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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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