# Probability estimation question.

Consider a random walk with initial point zero. So we have $$S_{n} = X_{1} + \dots + X_n$$. And we have two fixed numbers $$b > 0 > a$$. Now we want to show that $$(*) = \operatorname{P}(\{\forall n : a < S_{n} < b\}) \to 0$$ , i.e. our random walk cross the bound with full measure.

My attempt : $$\displaystyle (*) = \sum_{a < k < b} \binom{n}{(n+k)/2}p^{(n+k)/2} (1-p)^{(n-k)/2}$$.

Consider a function $$f(x) = x^{a}(1-x)^{n-a}$$, it's easy to see that function has maximum at point $$x = \frac{a}{n}$$. So we have $$\displaystyle (*) \le \sum_{a < k < b} \binom{n}{(n+k)/2} (\frac{n+k}{2n})^{(n+k)/2}(\frac{n-k}{2n})^{(n-k)/2} = \sum \binom{n}{(n+k)/2}\frac{1}{2^n} (1-\frac{k^2}{n^2})^{n/2} (\frac{1+\frac{k}{n}}{1-\frac{k}{n}})^{k/2}$$. Unfourtunatly my estimates are bad , because of if we are taking limit from both sides we have that rhs goes to $$1$$. Any help with estimations will be good!

If at any point there are $$(b-a)$$ consecutive indices where $$X_i=+1$$, then the random walk will certainly cross the top boundary if it has not crossed any boundaries already. Therefore, letting $$E_k=\{X_i=+1\text{ for all }i=k(b-a),k(b-a)+1,\dots,(k+1)(b-a)-1\}$$ then the events $$E_k$$ are independent, have probability less than $$1$$, and none of them must occur in order for $$\{a to occur.
• Then $\mathbb{P} (\{a < S_n < b\}) = 1 - \sum \mathbb{P}(E_{k})$ ? – openspace Feb 15 at 18:49
• $\sum P(E_k)=\infty$, so not quite. Try $P(a<S_n<b)\le \prod_k (1-P(E_k))$. @openspace – Mike Earnest Feb 15 at 19:42
• It's probability is $p^{(b-a)}$? – openspace Feb 16 at 17:18