gambler's ruin problem

gambler's ruin problem - 1 Gamblers Ruin Problem Consider a...

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1 Gambler’s Ruin Problem Consider a gambler who starts with an initial fortune of $1 and then on each successive gamble either wins $1 or loses $1 independent of the past with probabilities p and q = 1 - p respectively. Let R n denote the total fortune after the n th gamble. The gambler’s objective is to reach a total fortune of $ N , without first getting ruined (running out of money). If the gambler succeeds, then the gambler is said to win the game. In any case, the gambler stops playing after winning or getting ruined, whichever happens first. There is nothing special about starting with $1, more generally the gambler starts with $ i where 0 < i < N . While the game proceeds, { R n : n 0 } forms a simple random walk R n = Δ 1 + ··· + Δ n , R 0 = i, where { Δ n } forms an i.i.d. sequence of r.v.s. distributed as P (Δ = 1) = p, P (Δ = - 1) = q = 1 - p , and represents the earnings on the succesive gambles. Since the game stops when either R n = 0 or R n = N , let τ i = min { n 0 : R n ∈ { 0 ,N }| R 0 = i } , denote the time at which the game stops when R 0 = i . If R τ i = N , then the gambler wins, if R τ i = 0, then the gambler is ruined. Let P i = P ( R τ i = N ) denote the probability that the gambler wins when R 0 = i . Clearly P 0 = 0 and P N = 1 by definition, and we next proceed to compute P i , 1 i N - 1. The key idea is to condition on the outcome of the first gamble, Δ 1 = 1 or Δ 1 = - 1, yielding P i = pP i +1 + qP i - 1 . (1) The derivation of this recursion is as follows: If Δ 1 = 1, then the gambler’s total fortune increases to R 1 = i +1 and so by the Markov property the gambler will now win with probability P i +1 . Similarly, if Δ 1 = - 1, then the gambler’s fortune decreases to R 1 = i - 1 and so by the Markov property the gambler will now win with probability P i - 1 . The probabilities
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gambler's ruin problem - 1 Gamblers Ruin Problem Consider a...

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