Exercise Exercise Homework #6 Solutions Thursday 6 April 2006

Transcription

1 Unless otherwise stated, for the remainder of the solutions, define F m = σy 0,..., Y m We will show EY m = EY 0 using induction. m = 0 is obviously true. For base case m = : EY = EEY Y 0 = EY 0. Now assume the property is true for k < m. Applying the Mighty Conditioning property and inductive hypothesis, Exercise.. EY m = E EY m F m = EY m = EY 0 The second equality holds by the definition of martingale. Therefore, the property holds for all natural numbers. By definition of submartingale, EY m E EY m+ F m = EY m+ For the case of supermartingale, use its basic definition and flip the inequality in. Finally, use these inequalities in a similar inductive proof to show that EY m EY 0 and EY m EY 0, m N. By Mighty Conditioning Identity and definition of martingale Exercise.. EY n+m F n = E EY n+m F n+m F n = EY n+m F n Now keep repeating this argument until EY n+m F n = EY n+ F n and the result follows by applying the definition of martingale.

2 3 For a symmetric gambler s ruin, S n+ = S n + X n+ where Exercise..4 X i px = { / if x = ± 0 otherwise So EX i = 0 and EXi = i. Therefore, ES n σs 0,..., S n = S n + EX n σs 0,..., S n = S n Now for T n = S n n and T = σt 0,..., T n, ET n σt 0,..., T n = E S n + X n n T = E Sn n + + S n X n + Xn T = E Sn n + S n X n + Xn T = E T n + S n X n + Xn T = T n + S n 0 + EX n T = T n Let p k be the ruin probability given that we start from k and T be T = min{n : S n = 0 or S n = N} Assume S 0 = k. Optional stopping theorem says ES T = ES 0 = k. By this result, ES T = N p k + 0 p k = k p k = k/n Similarly it is also true that ES T T = k. By this result, k = EST T = ES T ET = N p k + 0 p k ET = N p k ET Next, do the Plug-N -Chug with p k i.e. k/n and solve for ET, whose solution turns out to be kn k.

3 4 For r i, EY r F i = Y i by Exercise... For r i, Exercise..5 EY r Y i = EEY r Y i F i = EY i EY r F i by Enhanced Scaling 3 = EY i 4 EY k Y j Y i = 0 for i j k. For the next part, expand Y k Y j and notice that EY k Y j F i = EEY k Y j F j F i = EY j EY k F j F i = EY j F i The three equalities follow from Mighty Conditioning Identity, Enhanced Scaling and Exercise... Expanding out Y k Y j as suggested: E Y k Y j F i = E Yk F i EY k Y j F i + E = E Yk F i E Yj F i Y j F i Take expectation on both sides of the previous claim. What we have is 0 E Y k Y j = E Yk E Y j We know the sequence {E Y n } is bounded by assumption and is nondecreasing. Therefore, it is a covergent sequence. Thus, as k, j, E Yk E Yj 0 and so E Yk Y j 0. {Y n } is Cauchy convergent in mean square. Using the result from Exercise 7.., convergence in mean square follows directly. 3

4 5 Apply Jensen s inequality and use definition of martingale to establish that µ{ey n+ F n } = µ{y n } Finally, argue the three given functions are convex by observing its plot, second derivative if it exists, or noting its epigraph, i.e. the set {x, y y fx}, is convex. 6 a This is a generalization of Problem 6 in Homework 3. EY n F n = n + r + b R n + Y n + R n Y n Exercise..6 Exercise.9.3 = R n n + r + b + Y n n + r + b = Y n On the right hand side of second equality, plug in the value of Y n in terms of R n. The last equality results after a little algebra. Next, notice that Y n. This leads to that Y n I { Yn a} I { a} sup E Y n I { Yn a} EI{ a} 0 n as a. Therefore, Y n is uniformly integrable. Then it follows that Y n converges almost surely and in mean. b Apply the optional stopping theorem. That is, EY T = Y 0 = /. Then, observe R T = T. Therefore, RT / = EY T = E T + T T + = E = E T + T + E = T + 4 c The maximal inequality gives that P max Y i 3/4 EY n 0 i n 3/4 = EY 0 3/4 = /3 The first equality came from Exercise... 4

5 7 Let F = σx 0,... X n. Exercise.9.8 R n+ EX n+ F = E 5 n + F = = 5 n + E R n+ F R n 5 n + X n + R n 5 n + X n = X n X n is the proportion of red cards remaining in the deck of cards. Suppose the strategy is to call Red Now at some arbitrary time T. Apply the optional stopping theorem to show that EX T = EX 0 = /. Thus, for any T 0, the probability of winning is /. 5