Clases 11-12: Integración estocástica.
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1 Clases 11-12: Integración estocástica. Fórmula de Itô * 3 de octubre de 217 Índice 1. Introduction to Stochastic integrals 1 2. Stochastic integration 2 3. Simulation of stochastic integrals: Euler scheme 6 4. Itô Formula 1 1. Introduction to Stochastic integrals With the purpose of constructing a large class of stochastic processes, we consider stochastic differential equations (SDE) driven by a Brownian motion. Given then two functions: b: R R, σ : R R, a driving Brownian motion {W (t): t T } and an independent random variable X (the initial condition) our purpose is to construct a process X {X(t): t T } such that, for t [, T ], the following equation is satisfied: Comments X(t) X + b(x(s))ds + The differential notation for an SDE is σ(x(s))dw (s). dx(t) b(x(t))dt + σ(x(t))dw (t), X() X * Notas preparadas por E. Mordecki para el curso de Simulación en procesos estocásticos 1
2 The integral b(x(s))ds is a usual Riemann integral, as we expect X(s) to be a continuous function As for fixed ω the trajectories of W are not smooth, we must define precisely the integral σ(x(s))dw (s) The three main properties to define the stochastic integral are: (A) If X is independent from W b a XdW (s) X b a dw (s), (B) The integral of the function 1 [a,b) is the increment of the process: 1 [a,b) dw (s) b a dw (s) W (b) W (a). (C) Linearity: (f + g)dw (s) 2. Stochastic integration Consider the class of processes that satisfy (A) and (B): H {h (h(s)) t T } fdw (s) + gdw (s). (A) h(t), W (t + h) W (t) are independent, t t + h T, (B) E(h(t)2 )dt <. Example: If E[f(W (t)) 2 ] K, then h(t) f(w (t)) H: In fact, f(w (t)), W (t + h) W (t) are independent, and E(f(W (t)) 2 )dt < KT. 2
3 Step processes The stochastic integral is first defined for a subclass of step processes in H: A step processes h is of the form h(t) h k 1 [tk,t k+1 )(t), k where t < t 1 < < t n T is a partition of [, T ], and (A) h k, W (t k + h) W (t k ) are independent for 1 k n, h >. (B) E(h 2 k ) <. For a step process h H we define the stochastic integral applying properties (A), (B) and (C): h(t)dw (t) (C) (A) (B) h k 1 [tk,t k+1 )(t)dw (t) k k h k 1 [tk,t k+1 )(t)dw (t) h k 1 [tk,t k+1 )(t)dw (t) k h k [W (t k+1 ) W (t k )]. k Notation: I(h) nt. h(t)dw (t). Properties The stochastic integral defined for step processes is a random variable that has the following properties (P1) E h(t)dw (t) (P2) Itô isometry: ( 2 T E h(t)dw (t)) E(h(t) 2 )dt. 3
4 Proof of (P1): We compute the expectation: ( ) T E h(t)dw (t) E ( ) h k [W (t k+1 ) W (t k )] k E (h k [W (t k+1 ) W (t k )]) (A) k E(h k )E[W (t k+1 ) W (t k )], k because W (t k+1 ) W (t k ) N(, t k+1 t k ). Proof of (P2): We first compute the square: ( 2 T h(t)dw (t)) ( k + 2 h k [W (t k+1 ) W (t k )] ) 2 h 2 k[w (t k+1 ) W (t k )] 2 k j<k h j h k [W (t j+1 ) W (t j )][W (t k+1 ) W (t k )] We have, by independence, as t j < t j+1 t k : E (h j h k [W (t j+1 ) W (t j )][W (t k+1 ) W (t k )]) E (h j h k [W (t j+1 ) W (t j )]) E[W (t k+1 ) W (t k )]. Furthermore, also by independence: E ( h 2 k[w (t k+1 ) W (t k )] 2) E(h 2 k)e[w (t k+1 ) W (t k )] 2 Summarizing, E(h 2 k)(t k+1 t k ) k+1 t k E(h(s) 2 )ds. ( 2 T E h(t)dw (t)) E (h k [W (t k+1 ) W (t k )]) 2 k k+1 k t k E(h(s) 2 )ds E(h(s) 2 )ds, 4
