Solution.

\(S_n\) is a martingale

\(E\left |S_n\right |\leq n<\infty \), and \[E\left (S_{n+1}\mid X_1,\dots ,X_n\right ) = S_n + E\left (X_{n+1}\right ) = S_n ,\] since \(X_{n+1}\) is independent of the past with mean zero.

\(S_n^{2}-n\) is a martingale

Expanding \(S_{n+1}^{2} = S_n^{2} + 2S_nX_{n+1} + X_{n+1}^{2}\) and noting \(X_{n+1}^{2}=1\) always, \[E\left (S_{n+1}^{2}\mid X_1,\dots ,X_n\right ) = S_n^{2} + 2S_n\,E\left (X_{n+1}\right ) + 1 = S_n^{2}+1 .\] Subtracting \(n+1\) from both sides, \[E\left (S_{n+1}^{2}-(n+1)\mid X_1,\dots ,X_n\right ) = S_n^{2}-n ,\] which is the martingale property.

The variance

A martingale has constant expectation, so \(E\left (S_n^{2}-n\right ) = E\left (S_0^{2}-0\right ) = 0\), giving \(E\left (S_n^{2}\right ) = n\). Since \(E(S_n)=0\), \[\Var (S_n) = n .\] The direct route — \(n\) independent terms each of variance one — gives the same answer, but the martingale argument is the one that survives when the steps are dependent.

Solution. The fortune is a martingale, so \(E(X_n) = X_0 = 10\) for every \(n\). Let \(p\) be the probability of reaching \(\$20\) first. At the stopping time the fortune is \(\$20\) with probability \(p\) and \(\$0\) with probability \(1-p\), and the optional stopping theorem — applicable because the fortune is bounded between \(0\) and \(20\) — gives \[10 = 20p + 0(1-p) \quad \Longrightarrow \quad p = \frac 12 .\]

The boundedness matters. Without an upper barrier the game is the doubling example in disguise: the fortune remains a martingale with constant expectation at every finite time, yet ruin occurs with probability one, and the expectation of the limit is not the limit of the expectations.