Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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, $\mathrm E(\phi(S_N)\mid T\gt n)\lt\phi(a)=\left(\frac{1-p}p\right)^a$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.