Stochastic Differential Equations Do not worry about your problems with mathematics, I assure you mine are far greater. Albert Einstein. Florian Herzog 2013 Stochastic Differential Equations (SDE) A ordinary differential equation (ODE) dx(t) = f(t; x) ; dx(t) = f(t; x)dt ; (1) dt with initial conditions x(0) = x0 can be written in integral form Z t x(t) = x0 + f(s; x(s))ds ; (2) 0 where x(t) = x(t; x0; t0) is the solution with initial conditions x(t0) = x0. An example is given as dx(t) = a(t)x(t) ; x(0) = x0 : (3) dt Stochastic Systems, 2013 2 Stochastic Differential Equations (SDE) When we take the ODE (3) and assume that a(t) is not a deterministic parameter but rather a stochastic parameter, we get a stochastic differential equation (SDE). The stochastic parameter a(t) is given as a(t) = f(t) + h(t)ξ(t) ; (4) where ξ(t) denotes a white noise process. Thus, we obtain dX(t) = f(t)X(t) + h(t)X(t)ξ(t) : (5) dt When we write (5) in the differential form and use dW (t) = ξ(t)dt, where dW (t) denotes differential form of the Brownian motion,we obtain: dX(t) = f(t)X(t)dt + h(t)X(t)dW (t) : (6) Stochastic Systems, 2013 3 Stochastic Differential Equations (SDE) In general an SDE is given as dX(t; !) = f(t; X(t; !))dt + g(t; X(t; !))dW (t; !) ; (7) where ! denotes that X = X(t; !) is a random variable and possesses the initial condition X(0;!) = X0 with probability one. As an example we have already encountered dY (t; !) = µ(t)dt + σ(t)dW (t; !) : Furthermore, f(t; X(t; !)) 2 R, g(t; X(t; !)) 2 R, and W (t; !) 2 R. Similar as in (2) we may write (7) as integral equation Z Z t t X(t; !) = X0 + f(s; X(s; !))ds + g(s; X(s; !))dW (s; !) : (8) 0 0 Stochastic Systems, 2013 4 Stochastic Integrals R T For the calculation of the stochastic integral 0 g(t; !)dW (t; !), we assume that g(t; !) is only changed at discrete time points ti (i = 1; 2; 3; :::; N − 1), where 0 = t0 < t1 < t2 < : : : < tN−1 < tN < T : We define the integral Z T S = g(t; !)dW (t; !) ; (9) 0 as the Riemannßum XN ( ) SN (!) = g(ti−1;!) W (ti;!) − W (ti−1;!) : (10) i=1 with N ! 1 : Stochastic Systems, 2013 5 Stochastic Integrals A random variable S is called the It^ointegral of a stochastic process g(t; !) with respect to the Brownian motion W (t; !) on the interval [0;T ] if [( XN ( )] lim E S − g(ti−1;!) W (ti;!) − (W (ti−1;!) = 0 ; (11) N!1 i=1 for each sequence of partitions (t0; t1; : : : ; tN ) of the interval [0;T ] such that maxi(ti − ti−1) ! 0. The limit in the above definition converges to the stochastic integral in the mean-square sense. Thus, the stochastic integral is a random variable, the samples of which depend on the individual realizations of the paths W (:; !). Stochastic Systems, 2013 6 Stochastic Integrals The simplest possible example is g(t) = c for all t. This is still a stochastic process, but a simple one. Taking the definition, we actually get Z T XN ( ) c dW (t; !) = c lim W (ti;!) − W (ti−1;!) N!1 0 i=1 = c lim [(W (t1;!)−W (t0;!)) + (W (t2;!)−W (t1;!)) + ::: N!1 +(W (tN ;!)−W (tN−1;!)) = c (W (T;!) − W (0;!)) ; where W (T;!) and W (0;!) are standard Gaussian random variables. With W (0;!) = 0, the last result becomes Z T c dW (t; !) = c W (T;!) : 0 Stochastic Systems, 2013 7 Stochastic Integrals R T Example: g(t; !) = W (t; !) 0 W (t; !) dW (t; !) = XN ( ) = lim W (ti−1;!) W (ti;!) − W (ti−1;!) N!1 i=1 h XN XN i 1 2 2 1 2 = lim (W (ti;!) − W (ti−1;!)) − (W (ti;!) − W (ti−1;!)) N!1 2 2 i=1 i=1 XN 1 2 1 2 = − lim (W (ti;!) − W (ti−1;!)) + W (T;!) ; (12) 2 N!1 2 i=1 where we have used the following algebraic relationship y(x − y) = yx − y2 + 1 2 − 1 2 1 2 − 1 2 − 1 − 2 2x 2x = 2x 2y 2(x y) . Stochastic Systems, 2013 8 Stochastic Integrals P N − 2 We take now a detailed look at :limN!1 i=1(W (ti;!) W (ti−1;!)) . XN XN 2 2 E[ lim (W (ti;!) − W (ti−1;!)) ] = lim E[(W (ti;!) − W (ti−1;!)) ] N!1 N!1 i=1 i=1 XN = lim (ti − ti−1) N!1 i=1 = T XN XN 2 2 Var[ lim (W (ti;!) − W (ti−1;!)) ] = lim Var[(W (ti;!) − W (ti−1;!)) ] N!1 N!1 i=1 i=1 XN 2 = 2 lim (ti − ti−1) : N!1 i=1 Stochastic Systems, 2013 9 Stochastic Integrals By reducing the partition, the variance becomes zero, XN XN 2 lim (ti − ti−1) ≤ max(ti − ti−1) lim (ti − ti−1) N!1 i N!1 i=1 i=1 = max(ti − ti−1) T i = 0 ; (13) P − ! N − 2 since ti−1 ti 0. Since the expected value of i=1(ti ti−1) is T and the variance becomes zero, we get XN 2 (W (ti;!) − W (ti−1;!)) = T (14) i=1 Stochastic Systems, 2013 10 Stochastic Integrals The stochastic integral has the solution Z T 1 1 W (t; !) dW (t; !) = W 2(T;!) − T (15) 0 2 2 This is inR contrast to our intuition from standard calculus. In the case of a deterministic T 1 2 −1 integral 0 x(t)dx(t) = 2x (t), whereas the It^ointegral differs by the term 2T . | This example shows that the rules of differentiation (in particular the chain rule) and integration need to be re-formulated in the stochastic calculus. Stochastic Systems, 2013 11 Stochastic Integrals Properties of It^oIntegrals. Z T E[ g(t; !) dW (t; !)] = 0 : 0 Proof: Z T XN ( ) E[ g(t; !)dW (t; !)] = E[ lim g(ti−1;!) W (ti;!) − W (ti−1;!) ] N!1 0 i=1 XN ( ) = lim E[g(ti−1;!)] E[ W (ti;!) − W (ti−1;!) ] N!1 i=1 = 0 : The expectation of stochastic integrals is zero. This is what we would expect anyway. Stochastic Systems, 2013 12 Stochastic Integrals Properties of It^oIntegrals. Z Z h T i T Var g(t; !)dW (t; !) = E[g2(t; !)]dt : (16) 0 0 Proof: Z Z h T i h T i Var g(t; !)dW (t; !) = E ( g(t; !)dW (t; !))2 0 0 [( XN ( ))2i = E lim g(ti−1;!) W (ti;!) − W (ti−1;!) N!1 i=1 Stochastic Systems, 2013 13 Stochastic Integrals XN XN = lim E[g(ti−1;!)g(tj−1;!) N!1 i=1 j=1 (W (ti;!) − W (ti−1;!))(W (tj;!) − W (tj−1;!))] XN ( )2 2 = lim E[g (ti−1;!)] E[ W (ti;!) − W (ti−1;!) ] N!1 i=1 XN 2 = lim E[g (ti−1;!)] (ti − ti−1) N!1 i=1 Z T = E[g2(t; !)]dt : (17) 0 Stochastic Systems, 2013 14 Stochastic Integrals The calculation of the variance of the It^oIntegrals shows two important properties: [( R )2i R h i • T T 2 E 0 g(t; !)dW (t; !) = 0 E g (t; !) dt R • T 2 1 0 E[g (t; !)]dt < The second property is the condition of existence for It^ointegrals. The next property is the linearity of It^ointegrals: Z T [a1 g1(t; !) + a2 g2(t; !)]dW (t; !) 0 Z Z T T = a1 g1(t; !)dW (t; !) + a2 g2(t; !)dW (t; !) ; (18) 0 0 for numbers a1; a2 and stochastic functions g1(t; !); g2(t; !). Stochastic Systems, 2013 15 It^o'slemma As mentioned shown in the second example, the rules of classical calculus are not valid for stochastic integrals and differential equations. It is the equivalent to the chain rule in classical calculus. The problem can be stated as follows: Given a stochastic differential equation dX(t) = f(t; X(t))dt + g(t; X(t))dW (t) ; (19) and another process Y (t) which is a function of X(t), Y (t) = ϕ(t; X(t)) ; where the function ϕ(t; X(t)) is continuously differentiable in t and twice continuously differentiable in X, find the stochastic differential equation for the process Y (t): dY (t) = f~(t; X(t))dt +g ~(t; X(t))dW (t) : Stochastic Systems, 2013 16 It^o'slemma In the case when we assume that g(t; X(t)) = 0, we know the result: the chain rule for standard calculus. The result is given by dy(t) = (ϕt(t; x) + ϕx(t; x)f(t; x))dt : (20) In the case of stochastic problems, we reason as follows: The Taylor expansion of ϕ(t; X(t)) yields 1 2 dY (t) = ϕt(t; X)dt + ϕtt(t; X)dt + ϕx(t; X)dX(t) 2 1 2 + ϕxx(t; X)(dX(t)) + h.o.t : (21) 2 Stochastic Systems, 2013 17 It^o'slemma We use (19) for dX(t) and get dY (t) = ϕt(t; X)dt + ϕx(t; X)[f(t; X(t))dt + g(t; X(t))dW (t)] ( 2 1 2 2 2 2 +ϕtt(t; X)dt + ϕxx(t; X) f (t; X(t))dt + g (t; X(t))dW (t) 2 ) +2f(t; X(t))g(t; X(t))dt dW (t) + h.o.t : (22) The differentials of higher order (dt; dW ) become fast zero, dt2 ! 0 and dtdW (t) ! 0.