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Lecture 21: Stochastic Differential EquationsIn this lecture, we study stochastic differential equations. See Chapter 9of  for a thorough treatment of the materials in this section.1. Stochastic differential equationsWe would like to solve differential equations of the formdX µ(t, X(t))dt σ(t, X(t))dB(t)for given functions a and b, and a Brownian motion B(t). A function (or apath) X is a solution to the differential equation above if it satisfiesˆ Tˆ TX(T ) µ(t, X(t))dt σ(t, X(t))dB(t).00Following is a quote from .Stochastic differential equations provide a link between probability theory and the much older and more developed fieldsof ordinary and partial differential equations. Wonderful consequences flow in both directions. The stochastic modelerbenefits from centuries of development of the physical sciences, and many classic results of mathematical physics (andeven pure mathematics) can be given new intuitive interpretations.We first state a result saying that SDEs can be solved.Theorem 1.1. (Existence and uniqueness) If the coefficients of the stochastic differential equationdX µ(t, X(t))dt σ(t, X(t))dB(t),with X(0) x0and0 t T,satisfy a space-variable Lipshictz condition µ(t, x) µ(t, y) 2 σ(t, x) σ(t, y) 2 K x y 2and the spatial growth condition µ(t, x) 2 σ(t, x) 2 K(1 x 2 ),then there is a continuous adapted solution X(t) such that (L2 bound). Moreover, if X(t) and Y (t) are both continuous solutions satisfying the L2 bound,thenP(X(t) Y (t) for all t [0, T ]) 1.The proof of this theorem is quite technical and can be found in .Thanks to this theorem, we know that most SDEs in fact have a solution.We now discuss some simple (but important) examples of SDEs which haveclosed form solutions.1

Lecture 2121.1. Coefficient matching method. One of the most natural, and mostimportant, stochastic differntial equations is given bydX(t) µX(t) dt σX(t) dB(t)with X(0) x0 0,where µ and σ 0 are constants.Let us pretend that we do not know the solution and suppose that weseek a solution of the form X(t) f (t, B(t)). For this candidate, we have f1 2f fdX (t) dt dB(t), t2 x2 xhence if we must have f1 2f fand σf µf . t2 x2 xThe second equation gives f (t, x) eσx g(t) . Using this in the first equationgivesσ2µf g 0 (t)f f.2σ20Therefore, g (t) µ 2 , and we see thatf (t, x) x0 eσx (µ σTherefore, X(t) x0 e(µ σ 2 /2)t σB(t)2 /2)t.1.2. Coefficient matching for product processes. Let α and σ be positive constants and consider the following SDEdX(t) αX(t)dt σdB(t)with X(0) x0 .Ornstein and Uhlenbeck first used (a version of) this equation to study thebehavior of gasses. It has been applied (or rediscovered) in a variety ofcontexts. This SDE exhibits the ‘mean reversion’ behavior (when α 0).Coefficient matching method failes for this SDE, so we try a different testfunctionˆ t X(t) a(t) x0 b(s)dB(s) ,0where a(0) 1. By differentiating each side we get,dX(t) a0 (t)X(t) dt a(t)b(t)dB(t),a(t)where we assume that a(t) 0 for all t. This should match the given SDE,so we must havea0 (t) α and σ a(t)b(t).a(t)Therefore, a(t) e αt and b(t) σeαt . From this, we see thatˆ tX(t) x0 e αt σeα(s t) dB(s).0

Lecture 2132. Numerical methodsMost PDE and SDE do not have closed form solutions. In this case wecan use numerical methods such as finite difference method, tree method,or Monte Carlo simulation to find an approximate solution. We will brieflydiscuss the some of the methods.2.1. Finite difference methods. Here is an example of using finite difference method in solving an ordinary differential equation.Example 2.1. Suppose we want to solve u0 (x) 5u(x) 2, u(0) 0 tocompute u(1).Step 1. Choose a small value of h. Our plan is to compute the (approximate) value of u at the points x 0, h, 2h, 3h, 4h, 5h, · · · , kh wherekh 1. We hope that the numerical value approaches the real value as wetake smaller values of h. (For example, take h 1/2)Step 2. Use Taylor’s formula to computeu((i 1)h) u(ih) h · u0 (ih)(2.1) u(ih) h · (5u(ih) 2).Note that u((i 1)h) can be computed approximately based on the valueof u(ih). Hence we can continue computing the values of u at our samplepoints.For h 1/2, we see that u( 12 ) u(0) 12 u0 (0) u(0) 21 (5 · u(0) 2) 1,and u(1) u( 21 ) 12 u0 ( 12 ) u( 12 ) 12 (5u( 12 ) 2) 92 .This method can easily be extended to partial differential equations. Forexample, when studying a function u(x, y) of two variables, we may computethe value of u at the intersection points of some fine grid, i.e., we choose somesmall real h, and compute the values u(ih, jh) for integers i 0, 1, 2, · · · andj 0, 1, 2, · · · .This method cannot be directly applied to solve SDEs. This is becausein SDEs, the equation corresponding to (2.1) involves random variables.2.2. Monte Carlo simulation. Monte Carlo simulation is a method thatis used to simulate a probability space by taking independent samples fromthe space according to the probability distribution.Monte Carlo simulation can be used to resolve this issue for SDEs. Suppose that we have a SDE of the formdf (t, Bt ) g(t, Bt )dBt h(t, Bt )dt.If we already know the path Bt , then the equation now becomes a PDE,hence one can use the finite difference method to solve it.Monte Carlo simulation (applied to this setting) involves the followingthree steps.Step 1: Choose a random sample path Bt according to the probabilitydistribution.

