MIT6_262S11_lec12

15 to show that xn and ij n are independent note that

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Unformatted text preview: the rv’s Xi in a process {Xn; n ≥ 1 have a finite number of p ossible sample values. Then any (possibly defective) stopping trial J can b e rep­ resented as a rooted tree where the trial at which each sample path stops is represented by a terminal node. Example: X is binary and stopping o ccurs when the pattern (1, 0) first o ccurs. ￿ ￿ 1 0 ￿ ￿ 1 0 1 0 ￿ ✈ ￿ ￿ ￿ ✈ ￿ ✈ ￿ ￿ ✈ ￿ ✈ ￿ ✈ ￿ ✘ ￿✘ 14 Wald’s equality Theorem (Wald’s equality) Let {Xn; n ≥ 1} b e a se­ quence of I ID rv’s, each of mean X . If J is a stop­ ping trial for {Xn; n ≥ 1} and if E [J ] < ∞, then the sum SJ = X1 + X2 + · · · + XJ at the stopping trial J satisfies E [SJ ] = X E [J ] Prf: SJ = X1IJ ≥1 + X2IJ ≥2 + · · · + XnIJ ≥n + · · · E [SJ ] = E ￿ ￿ n ￿ XnIJ ≥n = ￿ n ￿ E XnIJ ≥n ￿ The essence of the proof is to show that Xn and IJ ≥n are independent. 15 To show that Xn and IJ ≥n are independent, note that IJ ≥n = 1 − IJ<n. Also IJ<n is a function of X1, . . . , Xn−1. Since the Xi are I ID, Xn is independent of X1, . . . , Xn−1, and thus IJ<n, and thus of IJ ≥n. This is surprising, since Xn is certainly not indepen­ dent of IJ =n, nor of IJ =n+1, etc. The resolution of this ‘paradox’ is that, given that J ≥ n (i.e., that stopping has not o ccured b efore trial n), the trial at which stopping o ccurs depends on Xn, but whether or not J ≥ n o ccurs depends only on X1, . . . , Xn−1. Now we can finish the proof. 16 E [SJ ] = = ￿ n ￿ n =X =X ￿ E XnIJ ≥n ￿ ￿ E [Xn] E IJ ≥n ￿ n ￿ n ￿ E IJ ≥n ￿ ￿ Pr {J ≥ n} = X E [J ] In many applications, this gives us one equation in two quantities neither of which is known. Fre­ quently, E [SJ ] is easy to find and this solves for E [J ]. The following example shows, among other things, why E [J ] < ∞ is required for Wald’s equality. 17 Stop when you’re ahead Consider tossing a coin with probability of heads equal to p. $1 is b et on each toss and you win on heads, lose on tails. You stop when your winnings reach $1. If p > 1/2, your winnings (in the absence of stop­ ping) would grow without b ound, passing through 1, so J must b e a rv. SJ = 1 WP1, so E [SJ ] = 1. Thus, Wald says that E [J ] = 1/X = 2p1 1 . Let’s − verify this in another way. Note that J = 1 with probability p. If J > 1, i.e., if S1 = −1, then the only way to reach Sn = 1 is to go from S1 = −1 to Sm = 0 for some m (requiring J steps on average); J more steps on average then gets to 1. Thus J = 1 + (1 − p)2J = 2p1 1 . − 18 Next consider p < 1/2. It is still p ossible to win and stop (for example, J = 1 with probability p and J = 3 with probability p2(1 − p)). It is also p ossible to head South forever. Let θ = Pr {J < ∞}. Note that Pr {J = 1} = p. Given that J > 1, i.e., that S1 = −1, the event {J < ∞} requires that Sm − S1 = 1 for some m, and then Sn − Sm = 1 for some n > m. Each of these are independent events of probability θ, so θ = p + (1 − p)θ2 There are two solutions, θ = p/(1 − p) and θ = 1, which is impossible. Thus J is defective and Wald’s equation is inapplicable. 19 Finally consider p = 1/2. In the limit as p approaches 1/2 from b elow, Pr {J < ∞} = 1. We find other more convincing ways to see this later. However, as p approaches 1/2 from above, we see that E [J ] = ∞. Wald’s equality does not hold here, since E [J ] = ∞, and in fact does not make sense since X = 0. However, you make your $1 with probability 1 in a fair game and can continue to repeat the same feat. It takes an infinite time, however, and requires ac­ cess to an infinite capital. 20 MIT OpenCourseWare http://ocw.mit.edu 6.262 Discrete Stochastic Processes Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms ....
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