Let $\mathbb{P}\{W=1\}=p$ and $\mathbb{P}\{W=-1\}=(1-p)$. Current position is -1 $(X_0 = -1)$. Now, I want to know what is $\mathbb{E}[2^{T_{-1,0}}]$, where $T_{i,j}$ is the time to arrive $j$ from $i$ for the first time.
I was trying to calculate it as follows. \begin{align} \mathbb{E}[2^{T_{-1,0}}] & = \mathbb{E}[2^{T_{0,0}+1}]\mathbb{P}\{W=1\} +\mathbb{E}[2^{T_{-2,0}+1}]\mathbb{P}\{W=-1\} \\ & =2p + 2(1-p)\mathbb{E}[2^{T_{-1,0}}]^2 \end{align} Then, we have $\mathbb{E}[2^{T_{-1,0}}] = \frac{1+\sqrt{1-16p(1-p)}}{4(1-p)}$. But, I cannot understand why $\mathbb{E}[2^{T_{-1,0}}]$ increased with $p$. I think there must be an error. It doesn't make sense. Could you please let me know where is an error or How I can get $\mathbb{E}[2^{T_{-1,0}}]$.
There is nothing wrong except that you took the positive root.
You start with the random walk with increments given by $W \sim p\varepsilon_1 + (1-p) \varepsilon_0.$ Then using strong Markov property derive the equation $$ \varphi(p) = 2p + 2(1-p) \varphi(p)^2, $$ where $\varphi(p)$ is the expectation of $2^{T_{-1,0}}$ with the given distribution of $W.$ This equation is trivially true when $\varphi(p) = \infty.$ We also know that $\varphi(1) = 2$ and $\varphi,$ if finite at all, should be finite finite right on the left of the point $1.$ For those values of $p$ for which $\varphi(p)$ is finite, $\varphi$ is expressible using the quadratic formula as $$ \varphi(p) = \dfrac{1 \pm \sqrt{1 - 16p(1-p)}}{4(1-p)} $$ and this formula then holds fors $1 - 16p(1-p) \geq 0,$ numerical calculations then give the interval $p \in (0.933, 1].$ As of yet, we need to still choose the sign of $\varphi.$ Using the condition $\varphi(1) = 2$ (and the not so obvious fact that $\varphi$ should be continuous at $1$) we then see that (when you take the negative sign, the limit is 2; when you take the positive, the limit does not even exists) $$ \varphi(p) = \dfrac{1 - \sqrt{1 - 16p(1-p)}}{4(1-p)}. $$ This is a decreasing function on $(0.933, 1]$ with limit $2$ as $p \uparrow 1.$