Lecture 214Step 2: Use the sample path from Step 1 and finite difference method tosolve the SDE for the particular choice of sample path.Step 3: Repeat Step 1 and 2 many times.This gives a probability distribution of the random stochastic processf (t, Bt ). Monte Carlo simulation is based on the idea that the resultingprobability distribution of this method will converge to the distribution ofthe stochastic process that solves the SDE.2.3. Tree method. Tree method uses the idea that the Brownian motioncan be seen as a limit of a simple random walk. Suppose that we would liketo compute the value of f (t, Bt ) at some time t T , wheredf (t, Bt ) g(t, Bt )dBt h(t, Bt )dt.We begin by taking sample points t0 , t1 t2 , · · · of the time domain [0, T ].We replace the occurences of Bt in the SDE by a simple random walkwhich either goes up one step or down one step during each time interval[ti , ti 1 ] (where the step size is appropriately chosen). This gives an inductiveway to approximately find the probability distribution of f (T, BT ).Hull  illustrates how these methods are used in financial applications.3. Heat equationOur last topic of study is a well-known PDE, heat equation. It is wellknown that the Black-Scholes equation can be turned into a heat equationafter a suitable change of variables.Let u(x, t) be a function of two varaibles, space and time (denoted x andt) .The following differential equation is known as the one dimensional heatequation (diffusion equation): u 2u . t x2This is one of the few partial differential equations that is very well understood (and has a closed form solution).Example 3.1. Let u(x, t) represent the temparature in a long, thin, uniformbar of material whose sides are perfectly insulated so that its temperaturevaries only with distance x along the bar (and with time t). Then u(x, t)satisfies the heat equation (this is where the name of the equation comesfrom).Our goal is to solve various initial value problems for the heat equation.The initial values that we consider will be given asu(0, x) u0 (x)for some function u0 .(for x ),

Lecture 215Observation 1. Heat equation is linear, i.e., if u1 (x, t) and u2 (x, t)satisfies the heat equation, then (u1 u2 )(x, t) also satisfies the heat equation. More generally if we have a collection of solutions us (x, t) indexed by s R,then us (x, t) · c(s)ds is also a solution (as long as the integral exists and is differentiable up to appropriate order). This means that we cansuperimpose solutions of ‘easy’ initial value problems to obtain a solutionto a more general initial value problem.Observation 2. The ‘easy’ initial value problem we are going to use iswhen the initial value is given as a Dirac delta function. Let δ(x) be theDirac delta function and suppose that u0 δ so that we are solvingu(0, x) δ(x).The solution for this initial value problem is known to be12uδ (x, t) e x /(4t)2 πt(for x ,t 0).Note that the solution ’converges to’ the Dirac delta function as t tends tozero. Also note that for fixed value of t 0, this is a probability distributionfunction of the normal random variable.Exercise 3.2. Derive the solution above by using ξ t1/2 u(x, t),x ,tand U (ξ) and restating the heat equation as an ODE.Now suppose that a function u0 is given. We can understand u0 as afunction obtained by superimposing Dirac delta functions, i.e.,ˆ u0 (x) δ(x s)u0 (s)ds. Consider the following function obtained by superimposing the solutionsaccordingly:ˆ u(x, t) uδ (x s, t) · u0 (s)ds (such function need not be well-defined since the integration might not exist).However, if the function u0 is ‘reasonable’ then we can show that(3.1)ˆ ˆ 2 u uδ 2u uδ(x, t) (x s, t)·u0 (s)ds and(x, t) (x s, t)·u0 (s)ds.22 t x t xThis implies that u(x, t) satisfies the heat equation as well. Note thatu0 (x, 0) u0 (x), and hence as long as a ‘reasonable’ initial condition u0is given (so that u(x, t) is well-defined and (3.1) holds), we see that u(x, t)solves the initial value problem.

Lecture 216ReferencesP. Wilmott, S. Howison, J. Dewynne, The mathematics of financial derivativesS. Shreve, Stochastic calculus for finance II: continuous-time modelsM. Steele, Stochastic calculus and financial applicationsJ. Hull, Options, futures, and other derivatives