Distance from origin of biased random walk conditioned to be positive at time n Let $S_n$ be the position of a simple random walk on the integers started from $0$ that moves right with probability $p<1/2$. What is the asymptotic behavior of 
$$E[ S_n \mid S_n >0 ]$$
as $n \to \infty$?
This is a large deviation event, so I don't have  good intuition for how far the walk should be. I would believe anything between $O(1)$ and $O(n)$. However, my gut tells me that it is $O(\sqrt n)$ which is what would occur for $p=1/2$. 
Note that $S_k$ is allowed to be nonpositive for $k<n$, we are just conditioning on its location at time $n$. 
 A: This quantity turns out to be $O(1)$. It turns out the necessary calculations were already performed in this 1989 paper of Arratia and Gordon, who proved a more general theorem implying that the conditional law of $S_n$ given $S_n>0$ converges to a geometric distribution with explicit parameters (which we work out below).
Let $T_n$ be the number of rightwards moves made by the random walk in reaching $S_n$ after $n$ steps, so that $n-T_n$ leftward moves were made and thus $S_n=T_n-(n-T_n)=2T_n-n$ and $T_n\sim \textrm{Bin}(n,p)$.
The following is a direct consequence of Arratia and Gordon's Theorem 2. (They obtain asymptotics for the numerator and denominator, then remark in a paragraph before the theorem why the following is implied.)

Theorem. Let $0< p<\alpha<1$ be constants, and let $k_n$ be a sequence of integers tending to infinity such that $\lim_{n\to\infty}k_n/n=\alpha$. Let $r=p(1-\alpha)/\alpha(1-p)$. Let $T_n\sim \textrm{Bin}(n,p)$. Then for all integers $i\geq 0$,
  $$
\lim_{n\to\infty}\mathbb P(T_n=k_n+i\mid T_n\geq k_n)=r^i(1-r).
$$

Applying this theorem with $k_n=\lfloor n/2\rfloor+1$ and $\alpha=1/2$, we obtain that
$$
\lim_{n\to\infty}\mathbb P(T_n=\lfloor n/2\rfloor+i+1\mid T_n> \lfloor n/2\rfloor)=\frac{p^i}{(1-p)^i}\frac{1-2p}{1-p},\qquad i\geq 0.
$$
Summing these probabilities appropriately yields the limiting conditional expectation
$$
\lim_{n\to\infty}\mathbb E(T_n-\lfloor n/2\rfloor\mid T_n> n/2)=1+\frac{1-2p}{1-p}\sum_{i=0}^{\infty}\frac{ip^i}{(1-p)^i}=1+\frac{p}{1-2p}.
$$
Recalling that $S_n=2T_n-n$, we obtain that
$$
\lim_{n\text{ even}\to\infty}\mathbb E(S_n\mid S_n>0)=2+\frac{2p}{1-2p},
$$
and
$$
\lim_{n\text{ odd}\to\infty}\mathbb E(S_n\mid S_n>0)=1+\frac{2p}{1-2p}.
$$
