Transition probabilities for random walk on $\mathbb N$ with fixed endpoint Suppose that $W_n$ is a random walk on the nonnegative integers $\mathbb N$, that is,
$$P[W_{n}=i+1|W_{n-1}=i]=P[W_n=i-1|W_{n-1}=i]=1/2$$
whenever $i>0$ and
$$P[W_n=1|W_{n-1}=0]=1.$$

Is there an easy conceptual way to see that if $n<N$ and $i<i_N$,
then
$$P[W_n=i+1~|~W_{n-1}=i,W_N=i_N]\geq\frac12?$$
In words,
if we condition that $W_N=i_N$, and at time $n<N$ the chain is at a state smaller than the endpoint $i_N$, then the chain is more likely to move in the $+1$-direction, i.e., towards its endpoint?
I've tried with the Doob-h-transform but this leads to very nasty computations.
 A: I'm going to call $i_N = j$ for convenience.
The statement is immediately plausible since if $j>i$ and $W_N$ is fixed as $j$, the number of strictly positive paths from $(n,i-1)$ to $(N,j)$, will be less than the number of paths from $(n,i-1)$ to $(N,j)$ (which needs to take fewer upward steps, and is therefore more balanced between upward and downward steps.
Let's get a more solid reason, that is, a relatively simple proof, that the count of strictly non-negative paths from $(n,i+1)$ to $(N,j)$, will be more than the number of strictly non-negative paths from $(n,i-1)$ to $(N,j)$.  
For $j>i$, in order to go from $(n,i-1)$ to $(N,j)$, a path must cross $i$, that is, it must visit at least one point $(m,i)$ and we can choose the smallest such $m$.  Then for every path from $(n,i-1)$ to $(N,j)$, we can form a corresponding path from  $(n,i+1)$ to $(N,j)$ by negating every move time $m$, in the latter path until time $m$, and then matching every move thereafter.  Since in the time span $n\leq t\leq m$ the path from  $(n,i-1)$ to $(N,j)$ is never above the line at $i$, there is no possibility of being unable to do this replication because of the non-negativity constraint -- the corresponding path is always above $i$ until time $m$.
So there are at least as many paths from $(n,i+1)$ to $(N,j)$ as paths from $(n,i-1)$ to $(N,j)$, and in this situation there are in fact more:  Any path from $(n,i+1)$ to $(N,j)$ that never dips down to $i$ will have no corresponding path from $(n,i-1)$ to $(N,j)$, and with the non-negativity constraint there may also be paths from $(n,i+1)$ to $(N,j)$ that go "too high" before time $m$, forcing the corresponding path from $(n,i-1)$ to $(N,j)$ to go negative.
