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STAT 724/ECO 761 Problem Set Four Solutions Spring 2007 1. Consider the symmetric random walk on the integers: S0 = 0 and if Sn = i then the probability is p = 1/2 that Sn+1 = i + 1 and the probability is q = 1/2 that Sn+1 = i − 1. We showed in class that this symmetric random walk will eventually return to 0 with probability 1. Compute the expected time it takes to return to 0, and show that it is infinity. I.e. the symmetric random walk, starting at 0, will return to 0 with probability 1, but it will take infinitely long to do so, on average! (Hint: Use the theorem we proved in class that says F0 (s) = 1 − (1 − 4pqs2 )1/2 P n where F0 (s) = ∞ n=1 f0 (n)s is the generating function of the sequence defined by f0 (n) = P (S1 6= 0, . . . , Sn−1 6= 0, Sn = 0).) Solution: The quantity f0 (n) is the probability that the walk returns to 0 on the nth step, for P the first time. Thus the expected lenth of time to 0 return to 0 is given by ∞ n=1 nf0 (n). But this is exactly F0 (1) since we can differentiate a power series term by term with its radius of convergence. Since we have a formula for F0 (s) we can easily compute its derivative. The result 1 s is F00 (s) = − (1 − 4pqs2 )−1/2 (−8pqs) = √ since p = 1/2 and q = 1/2. 2 1 − s2 s goes to infinity as s goes to 1. Now, notice √ 1 − s2 2. The probability density function of a certain continuous random variable X is given by ( c(4x − 2x2 ) if 0 < x < 2, f (x) = 0 otherwise. a) what is the value of c? Solution: Z 2 11 8 c(4x − 2x2 ) dx = c[2x2 − 2/3x3 ]20 = c(8 − ) = c = 1 3 3 0 1 3 so c = . 8 b) Compute P (1/2 < X < 3/2). Solution: Compute Z 3/2 3 23 (4x − 2x2 ) dx = . 32 1/2 8 3. a) Calculate E[X], the expected value, for the random variable X of the preceding problem. Solution: Z Z 3 2 3 2 2 2 E(X) = x(4x − 2x ) dx = 4x − 2x3 dx = 8 0 8 0 3 4 3 1 4 2 3 32 x − x 0 = ( − 8) = 1 8 3 2 8 3 b) Calculate V ar(X) , the variance of X. Solution: Z 3 2 3 3 2 2 V ar(X) = 4x − 2x4 dx − 1 = x4 − x5 0 − 1 = 8 0 8 5 3 64 3 16 1 (16 − ) − 1 = ( ) − 1 = 8 5 8 5 5 4. Finish the calculation we started in class to show that if X has a normal distribution N (0, 1), i.e. if the density function of X looks like: 1 2 √ e−x /2 dx 2π then the mean of X is 0 and the variance of X is 1. 2 (HINT: to show E(X) = 0 use the fact that xe−x is an odd function, 2 and to show that V ar(X) = 1, use integration by parts to integrate x2 e−x .) 1 R ∞ 2 −x2 /2 Solution: We need to compute √ xe dx. We use integration 2π −∞ 2 by parts with u = x and dv = xe−x /2 dx. Note that with this choice of dv 2 we get v = −e−x 1 √ 2π 2 /2 . We get √ Z ∞ √ Z ∞ 2 2 2 2 2 −x2 /2 ∞ 2 −x /2 xe dx = √ −e−x /2 dx |0 − xe dx = √ −xe π 0 π −∞ 0 √ Z ∞ √ Z ∞ √ 2 21 1 2 2 =√ e−x /2 dx = √ e−x /2 dx = √ 2π = 1. π 0 π 2 −∞ 2π Z ∞ −x2 /2 5. Suppose that X1 , . . . , X20 are independent random variables that all have density function f (x) = 2x, 0 ≤ x ≤ 1. Let S = X1 + · · · + X20 . Use the central limit theorem to approximate P (S ≤ 10). Solution: Z 1 2 2 2x2 dx = x3 |10 = E(Xi ) = 3 3 0 Z 1 x4 1 E(Xi2 ) = 2x3 dx = |10 = 2 2 0 9 18 1 1 4 − = V ar(X) = − = 2 9 18 18 18 √ 1 so E(S) = 20 · 32 = 40 ; V ar(X) = 20 · 18 = 10 ; s.d.(S) = 310 . Also, 3 9 S = X1 + X2 + · · · + X20 is approximately normal by the Central Limit Theorem. We have p S − p40/3 10 − 40/3 √ √ ≤ = P (z ≤ −3.16) ∼ P (S ≤ 10) ∼ =P = 0. 10 3 10 3 6. Suppose that you have a chance to bet $5 on a fair game, meaning your chances are 1/2 of losing your $5 bet and 1/2 of getting your bet back plus $5 of winnings. Suppose you play this 50 times. Use the central limit theorem to estimate the probability that you will lose more than $75. Solution: Let xi be the r.v. which is the winnings on the ith play of the game. So xi = 5 with probability 1/2 and xi = −5 with probability 1/2. Clearly E(Xi ) = 0 and Var(X p i ) = 25. Let S = X1 + . . . X50 . Then E(S) = 0 and Var(S) = 1250, and Var(S) = 35.3553. We get P (S < −75) = P ( S −75 < ) = P (Z < −2.1213) = .0169. 35.3553 35.3553 3 7. In this problem we will reinforce the idea that the drift rate µ is tied in with the probability of an up or down movement in a binomial √ model. Consider the random walk which goes up or down by an amount σ ∆t with probability p or 1 − p, respectively, in the time interval ∆t, where p is given by √ µ ∆t p = 1/2 1 + . σ Show that as ∆t → 0 the resulting limiting process is a Brownian motion with drift rate µ and variance σ 2 . √ Solution: Start with a random walk √ that goes up by an amount ∆x = σ ∆t with probability p = 1/2(1 + µ ∆t/σ) and down by the same amount with probability 1 − p. Let ( +1 if it goes up on ith step Xi = −1 if it goes down on ith step so Xt = ∆x(X1 + X2 + . . . X[t/∆t] ). Now, √ µ ∆t E(Xi ) = (+1)p + (−1)(1 − p) = 2p − 1 = σ √ µ2 ∆t µ ∆t 2 Var(Xi ) = E(Xi2 ) − E(Xi )2 = 1 − =1− 2 σ σ √ √ √ t µ ∆t t µ ∆t E(Xt ) = ∆x = σ ∆t = µt ∆t σ ∆t σ t µ2 ∆t t µ2 ∆t Var(Xt ) = (∆x)2 1 − 2 = σ 2 ∆t 1 − 2 → σ 2 t. ∆t σ ∆t σ By the Central Limit Theorem, Xt is approximately normal, and as ∆t goes to zero, Xt approaches a normal distribution. By the above calculation we have Xt = µt + σBt where Bt stands for standard Brownian motion. 8. Use Ito’s Lemmma to show that Z T Z T 1 3 2 Zt dZt = ZT − Zt dt 3 0 0 4 Here Zt is a Weiner process. Solution: Let S = Z so that dS = dZ. Let g(S) = g(Z) = Z3 . Then 3 ∂g ∂g ∂2g Z3 = 0, = Z 2 , and ) = Zdt + Z 2 dZ = 2Z. By Ito’s Lemma, d( ∂t ∂Z ∂Z 2 3 RT Z3 R T so = 0 Z dt + 0 Z 2 dZ, whence the conclusion. 3 5