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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?

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Presumably $a\lt0\lt b$. Then this is a simple consequence of the finiteness of the integer interval $[a,b]$. –  Did Jan 12 '13 at 9:54
    
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 $X_1,\cdots, X_n$'? –  DinkyDoe Jan 12 '13 at 12:19
    
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...) –  Did Jan 12 '13 at 12:25
    
Aaargh..i should realy check better what im writing down . $ T = \inf\left\{n:S_n=a, \ or \ S_n=-b\right\}$. –  DinkyDoe Jan 12 '13 at 12:27
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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 $k\geqslant0$, $\mathbb P(T\gt kN)\leqslant(1-\epsilon)^k$.

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