Gaussian processes

Chapter 29 Gaussian processes Definition 29.1 (Gaussian Process) A Gaussian process X t , t  0 , is a stochastic process with the property that for every set of times 0  t 1  t 2  :::  t n , the set of random variables X t 1 ;Xt 2 ;::: ;Xt n  is jointly normally distributed. Remark 29.1 If X is a Gaussian process, then its distribution is determined by its mean function mt=IEXt and its covariance function s; t=IEXs,ms  X t , mt: Indeed, the joint density of X t 1 ;::: ;Xt n  is IP fX t 1  2 dx 1 ;::: ;Xt n  2 dx n g = 1 2  n=2 p det  exp n , 1 2 x , mt   ,1  x , mt T o dx 1 ::: dx n ; where  is the covariance matrix = 2 6 6 6 4 t 1 ;t 1  t 1 ;t 2  ::: t 1 ;t n  t 2 ;t 1  t 2 ;t 2  ::: t 2 ;t n  ::: ::: ::: ::: t n ;t 1  t n ;t 2  ::: t n ;t n  3 7 7 7 5 x is the row vector x 1 ;x 2 ;::: ;x n  , t is the row vector t 1 ;t 2 ;::: ;t n  ,and mt=mt 1 ;mt 2 ;::: ;mt n  . The moment generating function is IE exp  n X k=1 u k X t k   = exp n u  mt T + 1 2 u    u T o ; where u =u 1 ;u 2 ;::: ;u n  . 285 286 29.1 An example: Brownian Motion Brownian motion W is a Gaussian process with mt=0 and s; t=s^t . Indeed, if 0  s  t , then s; t=IEWsWt = IE h W sWt , Ws + W 2 s i = IEW s:IE W t , W s + IEW 2 s = IEW 2 s = s ^ t: To prove that a process is Gaussian, one must show that X t 1 ;::: ;Xt n  has either a density or a moment generating function of the appropriate form. We shall use the m.g.f., and shall cheat a bit by considering only two times, which we usually call s and t . We will want to show that IE exp fu 1 X s+u 2 Xtg = exp  u 1 m 1 + u 2 m 2 + 1 2 u 1 u 2  "  11  12  21  22 " u 1 u 2  : Theorem 1.69 (Integral w.r.t. a Brownian) Let W t be a Brownian motion and  t a nonran- dom function. Then X t= Z t 0 udW u is a Gaussian process with mt=0 and s; t= Z s^t 0  2 udu: Proof: (Sketch.) We have dX = dW: Therefore, de uX s = ue uX s  s dW s+ 1 2 u 2 e uX s  2 s ds; e uX s = e uX 0 + u Z s 0 e uX v  v  dW v  | z  Martingale + 1 2 u 2 Z s 0 e uX v  2 v  dv ; IEe uX s =1+ 1 2 u 2 Z s 0  2 vIEe uX v dv ; d ds IEe uX s = 1 2 u 2  2 sIEe uX s ; IEe uX s = e uX 0 exp  1 2 u 2 Z s 0  2 v  dv  (1.1) = exp  1 2 u 2 Z s 0  2 v  dv  : This shows that X s is normal with mean 0 and variance R s 0  2 v  dv . CHAPTER 29. Gaussian processes 287 Now let 0  st be given. Just as before, de uX t = ue uX t  t dW t+ 1 2 u 2 e uX t  2 t dt: Integrate from s to t to get e uX t = e uX s + u Z t s  v e uX v dW v + 1 2 u 2 Z t s  2 ve uX v dv : Take IE :::jF s conditional expectations and use the martingale property IE  Z t s  v e uX v dW v      F s  = IE  Z t 0  v e uX v dW v      F s  , Z s 0  v e uX v dW v  =0 to get IE  e uX t     F s  = e uX s + 1 2 u 2 Z t s  2 v IE  e uX v     F s  dv d dt IE  e uX t     F s  = 1 2 u 2  2 tIE  e uX t     F s  ; t  s: The solution to this ordinary differential equation with initial time s is IE  e uX t     F s  = e uX s exp  1 2 u 2 Z t s  2 v  dv  ; t  s: (1.2) We now compute the m.g.f. for X s;Xt ,where 0  s  t : IE  e u 1 X s+u 2 X t     F s  = e u 1 X s IE  e u 2 X t     F s   1.2  = e u 1 +u 2 X s exp  1 2 u 2 2 Z t s  2 v  dv  ; IE h e u 1 X s+u 2 X t i = IE  IE  e u 1 X s+u 2 X t     F s  = IE n e u 1 +u 2 X s o : exp  1 2 u 2 2 Z t s  2 v  dv  (1.1) = exp  1 2 u 1 + u 2  2 Z s 0  2 v  dv + 1 2 u 2 2 Z t s  2 v  dv  = exp  1 2 u 2 1 +2u 1 u 2  Z s 0  2 vdv + 1 2 u 2 2 Z t 0  2 v  dv  = exp  1 2 u 1 u 2  " R s 0  2 R s 0  2 R s 0  2 R t 0  2 " u 1 u 2  : This shows that X s;Xt is jointly normal with IEXs=IEXt= 0 , IEX 2 s= Z s 0  2 vdv ; IEX 2 t= Z t 0  2 vdv; IE X sX t = Z s 0  2 v  dv : 288 Remark 29.2 The hard part of the above argument, and the reason we use moment generating