Local time for conditioned simple random walk Consider simple random walk started at $1$ on the path $[0,M]$. Let $\tau_x$ be the time to reach $x$ and condition on the event $\tau_M < \tau_0$ (the walk hits $M$ before reaching $0$). Conditioned on this event, let $V_k$ be the number of visits to $k$ up to time $\tau_M$. What is $\mathbb E V_k$? 
All we need is a bound independent of $M$. Something like there exists $C>0$ such that
$$\sup_{M >0 }\Bigl[ \max_{0< k < M} \mathbb E V_k \Bigr]< C.$$
 A: To second Anthony Quas's answer, here's a simple way to calculate $\mathbb{E}V_k$. Notice that conditional on $\tau_M<\tau_0$ your walk will visit $k$ at least once. Once we are at $k$, the number of consecutive visits is a positive geometric random variable, since the walk is a Markov process. The parameter is $1-\mathbb{P}(V_k=1 | \tau_M<\tau_0)$, and therefore its expectation value is 
$$\mathbb{E}[V_k]=\frac{1}{\mathbb{P}(V_k=1 | \tau_M<\tau_0)} = \frac{\mathbb{P}(\tau_M < \tau_0)}{\mathbb{P}(\tau_k<\tau_0)\frac{1}{2}\mathbb{P}(\tau_{M-k}<\tau_0)},$$
since the walk has to visit $k$ before $0$, step to $k+1$, and then visit $M$ before returning to $k$. Since $\mathbb{P}(\tau_M < \tau_0) = 1/M$ (ballot problem) we find $$\mathbb{E}[V_k]= 2 k (1-k/M)$$.
A: Surely you can't hope for a bound that's independent of $M$. Imagine that $M$ is huge and let $K\ll M$ be large. Then consider what happens after you first hit $M-K$. You're seeing approximately a simple symmetric random walk so it should take $K^2$ steps to reach $M$. There should be around $K$ of these at $M-K$. 
Or are my heuristics way off?
