# Conditional probability involving symmetric random walk on Z with return

Let $\zeta = (\zeta_{0},\zeta_{1}, \ldots,\zeta_{2n} )$ be the trajectory of a simple symmetric random on $\mathbb{Z}$. Find $\lim_{n \to \infty}{P (a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b\ | \ \zeta_{0} = \zeta_{2n}=0 })$, where $a$ and $b$ are fixed.

I can't figure how to compute $P (a \leq \frac{\zeta_{n}}{\sqrt{n}}\leq b, \zeta_{0} = \zeta_{2n}=0 )$. I know how to do $P (a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b)$ with Moivre-Laplace and how to get the asymptotics for the probability of return at the $2n$-step... So, I need your help..

Thank you!

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The walk has to go from $0$ to $\zeta_n$ and then from $\zeta_n$ to $0$. You know the asymptotic probability density for a random walk of length $n$ to travel by $c\sqrt n$. The probability of doing that twice is just the square, and the conditional probability you want is the integral of that square from $a$ to $b$ over the integral of that square from $-\infty$ to $\infty$.
Oh wow, you're right. I think I get it but... However, I'm having a little trouble understanding your last sentence. Shouldn't my conditional probability be just $\frac{P (a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b)^{2}}{P(\zeta_{2n} = 0) }$ ? – Dquik Oct 14 '12 at 7:02
@Dquik: No -- by squaring the probability, you're allowing any combination of going to some place in $[a,b]$ and then going to some other place in $[a,b]$; but you have to go to the same place (reflected) to get back to $\zeta_{2n}=0$. So you're first integrating, then squaring, but you need to first square, then integrate to get only the diagonal terms. – joriki Oct 14 '12 at 10:36
@Dquik: You propose to square the probability $P (a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b)$. That gives you the probability of the event $a \leq \frac{\zeta_{n}}{\sqrt{n}} \leq b$ happening twice independently. But that's not what you want -- it includes all sorts of combinations, for instance moving up by $a$ and then moving down by $b$, or vice versa. You want to move up by $x$ and then move down by $x$ in order to get back to $0$, so you want $P(x)^2$ summed over all $x$, not $P(x)$ summed over all $x$ and then squared, which is what you wrote. – joriki Oct 14 '12 at 17:09
Oh, now I understood what you meant. So, the numerator should be $\sum_{a \leq z \leq b} {P^2 (\frac{\zeta_{n}}{\sqrt{n}} } = z)$. However, why isn't the numerator just $P(\zeta_{2n}=0$? (which is just a binomial?) Will you please be willing to put a more complete answer so that we don't make this comments discussion too long? As you can see I'm new to probabilities, so it would be really helpful... – Dquik Oct 15 '12 at 1:21