functions, is to prove the normality. The computation of means and variances does not require the use of moment generating functions. Indeed, X t= Z t 0 udW u is a martingale and X 0 = 0 ,so mt=IEXt= 0 8t  0: For fixed s  0 , IEX 2 s= Z s 0  2 vdv by the Itˆoisometry.For 0  s  t , IE X sX t , X s=IE  IE  XsX t , X s     F s  = IE 2 6 6 6 4 X s  IE  X t     F s  , X s  | z  0 3 7 7 7 5 =0: Therefore, IE X sX t = IE X sX t , X s + X 2 s = IEX 2 s= Z s 0  2 vdv : If  were a stochastic proess, the Itˆoisometrysays IEX 2 s= Z s 0 IE 2 v dv and the same argument used above shows that for 0  s  t , IE X sX t = IEX 2 s= Z s 0 IE 2 v dv : However, when  is stochastic, X is not necessarily a Gaussian process, so its distribution is not determined from its mean and covariance functions. Remark 29.3 When  is nonrandom, X t= Z t 0 udW u is also Markov. We proved this before, but note again that the Markov property follows immediately from (1.2). The equation (1.2) says that conditioned on F s , the distribution of X t depends only on X s ; in fact, X t is normal with mean X s and variance R t s  2 v  dv . CHAPTER 29. Gaussian processes 289 y t y = z z s z s v = z v y z t s y = z v (a) (b) (c) Figure 29.1: Range of values of y; z; v for the integrals in the proof of Theorem 1.70. Theorem 1.70 Let W t be a Brownian motion, and let  t and ht be nonrandom functions. Define X t= Z t 0 udW u; Y t= Z t 0 huXudu: Then Y is a Gaussian process with mean function m Y t=0 and covariance function  Y s; t= Z s^t 0  2 v  Z s v hydy  Z t v hy dy  dv : (1.3) Proof: (Partial) Computation of  Y s; t :Let 0  s  t be given. It is shown in a homework problem that Y s;Yt is a jointly normal pair of random variables. Here we observe that m Y t=IEY t= Z t 0 huIEXu du =0; and we verify that (1.3) holds. 290 We have  Y s; t=IEYsYt = IE  Z s 0 hy X y  dy : Z t 0 hz X z  dz  = IE Z s 0 Z t 0 hy hz X y X z  dy dz = Z s 0 Z t 0 hy hz IE X y X z  dy dz = Z s 0 Z t 0 hy hz  Z y^z 0  2 v  dv dy dz = Z s 0 Z t z hy hz   Z z 0  2 v  dv  dy dz + Z s 0 Z s y hy hz   Z y 0  2 v  dv  dz dy (See Fig. 29.1(a)) = Z s 0 hz   Z t z hy  dy  Z z 0  2 v dv  dz + Z s 0 hy   Z s y hz  dz  Z y 0  2 v dv  dy = Z s 0 Z z 0 hz  2 v   Z t z hy  dy  dv dz + Z s 0 Z y 0 hy  2 v   Z s y hz  dz  dv dy = Z s 0 Z s v hz  2 v   Z t z hy  dy  dz dv + Z s 0 Z s v hy  2 v   Z s y hz  dz  dy dv (See Fig. 29.1(b)) = Z s 0  2 v   Z s v Z t z hy hz  dy dz  dv + Z s 0  2 v   Z s v Z s y hy hz  dz dy  dv = Z s 0  2 v   Z s v Z t v hy hz  dy dz  dv (See Fig. 29.1(c)) = Z s 0  2 v   Z s v hy  dy  Z t v hz dz  dv = Z s 0  2 v   Z s v hy  dy  Z t v hy dy  dv Remark 29.4 Unlike the process X t= R t 0 udW u , the process Y t= R t 0 Xudu is CHAPTER 29. Gaussian processes 291 neither Markov nor a martingale. For 0  st , IEYtjF s = Z s 0 huX u du + IE  Z t s huX u du     F s  = Y s+ Z t s huIEXu     Fs du = Y s+ Z t s huXs du = Y s+ Xs Z t s hu du; where we have used the fact that X is a martingale. The conditional expectation IE Y tjF s is not equal to Y s , nor is it a function of Y s alone. 292 [...]... Zvs v Z t Z0s 2 v  = hy  dy hz  dz dv Zvs Zvt Z0s 2 v  = hy  dy hy  dy dv 0 v v Remark 29.4 Unlike the process X t = R t u dW u, the process Y t = R t X u du is 0 0 CHAPTER 29 Gaussian processes 291 neither Markov nor a martingale For 0  s IE Y tjF s = Zs 0 t, huX u du + IE = Y s + = Y s + Zt s Zt s Z t s huX u du F s  huIE X u F s du huX s du = Y . Chapter 29 Gaussian processes Definition 29.1 (Gaussian Process) A Gaussian process X t , t  0 , is a stochastic. s is normal with mean 0 and variance R s 0  2 v  dv . CHAPTER 29. Gaussian processes 287 Now let 0  st be given. Just as before, de uX t = ue