plan the consumer borrows 400 in period 0 and pays back 440 in period 1 The consumption plan 0001 800 220 also satis es the budget constraint To achieve this consumption plan the consumer borrows 700 in period 0 and pays back 770 in period 1 12 Optimal consumption plan Which consumption plan will the consumer choose To determine the optimal consumption plan we assume existence of a utility function Uco 01 such that the consumer prefers 05 0 to 0 a if and only if We of gt T103376 4 If the utility values are equal then the consumer is said to be indifferent to the two con sumption plans Under reasonable assumptions on consumer preferences existence of a utility function that re ects the consumer preferences is guaranteed to exist For the remainder of this handout we shall assume that 1030701 M Bx 5 Example 2 We continue with our example We shall set 3 06 The utility of the consump tion plan 0001 500 550 is 3643 the utility of the consumption plan 0001 800 220 is 3718 and the utility of the original endowment 0001 100 990 is 2888 Of these three choices the consumer prefers 800 220 The ability to borrow de nitely helps our consumer Formally the consumer s optimization problem is MAX Uc0c1 Co 11 W0 6 De ne 6160 3 17 W0 CO7 1030 U007 01030 M xC col 8 The consumer s optimization problem can be equivalently expressed as 1 V 030 2 0 g 30 Note how the parameter 3 serves as a discount factor on future consumption Smaller values of 3 imply a larger discount factor on future consumption which implies that our consumer prefers more consumption today Consider for example the two extreme values for 3 namely 0 and 00 How do we obtain the optimal rst period consumption of and hence the optimal second period consumption 0103 The form of the utility function implies that consumption in both periods must be positive This is because the derivative of at zero is in nite Consequently the optimal choice for co necessarily lies strictly between 0 and W0 Accordingly to nd the optimal choice of 00 we set the derivative of to zero Using the chain rule optimality conditions imply that o i ltc3gt 81 81 e 1 L1 lt10 7 360 861 leg 7 2M 7 2 CICE39 2 It follows directly from 10 that the optimal consumption plan necessarily satis es the condition Ci 51T7 11 or equivalently a 621 rgt2c3 lt12 Substituting 12 into the budget constraint 3 and solving for of and of the optimal con sumption plan is 1 if 17 CJlW1lllJquotPJ 0 13 321 N2 1 cl 1321TWo7 01Wo 14 The constants p0 and p1 are independent of present wealth W0 that is they are known param eters strictly determined by the two discount factors 6 and r We shall use this important fact later Note further that the optimal utility is of the form UWo WT WNW 15 Example 5 In our example of 0716W0 of 0312W0 and UW0 1182xW0 Here Thus 03 716 of 312 and U1000 3736 Due to the availability of a capital market at which to borrow or loan the consumer has increased his utility by almost 30 above the level corresponding to the initial endowment U100 990 To obtain the optimal consumption plan the consumer must borrow bf 716 7 100 616 today pay back 678 tomorrow thereby leaving him with 990 7 678 312 to consume in the nal period 2 Consumption and Investment with a Perfect Capital Market We now consider a world in which the consumer has the opportunity to invest 0 today in production from which he will receive lo tomorrow The function lo encapsulates the return on the investment opportunities in production available to the consumer For example the consumer may wish to obtain an education while he is young expecting a return on this investment in his working years It is generally assumed that f0 0 ii is strictly increasing more investment leads to more return and iii exhibits diminishing returns in that the marginal return on an incremental rise in investment declines as the total investment increases When 1 is differentiable these assumptions imply the rst derivative is positive and the second derivative is negative Such a function is called concave To ensure at least some investment will be made to make our subsequent calculations easier we also assume that the derivative f 0 is in nite The equations governing the consumer s feasible actions today and tomorrow are now 60IO m0b0 Cl m1 f0 7 1 Tbo 17 Rearranging terms as we did before the new budget constraint is f107 1r C1 1r co w0 l 10 W000 18 The new optimization problem facing the consumer is 030 2 0 g 30 W010 0 Z An examination of 18 and 19 reveals a fundamental property All consumers regardless of their utility function should rst determine the optimal investment plan to increase their wealth That is they should select the value of 0 to maximize WOUO This is achieved by equating the marginal return of investment WHO to 1 r When the function lo represents the investment opportunities for a rm in which consumers hold stock then each consumer that holds stock should insist that the rm optimize its investment opportunity regardless of each consumer s different desires for consumption today versus tomorrow Because there ex ists a capital market for each