5 concluding the proof of (P2). This property is an isometry, because it can be stated as: I(h) 2 L 2 (Ω) h 2 L 2 (Ω [,T ]), where I(h) 2 L 2 (Ω) E(I(h)2 ), h 2 L 2 (Ω [,T ]) E(h(s) 2 )ds, This makes possible to extend the integral to the whole set H by approximation: Given h H we find a sequence of steps processes (h n ) such that h n h 2 L 2 (Ω [,T ]), (l ). We define I(h) lím I(h n ) as n. It is necessary to prove that I(h n ) is a Cauchy sequence in L 2 (Ω) to prove that the limit exists. Comments I(h) h(t)dw (t), is a random variable. For t T we define I(h, t) to obtain a stochastic process. 1 [,t) h(t)dw (t) nt. Covariance of two stochastic integrals h(s)dw (s), Proposition 1. Consider two process h and g in the class H. Then ( ) T E g(t)dw (t) h(t)dw (t) E(g(t)h(t))dt. The proof is based on the polarization identity: Demostración. Observe that I(h + g) I(h) + I(g), E(I(f) 2 ) E(f(t)2 )dt. ab 1 4 [(a + b)2 (a b) 2 ]. 5
6 Then 4E(I(g)I(h)) E(I(h + g) 2 I(h g) 2 ) 4 E((h + g) 2 (h g) 2 )dt E(hg)dt. 3. Simulation of stochastic integrals: Euler scheme The definition of a stochastic integral suggest a way to simulate an approximation of the solution. Given a process h, we choose n and we define a mesh t n k kt/n, k,..., n. h(t n i )[W (t n i+1) W (t n i )]. k It is very important to estimate the integrand in t n i and to take the increment of W in [t n i, tn i+1 ] to have the independence. Otherwise a wrong result is obtained. Example: Wiener integrals Suppose that h(t) is a deterministic continuous function. Then, the approximating processes are also deterministic: So, we must find the limit of I n h n h(t n i )1 [t n i,t n i+1 ). k h n dw (t) h(t n i )[W (t n i+1) W (t n i )]. k As I n is a sum of independent normal variables, it is normal. Furthermore (applying the properties or computing directly), we obtain that We conclude that E(I n ), var(i n ) h(t n i ) 2 (t n i+1 t n i ) k h(t)dw (t) N (, h(t) 2 dt ). h(t) 2 dt. 6
7 Simulation of a Wiener integral Consider f(t) t over [, 1]. According to our results: 1 tdw (t) N (, 1 ( ) t) 2 dt ( N, 1 ) tdt N (, 1/2). We check this result by simulation, writing the following code: # Wiener integral of sqrt(t) n<-1e3 # discretization to simulate W m<-1e3 # nr of variates steps<-seq(,1-1/n,1/n) integral<-rep(,m) for(i in 1:m){ integral[i]<-sum(sqrt(steps)*rnorm(n,,sqrt(1/n))) } >mean(integral) # Theoretical mean is [1] >var(integral) # Theoretical variance is 1/2 [1] We now plot our results and compare with the theoretical density N(, 1/2): >hist(integral,freqfalse) >curve(dnorm(x,mean,sdsqrt(1/2)),addtrue) Histogram of integral Density integral Comparison of distributions To perform a statistical comparison of our results, we compare the distributions: plot(ecdf(integral),type"l") curve(pnorm(x,,sqrt(1/2)), addtrue, col red") 7
8 seq(, 1, length m) sort(integral) To assess whether the empirical distribution (black) is not far from the theoretical (red) a Kolmogorov-Smirnov test should be performed. >ks.test(integral,"pnorm",,sqrt(1/2)) The result is One-sample Kolmogorov-Smirnov test data: integral D.2565, p-value.5562 alternative hypothesis: two-sided Here we reject when our statistic D is such that md > k 1,36, and this is true if and only if p value <,5. Simulation of a stochastic integral Suppose h(t) f(w (t)) W (t) 2. The R code is # Stochastic integral of f(w(t)) n<-1e3 # for the grid of BM t<-1 f<-function(x) x^2 m<-1e4 # generate m samples of the integral integral<-rep(,m) for(i in 1:m){ increments<-rnorm(n,,sqrt(t/n)) bm<-c(,cumsum(increments[1:(n-1)])) integral[i]<-sum(f(bm)*increments) } 8
