I'm reading the course notes of a probability course about martingales currently and I'm trying to solve some of the exercises, however I'm very much stuck with the following exercise:
Let $\left\{ X_{n}\right\} _{n=1}^{\infty}$ be a simple random walk on $\mathbb{Z}$ started at $X_{0}=a$ . Given $b\in\mathbb{Z}$ denote by $T_{b}$ the first hitting time of $b$ , show that $\mathbb{E}_{a}\left[T_{b}|\, T_{b}<T_{0}\right]=\frac{b^{2}-a^{2}}{3}$ .
Hint: Show that $M_{n}:=X_{n}^{3}-3nX_{n}$ is a martingale and use the fact that $\left|M_{\min\left(T_{b},n\right)}\right|\leq b^{3}+3bT_{b}$
I've shown that indeed that is a martingale but I have no idea how to proceed, help would be appreciated.
Thanks!
Optional samping theorem holds for every bounded stopping time.
Consider the stopping time $\min(T_0,T_b,n)$ where $T_0$ is the first time the Markov Chain hits $0$, then
$|M_{\min(T_0,T_b,n)}|\leq b^3+3b\min(T_0,T_b)$, the right handside is integrable.
Optional Sampling theorem says
$\mathbb{E}_a M_{\min(T_0,T_b,n)} = \mathbb{E}_a M_0$
since the right hand side is bounded by an integrable random varaible, we take $n$ to infinity and get
$\mathbb{E}_a M_{\min(T_0,T_b) }=a^3$
so the left handside equals to
$\mathbb{P}_a(T_b<T_0)(b^3 - 3b \mathbb{E}_aT_b1_{\{T_b<T_0\}}) + \mathbb{P}_a(T_0<T_b)(0) = a^3$
Note that $X$ is also a martingale, so $\mathbb{P}_a(T_b<T_0)=a/b$ (apply OTS to $X$). This means
$$\frac{b^2-a^2}{3} = \mathbb{E}_aT_b1_{\{T_b<T_0\}}$$