Expectation of $T^2$ where $T$ is the absorption time at ${a,−a}$ of a simple random walk $\{S_n\}$

Question

I asked a very similar question before:

Expectation of $TS_T$ where $T$ is the absorption time at $\{a,-a\}$ of a simple symmetric random walk $\{S_n\}$

But this time I have an ASYMMETRIC random walk, and I want the expectation of $T^2$. Well known martingales don't seem to work.

Assume the steps are $1$ with probability $p$ and $-1$ with probability $1-p$.

Answer 1

Let $q=1-p$. One can show that for every real number $s$ small enough, $$ \mathrm E_0(s^{T})=s^a\,\frac{(2p)^a+(2q)^a}{(1+v(s))^a+(1-v(s))^a},\qquad v(s)=\sqrt{1-4pqs^2}. $$ Let $n$ denote a positive integer. Differentiating this identity $n$ times with respect to $s$ and plugging $s=1$ in the result yields $\mathrm E_0(T(T-1)\cdots(T-n+1))$. Alternatively, plugging $s=\mathrm e^x$ in this identity for $x$ small enough and expanding the resulting identity up to the order of $x^n$ when $x\to0$ yields $\mathrm E_0(T^k)$ for every positive integer $k\leqslant n$. For example, $$ \mathrm E_0(T)=\frac{a}{p-q}\,\frac{p ^a-q^a}{p^a+q^a}. $$