Homework 10 (Stats 620, Winter 2017) Due Tuesday April 18, in class Questions are derived from problems in Stochastic Processes by S. Ross.

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Homework (Sas 6, Winer 7 Due Tuesday April 8, in class Quesions are derived from problems in Sochasic Processes by S. Ross.. A sochasic process {X(, } is said o be saionary if X(,..., X( n has he same join disribuion as X( + a,..., X( n + a for all n, a,,..., n. (a Prove ha a necessary and sufficien condiion for a Gaussian process o be saionary is ha Cov((X(s, X( depends only on s, s, and E[X(] c. (b Le {X(, } be Brownian moion and define V ( e α/ X(αe α. Show ha {V (, } is a saionary Gaussian process. I is called he Ornsein-Uhlenbeck process. Soluion: If he Gaussian process is saionary hen for > s ( ( X( d X( s X(s X( E[X(s] E[X(] for all s and Cov(X(, X(s Cov(X( s, X( for all < s. Now, assume E[X(] c and Cov(X(, X(s h( s. For any T (,, k define vecor X T (X(,, X( k. Le T ( a,, k a. If {X(} is a Gaussian process hen boh X T and X T are mulivariae normal and i suffices o show ha hey have he same mean and covariance. This follows direcly from he fac ha hey have he same elemenwise mean c and he equal pair-wise covariances, Cov(X( i a, X( j a h( i j Cov(X( i, X( j. (b Since all finie dimensional disribuions of {V (} are Normal, i is a Gaussian process. from par (a i suffices o show he following: (a E[V (] e α/ E[X(αe α ]. E[V (] is consan. (b For s, Cov(V (s, V ( e α(+s/ Cov(X(αe αs, X(αe α e α(+s/ αe αs αe α( s/, which depends only on s.. Le X( be sandard Brownian moion. Find he disribuion of: (a X(. (b min s X(s (c max s X(s X(

Hin: all hree pars have he same answer. Soluion: (ale Y ( X(. For y F Y (y P( X( y P( y X( y y (b Le Y ( min s X(s. For y F Y (y P(Y ( y P( min P(T y π ( x dx π π (c Le Y max s X(s and X X( hen s y/ ( u X(s y π ( u du F (x, y P(X x, Y y P(X x P(Y > y, X x. ( u Le Φ and φ be he disribuion and densiy funcions respecively of a sandard normal random variable. Using resuls derived in class, ( ( x x y F (x, y Φ Φ, y x, y >., x y F (x, y ( x y φ. Since he Jacobian for he ransformaion V Y X, W Y is of uni modulus, he densiy of (V, W is given by f(v, w ( v w φ, v, w. ( P(Y X y P(V y y π φ ( v y ( v w φ dwdv dv ( ( u

3. Le M( max s X(s where X( is sandard Brownian moion. Show ha P{M( > a M( X(} e a /, a >. Hin: One approach is oulined below. There may be oher ways. (i Differeniae he ression P (M( > y, B( < x y x π e u / du o find he join densiy of M( and B(. (ii Transform variables o find he join densiy of M( and M( B(. This involves using he Jacobian formula (e.g. Ross, A Firs Course in Probabiliy, 6h ediion, Secion 6.7: If X and X have join densiy f X X, Y g (X, X, Y g (X, X, X h (Y, Y and X h (Y, Y, hen (supposing suiable regulariy f Y Y (y, y f X X (h (y, y, h (y, y J(h (y, y, h (y, y where J is he marix deerminan (Jacobian given by J(x, x g / x g / x g / x g / x (iii Find he condiional densiy of M( given M( B(. Soluion: V max X(s X( and W max X(s. s s The join densiy of (V, W is given by equaion (, and he marginal densiy of V follows from equaion (: ( v w f V (v φ dw ( v φ. The condiional densiy of W given V is f(, w/f V (, which gives P(W a V a f(, w f V ( φ( a/ dw. φ( P(W > a V P(W a V e a. 4. For a Brownian moion process wih drif coefficien, le f(x E [ime o hi eiher A or B X x], 3

where A >, B >, B < x < A. (a Derive a differenial equaion for f(x. (b Solve his equaion. (c Use a limiing random walk argumen (see Problem 4. of Chaper 4 o verify he soluion in par (b. Soluion: (a Noe ha he condiional disribuion of process {Y ( X( + h : X(h x} is he same as disribuion of {X( : X( x}. if T (x ime o hi eiher A or B given X( x, hen T (x h + T (X(h + o(h. for Y X(h X(, From he Taylor series ansion i follows ha, f(x E[T (x] h + E[f(x + Y ] + o(h. f(x h + E[f(x + f (xy + f (xy / + ] + o(h, h + f (xh + f (x( h + h/ + o(h. Dividing he equaion above by h on boh side, we obain Tha is, leing h, + f (x + f (x( h + / o(h/h, (b Le v + f (x. Equaion (3 becomes v c e x + f (x. This gives Finally + f (x + f (x/. (3 f(x v + dv dx f (x c e x ( c e x Using he boundary condiions f(a f( B, equaion (4 gives f(x A + B ( e x e A e A e B 4. x + c. (4 + A x. (5

(c From he class noes for T T (, Here where θ n saisfies E[T ] E[T (n ] lim n n AP A B( P A lim n n E[(Y + / n/ n] lim n P A n AP A B( P A /n. e θnb e θna, (6 e θnb E[e θn(y +/ n/ n ]. E[e θny / n ] e θn/n. Since Y is sandard normal. By he momen generaing funcion e θ n/(n e θn/n, i follows ha θ n. Subsiuing in (6, we obain P A E[T ] A + B which is he same as f( obained earlier. eb e A e B. ( e B e A e B B, Recommended reading: Secions 8.3, 8.4, 8.5. Supplemenary exercises: 8.3, 8.4, 8.6, 8.6 Opional, bu recommended. Do no urn in soluions hey are in he back of he book. 5