Stochastic

# 10 infinite state spaces using g 0 0 and summing we

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Unformatted text preview: T = 1 =4 ⇡ (T T ) However, this time if we start with a T at time 1 and a T at time 0, then a T at time 1 will give us a T T and a return to T T at time 1; while if we get a H at time 1, then we have wasted 1 turn and we have nothing that can help us later, so 1 1 4 = ET T TT T = · 1 + · (1 + ETT T ) 2 2 Solving gives ETT T = 6, so it takes longer to observe T T . The reason for this, which can be seen in the last equation, is that once we have one T T , we will get another one with probability 1/2, while occurrences of HT cannot overlap. In the Exercise 1.59 we will consider waiting times for three coin patterns. The most interesting of these is ETHT H = ETT HT . Example 1.49. Duration of fair games. Consider the gambler’s ruin chain in which p(i, i + 1) = p(i, i 1) = 1/2. Let ⌧ = min{n : Xn 62 (0, N )}. We claim that Ex ⌧ = x(N x) (1.26) To see what formula (1.26) says, consider matching pennies. There N = 25 and x = 15, so the game will take 15 · 10 = 150 ﬂips on the average. If there are twice as many coins, N = 50 and x = 30, then the game takes 30 · 20 = 600 ﬂips on the average, or four times as long. There ar...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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