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Assume that a gambler plays a fair game where he can win or lose 1 dollar in each round. His initial stock is 200 dollars. He decides a priory to stop gambling at the moment when he either has 500 dollars or 0 dollar in his stock. Time is counted by the number of rounds played.

a) Show that the probability that he will never stop gambling is zero,

b) Compute the probability that at the time when he stops gambling he has 500 dollars, and the probability that he has zero dollars, and

c) Compute the expected time till the gambler stops playing. {Hint: make use of the martingale Y = (X^2)-n , where X is the players stock at time n and use the optional stop time theorem for martingales}

Please help with it. I tried to read some theory but I didn't find any solution.

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a) We know that for $k>300$ that the transition probability $p_{i,0}(k) + p_{i,500}(k)>0, \forall i\in \{1,...,499\}$ and so the game ends. So lets take some state $i\in \{0,...,500\}$ where this probability is the lowest and define this probability as $\hat{p}_k$. Then we find \begin{equation} 0 < \mathbb{P}(\text{Game does not end in $n\cdot k$ steps}|X_0 = 200) <(1-\hat{p}_k)^n\xrightarrow{n\to\infty}0. \end{equation} b) Let $p_k$ be the probability that we end up with 0 dollars if we start with $k$ dollars: \begin{equation} p_k = \frac{1}{2}\cdot p_{k+1} + \frac{1}{2}\cdot p_{k-1}, \text{ if }1\le k\le 499, \end{equation} with boundary conditions $p_0 = 1$ and $p_{500} = 0$. For such difference equations we look for a solution f the form $p_k = \theta^k$. So if we substitute this in our equation for $p_k$ and divide by $\theta^{k-1}$ we find the equation \begin{equation} \frac{1}{2}\theta^2 - \theta + \frac{1}{2} = 0, \end{equation} which has roots $\theta_1=\theta_2 = 1$. As both our solutions are the same our general solution is of the form $p_k = A_1 + A_2k$. So if we use our boundary conditions $p_0 = 1$ and $p_{500} = 0$ and take $k=200$ we get the solution $p_{200} = \frac{3}{5}$. So the probability of ending up with 500 dollars will be $\frac{2}{5}$.

c) First we have to show $Y_n$ is martingale. \begin{equation} \mathbb{E}[Y_{n+1}|X_1,...,X_n] = \frac{1}{2}(X_n-1)^2 + \frac{1}{2}(X_n +1)^2 - (n+1) = Y_n, \end{equation} and $\mathbb{E}(Y_n)<\infty$ as $X<500$, so $Y_n$ is martingale.

Now show that the conditions for the Optional stopping Theorem hold and use $\mathbb{E}(Y_T) = \mathbb{E}(Y_0)$ to find $\mathbb{E}(T)$

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    $\begingroup$ +1 though there is an easier approach to (b): it is a fair game, so the expected position at every step is constant. So you have $200q_{200}=500q_{500}+0q_0$, where $q_k$ is the probability of winning $500$starting with $k$, making $q_{200}=\frac{200}{500}=\frac25$ $\endgroup$ – Henry Jun 17 at 17:56

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