Determining the expected stopping time for certain values?

Question

Suppose a symmetric random walk. I would like to calculate the expected stopping time the walk reaches the value of $-3$ or $5$, that is, $E[S]$ for $$S=\min\{{t:S_{t}=-3 \, \text{or} \, S_{t}=5}\}.$$ I know that $E[S_{t}]=\sum_k E(X_{k}1\{{t≥k}\})$ and that is equal to $0$ if the stopping time is bounded. Where $1$ is the indicator function and $X_{k}$ is the position in the walk at time $k<t$. Any hints?

Answer 1

Hints: Denote by $(X_n)_{n \in \mathbb{N}}$ the symmetric random walk.

1. Show that $M_n := X_n^2-n$ is a martingale.
2. By the optional stopping theorem, $$\mathbb{E}(M_{n \wedge S}) = \mathbb{E}(M_0)=0.$$ Using the dominated convergence theorem (hint: $|M_{S \wedge n}| \leq 5$) show that $$\mathbb{E}(M_S) = \lim_{n \to \infty} \mathbb{E}(M_{n \wedge S})=0,$$ i.e. $$\mathbb{E}(X_S^2) = \mathbb{E}(S). \tag{1}$$
3. Since $(X_n)_{n \in \mathbb{N}}$ is a martingale, we find with a very similar reasoning that $$\mathbb{E}(X_S) = \lim_{n \to \infty} \mathbb{E}(X_{n \wedge S})=0. \tag{2}$$
4. Write $X_S = -3 \cdot 1_{\{X_S=-3\}} + 5 \cdot 1_{\{X_S=5\}}$. Then $(2)$ shows $$-3 \mathbb{P}(X_S=-3) + 5 \underbrace{\mathbb{P}(X_S=5)}_{1-\mathbb{P}(X_S=-3)} = 0.$$ Use this to calculate $\mathbb{P}(X_S=-3)$ and $\mathbb{P}(X_S=5)$.
5. By $(1)$, $$\mathbb{E}(S) = \mathbb{E}(X_S^2) = 9 \mathbb{P}(X_S=-3) + 25 \mathbb{P}(X_S=5).$$ Plugging in $\mathbb{P}(X_S=-3)$ and $\mathbb{P}(X_S=5)$ gives $\mathbb{E}(S)$.