2
$\begingroup$

Hoi, Let $X_1,X_2...$ iiD random variables with $\mathbb{P}(X_n=1)=p$, $\mathbb{P}(X_n=-1)=1-p$ and let $0<a<b$.

Then let $S_n=X_1+\cdots +X_n$ and $T= \inf\left\{n: S_n=a, \ \text{or} \ S_n = -b\right\}$ and $\mathcal{F_n}= \sigma\left\{X_1,\cdots X_n\right\}$

I want to show there exists $N,\epsilon$ such that $\mathbb{P}(T\leq n+N |\mathcal{F}_n)>\epsilon $

I'm not sure what to show here.

Also i want to show there exists $N,\epsilon$ such that $\mathbb{P}(T>kN)\leq (1-\epsilon)^k$ which apparantly follows. Can someone provide me some insight into what to show?

$\endgroup$
  • $\begingroup$ Presumably $a\lt0\lt b$. Then this is a simple consequence of the finiteness of the integer interval $[a,b]$. $\endgroup$ – Did Jan 12 '13 at 9:54
  • $\begingroup$ Im sorry. The only important detail i forgot to mention $0<a<b$. Also, i dont fully understand the conditional probability...should i see this as 'knowing the outcome of $X_1,\cdots, X_n$'? $\endgroup$ – DinkyDoe Jan 12 '13 at 12:19
  • $\begingroup$ If 0<a<b then T is the first hitting time of a, no? Then what is the use of b? (And the rest of the exercise becomes wrong, unless one assumes that p>1/2...) $\endgroup$ – Did Jan 12 '13 at 12:25
  • $\begingroup$ Aaargh..i should realy check better what im writing down . $ T = \inf\left\{n:S_n=a, \ or \ S_n=-b\right\}$. $\endgroup$ – DinkyDoe Jan 12 '13 at 12:27
1
$\begingroup$

Hints: Starting from any point in $(-b,a)$, to perform $N=a+b$ steps in the $+1$ direction implies that one left $(-b,a)$ before time $N$, hence $\mathbb P(T\geqslant n+N\mid\mathcal F_n)\leqslant1-\epsilon$ with $\epsilon=p^N$. Now, if $T\gt (k+1)N$, then $T\gt kN$ and at time $kN$ one is at a point in $(-b,a)$, hence $\mathbb P(T\gt(k+1)N\mid T\gt kN)\leqslant1-\epsilon$ for every $k\geqslant0$, which implies that, for every $k\geqslant0$, $\mathbb P(T\gt kN)\leqslant(1-\epsilon)^k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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