Computing expected minimum of stopping time and n with simple random walk

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Let $(S_n)$ be an elementary random walk, ie. $S_n = \sum_{i=1}^n X_i$ where $P(X_i = 1) = P(X_i = -1) = \frac12$.

Let $T = \inf \{ n : S_n \in \{-2,2\}\}$. $T$ is clearly a stopping time and is almost surely finite. We use the notation $T \land n$ to mean $\min\{T,n\}$.

I need to prove that $E(T \land n) = 4P(T \leq n) + K$, where $K$ is an error term that is bounded by $P(T > n)$.

I've tried the following so far:

$$ E[T \land n] = E[T \mathbb1_{\{T \leq n\}} + n \mathbb1_{\{T > n\}}] = E[T \mathbb1_{\{T \leq n\}}] + nP(T > n) $$

But I can't work out how to proceed. The second term on the RHS is $nP(T>n)$ which is not bounded by $P(T > n)$, and I can't see where the factor of $4$ could come from.

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The key was to consider the Martingale $M_n = S_n^2 - n$. Applying the optional stopping theorem to the stopped Martingale $M_{n \land T}$ gives that

$$0 = EM_0 = EM_{T \land n} = ES_{T \land n} - E[T \land n]$$

and solving $ES_{T \land n}$ by considering cases.