The probability $\varepsilon_{r,2n+1}$ that upto epoch 2n+1 there occur exactly r changes of sign equals $2p_{2n+1,2r+1}$
Basically
$\varepsilon_{r,2n+1} = 2P(S_{2n+1} = 2r+1)$
This is a theorem in a book named "An Introduction to Probability Theory and its Applications" Vol 1 by William Feller, Chpt 3, Section 5.
The proof according to the book is given like this
"We begin by rephrasing the theorem in a more convenient form. If the first step leads to the point $(1,1)$ we take this point as the origin of a new coordinate system. To a crossing of the horizontal axis in the old system there now corresponds a crossing of the line below the new axis, i.e, a crossing of level $-1$. An analogous procedure is applicable when $S_1 = -1$, and it is thus seen that the theorem is fully equivalent to the following proposition : The probability that up to epoch $2n$ the level -1 is crossed exactly $r$ times equals $2p_{2n+1,2r+1}$."
What follows from this is that it take the case where $r=0$ i.e, it's the case where the level -1 is not crossed, and from this case the hypothesis is proved for $r=0$. I had no problem understanding this case
The difficulty came when it took the case for $r=1$. That is the given path now crosses level -1 once. Here's how the proof has been written in that book.
"Next let $r=1$. A path that crosses the level $-1$ at epoch $2v-1$ may be decomposed into the sectiom from $(0,0)$ to $(2v,-2)$ and a path of length $2n-2v$ starting at $(2v,-2)$. To the latter section we apply the result for $r=0$ but interchanging the roles of plus and minus. We conclude that the number of paths of length $2n-2v$ starting at $(2v,-2)$ and not crossing the level $-1$ equals the number of paths from $(2v,-2)$ to $(2n+1,-3)$." This follows from the first case of $r=0$. Everything is fine till here.
However the main problem that I couldn't grasp is the statement that follows afterwards "But each path of this kind combines with the initial section to a path from $(0,0)$ to $(2n+1,-3)$. It follows that the number of path of length $2n$ that cross the level $-1$ exactly once equals the number of paths from the origin to $(2n+1,-3)$.
So from what I can understand, is that they are combining paths from $(2v,-2)$ to $(2n+1,-3)$ (the number of such paths equals number of paths of length $2n-2v$ from $(2v,-2)$ not crossing the level $-1$) with the paths from $(0,0)$ to $(2v,-2)$, and then claiming that the number of such combined paths equals the number of paths from $(0,0)$ to $(2n+1,-3)$. First of all how the heck is this possible? Of course paths from $(0,0)$ to $(2n+1,-3)$ will cross levels $-1$ and $-2$. But how does it guarantee that there is no path from $(0,0)$ to $(2v,-2)$ that would cross level $-1$ at epoch other that epoch $2v-1$ ? After all, we are talking about paths upto epoch 2n that crosses level $-1$ once. Can someone elaborately explain this case when $r=1$ ?
Clearly each path from $(0,0)$ to $(2v,-2)$ to $(2n+1,-3)$ gives a path from $(0,0)$ to $(2n+1,-3)$. So you are asking to show the opposite direction: given a path from $(0,0)$ to $(2n+1,-3)$, show that it can be decomposed as a path from $(0,0)$ to $(2v,-2)$ to $(2n+1,-3)$ such that it crosses level $-1$ exactly once in the first segment of the decomposition.
Remember that we have the freedom to determine $v$; this is the key in showing this. Given a path from $(0,0)$ to $(2n+1,-3)$, there will be a first stage at which the path crosses level $-1$. Then at some even time it ends up at level $-2$ for the first time; let $v$ be such that $(2v,-2)$ is the first time the path reaches level $-2$. Then clearly the path from $(0,0)$ to $(2v,-2)$ crosses level $-1$ exactly once, and the rest of the path is some path from $(2v,-2)$ to $(2n+1,-3)$.