Stochastic processes: martingales

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Let Y_{n} be the gambler's winnings after n games. Clearly, Y_{n} is a martingale. We introduce a new stochastic process Z_{n}, where Z_{n}={Y_{n}}^2-n. It can be shown that Z_{n} is a martingale with respect to Y_{n}. (Can you try to show this?)

Let N be the random variable for the step where the gambler's winnings first reach A or -B. Then, we have E(Z_{N})=E({Y_{N}}^2)-E(N). By applying the Martingale Stopping Theorem (first check the necessary conditions are satisfied), we can show E(Z_{N})=0.

This leaves us with E(N)=E({Y_{N}}^2). To determine E({Y_{N}}^2), use the definition of expectation, and observe that Y_{N} can only take the values A or -B. To calculate the relevant probabilities, apply some formulae related to stopping times of Markov Chains (with stationary transition probabilities). We are now able to compute E({Y_{N}}^2), which will be equal to E(N).
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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