consumer to borrow or lend each consumer can redistribute the increases in wealth as they desire This principle in various forms is known as the Fisher Separation Theorem of Finance Example 4 Suppose lo 33 Now mo 337mm and so the optimal choice for investment is 3 225 The additional wealth created through investment equals 495117 225 225 so that W0225 1225 From 13 and 14 the optimal consumption plan is c3 877 and cf 382 with U1225 4134 The utility has increased by about 107 which also corresponds to 10012251000 7 1 To obtain the optimal consumption plan the consumer must borrow b3 877 225 7 100 1002 today pay back 1103 tomorrow thereby leaving him with 990 495 7 1102 382 to consume in the nal period 3 Consumption and Investment Without a Capital Market We now consider the situation in which the consumer has investment opportunities as described in the previous section but no longer has the opportunity to borrow or loan ie b0 0 We shall see that without access to a capital market our consumer is far worse off The equations governing the consumer s feasible actions today and tomorrow are now 30 0 m0 C1 m1f10 21 The new budget constraint can be represented as 6160 m1 fm0 7 60 22 The optimality conditions 10 imply that E fIo 23 or equivalently that the optimal choice for 0 must satisfy the identity m1 1 f 0 I 24 m0 7 IO Bf o After substituting the speci c choice for and performing simple algebra the optimal choice for 0 must satisfy the identify 9801 990 33W A 7 9801 25 0 Since the left hand side of 25 is an increasing function of 0 that is nite when 0 0 and the right hand side of 25 is a decreasing function of 0 that is in nite when 0 0 a unique solution exists which can be obtained by bisection search Alternatively identity 25 can be transformed into a cubic equation which has a closed form solution The optimal value 3 is about 825 with a corresponding consumption plan of of 9175 and 051 1085 with U 2934 The utility has dropped considerably to almost the level corresponding to the original endowment 4 Homework Problems 1 Jones is endowed with money m0 55 000 today and m1 88 000 tomorrow He desires to consume co 80 000 today and 01 66 000 tomorrow a If there is no opportunity to borrow or lend can Jones achieve his consumption objective Explain A T V Suppose there is a perfect capital market in which Jones may borrow or lend as much as he desires at the market interest rate of 10 Can Jones now achieve his consumption objective Explain A O V Suppose that in addition to a perfect capital market Jones has an opportunity to invest Io 100000 today and receive fIo tomorrow Determine the minimum value for fIO for which Jones will be able to exactly achieve his consumption objective A CL V Explain exactly what Jones must do using the capital market and investment oppor tunity available to him so that he may exactly achieve his consumption objective 2 Jones is endowed with money m0 40 000 today and m1 99 000 tomorrow He desires to consume co 100 000 today and 01 55 000 tomorrow a If there is no opportunity to borrow or lend can Jones achieve his consumption objective Explain b Suppose there is a perfect capital market in which Jones may borrow or lend as much as he desires at the market interest rate of 10 Can Jones now achieve his consumption objective Explain c Suppose that in addition to a perfect capital market Jones has an opportunity to invest IO 50 000 today and receive lo tomorrow Determine the minimum value for lo for which Jones will be able to exactly achieve his consumption objective d Explain exactly what Jones must do using the capital market and investment oppor tunity available to him so that he may exactly achieve his consumption objective OJ Smith s utility function is UCO01 ln co 04 ln 1 Smith is endowed with money m0 90 000 today and m1 500000 tomorrow There is a perfect capital market for borrowing and lending at the market rate of interest of 25 per period a Determine the optimal consumption plan for Smith Explain exactly what Smith must do each period to achieve his optimal consumption plan Recall that ln z 1 Smith has an opportunity to invest IO 80 000 today and receive 10 135 000 tomorrow Should he take advantage of this opportunity If not explain why not If so explain exactly what he should do each period to achieve his new optimal consumption plan A T V A O V lf Smith no longer has access to the capital market he cannot borrow or lend then should he take advantage of the investment opportunity presented in b Explain your reasoning 5 Homework Solutions Problem 1 a No Desired consumption 00 exceeds his endowment mm b No The PV of his wealth 55 8811 135 is less than the PV of his desired consumption 80 6611140 Jones needs to add 5 to the PV of his wealth Consequently ff 7100 5 which means that the smallest value for lo is 1155 0 9 In the rst period Jones consumes 80 invests 100 and borrows 125 In the second period his endowment of 88 plus his investment payout of 1155 generates a supply of 2035 of which 1375 is needed to pay back the loan thereby leaving 66 for consumption as desired