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

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!

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 $$\mathbb{E}(T_b)=\infty$$).

• Oof! That's a lot. Thank you so much! – Gengar Mar 6 at 18:14
• @Gengar You are welcome. Let me know if you get stuck – saz Mar 6 at 18:15