9 Remember f(w (t n i ))[W (t n i+1) W (t n i )]. k We obtain the results: >mean(integral) [1] >var(integral) [1] We compute the theoretical variance with the isometry property. First, if Z N(, 1): E(W (t) 4 ) E( tz) 4 t 2 E(Z 4 ) 3t 2. Then Theoretical example ( 1 2 E W (t) 2 dw (t)) 1 1 E(W (t) 4 )dt 3t 2 1. Consider h(t) W (t). It is not difficult to verify that h H. We want to integrate W (t)dw (t). To define the approximating process, denote t n i processes are defined by So, we must find the limit of We have h n W (t n i )1 [t n i,t n i+1 ). k h n dw (t) W (t n i )[W (t n i+1) W (t n i )]. k T i/n. The approximating W (t n i )[W (t n i+1) W (t n i )] W (t n i+1)w (t n i ) W (t n i ) [W (tn i+1) W (t n i )] [W (tn i+1) 2 W (t n i ) 2 ] 9
10 When we sum, the second term is telescopic: W (t n i )[W (t n i+1) W (t n i )] W (t n i+1)w (t n i ) W (t n i ) 2. k We have seen that We then conclude that 1 [W (t n 2 i+1) W (t n i )] W (T )2. k [W (t n i+1) W (t n i )] 2 T, k W (t)dw (t) 1 2 Observe, that if x(t) is a differentiable function, ( W 2 (T ) T ). x(t)dx(t) 1 2 x2 (T ). The extra term T is a characteristic of the stochastic integral, discovered by Kiyoshi Itô in the decade of Itô Formula Theorem 1. Let {W (t): t } be a Brownian motion, and consider a smooth 1 function f f(t, x): [, ) R R. Then f(t, W (t)) f(, W ()) f (s, W (s))dw (s) x + ( f t (s, W (s)) ) f (s, W (s) ds. x2 If f(t, x) f(x) (i.e. the function does not depend on time) the formula takes the simpler form f(w (t)) f(w ()) f (W (s))dw (s) By smooth we mean f(t) differentiable in t and twice differentiable in x. f (W (s))ds. 1
11 On the proof of Itô formula. We explain more in detail the proof for f(t, x) f(x) (i.e. the function does not depend on time). The key moment is the Taylor expansion: f(w (t + h)) f(w (h)) f (W (t))[w (t + h) W (t)] f (W (t))[w (t + h) W (t)] 2 + o([w (t + h) W (t)] 2 ). Then, due to the property of the quadratic variation, when h is small giving the second integral in the formula. Example If f(t, x) x 2, then [W (t + h) W (t)] 2 h. (1) and Itô formula gives f t, f x 2x, 2 f x 2 2, W (t) 2 2 W (s)w (s) + ds 2 If t 1: 1 W (s)w (s) 1 2 (W (1)2 1), our previous result. Example: Geometric Brownian motion If f(t, x) S e σx+µt, then S(t) f(t, W (t)) S e σw (t)+µt, W (s)w (s) + t, is the Geometric Brownian motion. To apply Itô Formula, we compute: and f t µf, f(t, W (t)) f(, W () f x σf, 2 f x 2 σ2 f. (µ + 12 σ2 ) f(s, W (s))ds + σf(s, W (s))dw (s). 11
12 As f(t, W (t)) S(t), we obtain S(t) S() (µ + 12 σ2 ) S(s)ds + This same expression, in differential form, is σs(s)dw (s). ds(t) rs(t)dt + σs(t)dw (t), S() S. (2) where we used that µ r σ 2 /2. So, as the the two assets in Black-Scholes model satisfy the equations { db(t) B(t)(rdt), B() B, ds(t) S(t)(rdt + σdw (t)), S() S. The second is a modification of the first + noise. 12
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