Random walk : probability of reaching value $i$ without passing by negative value $j$ This is just some question that popped out of nowhere while starting studying random walks, and I don't really know how to approach this.
Say I have a random walk that starts at zero, and goes up or down by one at each step with equal probability.
For some integer $i$, we stop the walk whenever it reaches either $i$ or $-i$.  
Suppose we are given that the walk stopped by reaching $i$.
I'm interested in the minimum value the walk passed through.
In other words, for some $0 \geq j > -i$, what it the probability that the walk took value $j$ at some point, but not $j - 1$ ?  
 A: To compute $P(j\cap i)$, the probability of reaching $i$ having reached $j$ as minimum, consider playing two games after each other:


*

*$G$ Is the game that is won by starting from zero, ending at $j$ without having reached $i$

*$H$ is the game that is won by starting from $j$, ending at $i$ without having reached $j-1$


For the first one we have
$$
P(G)=\frac{i}{i+|j|}
$$
for the second we have
$$
P(H)=\frac{1}{1+i+|j|}
$$
and so
$$
P(j\cap i)=P(H)P(G)=\frac{i}{(i+|j|)(1+i+|j|)}
$$
and as I understood it, you asked for the conditional probability $P(j\mid i)$ of having reached $j$ as a minimum given that we end at $i$. This should then be
$$
P(j\mid i)=\frac{P(j\cap i)}{P(i)}=\frac{2i}{(i+|j|)(1+i+|j|)}
$$
since $P(i)=\frac12$ if I am not mistaken. This formula should work for $-i<j\leq 0$ then. I calculated a table of probabilities for the example $i=10$ using Wolfram|Alpha. It looks plausible. For one thing, the probabilities sum to $1$, and of course they decrease with $j$ as they should - which is evident from the formula.
