Consider a random walk $S\_n=\sum\_{i=1}^n X\_i$ where $P(X\_i=+1)=P(X\_i=-1)=1/2$ with $n$ large. By Chernoff's bound we know that, for example, $\sum\_{i=1}^{n/2} X\_i=O(\sqrt{n})$ with high probability.


Now say I told you that $S\_n=0$, or more generally that $S\_n$ is ``close to'' its expectation of 0, e.g. that $S\_n=O(\sqrt{n})$. Does it still hold that $\sum\_{i=1}^{n/2} X\_i=O(\sqrt{n})$ with high probability?


Morally speaking I feel like this should be obvious since if $S\_n=0$, then for example $X\_1=-1$ makes it slightly more likely that $X\_2=+1$, and similarly the farther the partial sum gets away from 0 the stronger a force that should be pushing it back to the origin. 


Even more generally, is there some sort of ``theory of small deviations'' which says that conditional on a typical event happening that the event happened because of the typical thing you would expect?



Concentration of closed random walks
I believe the following coupling argument shows that (in particular) if we specify that the random walk ends at 0 then halfway through the walk the probability that we're within distance $\lambda$ of the origin is at least as large as the probability that an unconditional walk is within distance $\lambda$ of the origin. This argument might be the same morally as the one made by Taro, but I feel a little more comfortable with things being made more explicit.


Let $c$ be an integer with $-n\le c\le n$ (it will be simplest to think of the case $c=0$. We wish to define two random walks $S\_{n,n}$ and $T\_{n,n}$ of length $n$ on $\mathbb{Z}$, where $S\_{m,n}$ and $T\_{m,n}$ will denote the position of these walks at step $m$. We let $S\_{m,n}$ be distributed such that $S\_{0,n}=0$ and $S\_{m,n}=S\_{m-1,n}\pm 1$ with each sign being equally likely and independent of all previous decisions. We wish to define $T\_{n,n}$ such that (1) it has the same distribution as uniformly at random choosing a walk of length $n$ from the origin to $c$ and (2) we have $|T\_{m,n}-c|\le |S\_{m,n}-c|$ for all $m$, which in particular imply what we're looking for when we take $c=0$ and $m=n/2$.


To this end, let $\{X\_i\}\_{i=1}^{n}$ be iid random variables that are distributed uniformly on $[0,1]$. We will say that a random walk $R\_{m,n}$ on $\mathbb{Z}$ deviates from $c$ if $|R\_{m,n}-c|>|R\_{m-1,n}-c|$ and that $R\_{m,n}$ doesn't deviate if the inequality is reversed. With this all defined, let $T\_{0,n}=0$. If $X\_i\le 1-\frac{|T\_{m,n}-c|}{n-m+1}$, then choose $T\_{m,n}$ to deviate from $c$ if and only if $S\_{m,n}$ deviates from $c$. Otherwise, choose $T\_{m,n}$ to not deviate from $c$.


It's not hard to see that with this (2) is maintained since $T\_{m,n}$ can only deviate from $c$ when $S\_{m,n}$ deviates from $c$. To see (1), let $R\_{n,n}$ be distributed as in (1) and assume that $R\_{m-1,n}=d$, say with $d\ge c$. At $R\_{m-1,n}$ the random walk has a total of $n-m+1+(d-c)$ steps to the left it must eventually take and $n-m+1+(c-d)$ steps to the right it must eventually take, and $R\_{m,n}$ can be thought of as choosing to take one of these left steps with probability $\frac{n-m+1+(d-c)}{2n-2m+2}$. But by the way we defined $T\_{m,n}$, the probability that it takes a step to the left given $T\_{m-1,n}=d$ is exactly $$\frac{1}{2}(1-\frac{d-c}{n-m+1})+\frac{d-c}{n-m+1}=\frac{n-m+1+d-c}{2n-2m+2}$$ as desired.


