Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 26 / 1

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.

For $0<x<r$, the exit time is at least one. Since $\{\eta \ge n\}=\{\eta >n-1\}\in {\mathcal{F}}_{n-1}$, this event is independent of $X_n$. The Tonelli theorem givesTherefore the integrability of a stopped random-walk increment bound isIt is the survival event, not the exit-at-$n$ event, that is independent of the next increment. The selected exit increment need not have the same distribution or mean as $X_1$.

For $r\le x$, the printed variable $X_{\eta }=X_0$ is undefined because the increment sequence starts at one. Either restrict this part to $0<x<r$, or make the harmless additional convention $X_0=0$. With that convention the conclusion also holds in the immediate-exit case. The integrability assertion needs this indexing qualification.

The random walk $S_n$ is a martingale, since its integrable increments are independent of the past and have mean zero. For the bounded stopping time $n\wedge \eta$, the bounded optional stopping theorem, proved in the next question, gives $\mathbb{E}{S}_{n\wedge \eta }=x$.

If $0<x<r$, the value immediately before exit lies in $(0,r)$, soBefore exit the stopped value has absolute value less than $r$, and after exit it equals $S_{\eta }$. Consequently $|S_{n\wedge \eta }|\le r+|X_{\eta }|$ for every $n$. This is an integrable dominating random variable by the preceding part. Since $\eta <\infty$ with probability one, the dominated convergence theorem givesFor $r\le x$, $S_{\eta }=S_0=x$ directly, without any convention about $X_0$. Thus the expected stopped position is well-defined and has the stated value for every $r>0$.