# About asymmetric simple random walk

Let $\left\{ S_{n}\right\} _{n\geq0}$ be the (asymmetric) simple random walk such that $S_{n+1}=S_{n}+\xi_{n+1}$ where $\xi_{n}\in\{-1,1\}$ and $\left\{ \xi_{n}\right\} _{n\geq0}$ are iid. Moreover, $\mathbb{P}(\xi_{n}=1)=p$ and $\mathbb{P}(\xi_{n}=-1)=1-p$ and ${\displaystyle p>\frac{1}{2}}.$ For $x\in\mathbb{Z}$ we define $T_{x}=\inf\left\{ n:S_{n}=x\right\}$ and $\phi(x)=\left(\frac{1-p}{p}\right)^{x}.$

Then for $a<0<b$, Show that $E\left[\phi(S_{N})\bigg|T>N\right]\leq\left(\frac{1-p}{p}\right)^{b}+\left(\frac{p}{1-p}\right)^{a}$where $T=T_{a}\wedge T_{b}$.

My thoughts:

${\displaystyle \phi(S_{N})={\displaystyle \left(\frac{1-p}{p}\right)^{S_{N}}}=\left(\frac{1-p}{p}\right)^{\sum_{i=1}^{N}\xi_{i}}=\prod_{i=1}^{N}\left(\frac{1-p}{p}\right)^{\xi_{i}}}$

Since $\left\{ \xi_{n}\right\} _{n\geq0}$ are iid, I somehow guess $E\left[\phi(S_{N})\bigg|T>N\right]=1$, but I am not sure.

Even if I am right, I can not show $\left(\frac{1-p}{p}\right)^{b}+\left(\frac{p}{1-p}\right)^{a}\geq1$.

I only know that $T>N\Rightarrow a<S_{N}<b$ and $0<\left(\frac{1-p}{p}\right)<1$.

Thank you!

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On the event $[T\gt N]$, $a\lt S_N\lt b$ hence $\phi(S_N)\lt \phi(a)$ almost surely, since $\phi$ is decreasing. In particular, $\mathrm E(\phi(S_N)\mid T\gt n)\lt\phi(a)=\left(\frac{1-p}p\right)^a$.

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Are you sure about the upper bound you are asked to prove? –  Did Apr 15 '12 at 19:56
I agree with your argument, but that is not the question here. I double checked the question. It is exactly the same as the textbook. But frankly speaking I find this exercise annoying, because later we only need the conclusion that it is bounded. So the upper bound such as yours is sufficient for the later use. Anyway, I can not argue that the exercise is wrong. Thank you! –  user16859 Apr 15 '12 at 21:02