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 $\{a<S_n<b\text{ for all }n\}$ to occur.

Can you conclude?

  • $\begingroup$ Then $\mathbb{P} (\{a < S_n < b\}) = 1 - \sum \mathbb{P}(E_{k})$ ? $\endgroup$ – openspace Feb 15 at 18:49
  • $\begingroup$ $\sum P(E_k)=\infty$, so not quite. Try $P(a<S_n<b)\le \prod_k (1-P(E_k))$. @openspace $\endgroup$ – Mike Earnest Feb 15 at 19:42
  • $\begingroup$ oh, yes. Then it will be something goes to zero on the right side $\endgroup$ – openspace Feb 15 at 20:40
  • $\begingroup$ But why probability of this events should be greater than zero? $\endgroup$ – openspace Feb 16 at 10:44
  • 1
    $\begingroup$ It's probability is $p^{(b-a)}$? $\endgroup$ – openspace Feb 16 at 17:18

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