Computing $E(T_b^2)$ for asymmetric random walk

Question

Let $S$ be an asymmetric random walk with $p=P(X_1=1)>1/2$. Define $T_b=\inf\{n:S_n=b\}$. Prove that $\text{var}(T_b)=\frac{4bpq}{(p-q)^3}$ where $q=1-p$.

We know $$\text{var}(T_b)=ET_b^2-(ET_b)^2.$$ By Theorem 4.8.9 in Durrett we have that $ET_b=b/(2p-1)=b/(p-q)$. So we have $(ET_b)^2$.

I can follow how Durrett computes $ET_b$ but I cannot wrap my brain around on how to start computing $ET_b^2$. Any help would be much appreciated. Thank you in advance!

Answer 1

Hints:

1. Check that $$M_n := S_n-n \mathbb{E}(X_1)$$ and $$N_n := (S_n-n \mathbb{E}(X_1)) ^2- n \text{var}(X_1)$$ are martingales.
2. Apply the optional stopping theorem to show that $$\mathbb{E} \big[ (S_{n \wedge T_b}-(n \wedge T_b) \mathbb{E}(X_1))^2 \big] = \text{var}(X_1) \mathbb{E}(n \wedge T_b), \tag{1}$$ i.e. $$\mathbb{E}(M_{n \wedge T_b}^2) = \text{var}(X_1) \mathbb{E}(n \wedge T_b). \tag{2}$$
3. Show that $$\mathbb{E}(M_{n \wedge T_b} M_{m \wedge T_b}) = \mathbb{E}(M_{m \wedge T_b}^2) \quad \text{for all $m \leq n$}. \tag{3}$$ Use $(2)$ and $\mathbb{E}(T_b)<\infty$ to deduce that $$\mathbb{E}((M_{n \wedge T_b}-M_{m \wedge T_b})^2) \xrightarrow[]{n,m \to \infty} 0. \tag{4}$$
4. Conclude from Step 3 that $M_{n \wedge T_b} \to M_{T_b}$ in $L^2$.
5. Combine Step 4 with $(2)$ to prove that $$\mathbb{E}(M_{T_b}^2) = \text{var}(X_1) \mathbb{E}(T_b),$$ i.e. $$\mathbb{E} \big[ ( S_{T_b}- T_b \mathbb{E}(X_1))^2 \big] = \text{var}(X_1) \mathbb{E}(T_b).$$
6. As $S_{T_b}=b$ it follows that $$\mathbb{E} \big[ ( b- T_b \mathbb{E}(X_1))^2 \big] = \text{var}(X_1) \mathbb{E}(T_b).$$ Expand the square on the left-hand side and use the fact that $\mathbb{E}(T_b) = b/(p-q)$ to compute $\mathbb{E}(T_b^2)$.

Remark: Note that it is crucial that $(S_n)_{n \in \mathbb{N}}$ is an asymmetric random walk with $p>1/2$. For symmetric random walks or asymmetric ranndom walks with $p<1/2$ the above reasoning does not work (e.g. because $\mathbb{E}(T_b)=\infty$).