Suppose that $(X_{n}:n\in\mathbb{N})$ is a $\pm1\mbox{-valued sequence.}$ Let $p\in(0,1)$ and $p=\mathbb{P}(X_{i}=1)\mbox{ and}\mathbb{P}(X_{i}=-1)=1-p=q$ . Define the simple random walk $S=(S_{n})$ , $S_{n}=S_{0}+\sum_{i=1}^{n}X_{i}$ , with $S_{0}=0$ . $N_{n}(0,b)$ denotes the number of steps that takes to go from the point $0$ at time $0$ to the point $b$ at time $n$ . Recall the Ballot theorem that states for $b>0$ the number of paths from $(0,0)$ to $(n,b)$ which do not revisit the $x-\mbox{axis}$ (origin) equals $\frac{b}{n}N_{n}(0,b)$ . I wrote in my notes that the following equality is justified by the ballot theorem, but I can't understand why. Any help would be wonderful!
$p([P(S_{n-1}=b-1)-(\frac{q}{p})\mathbb{P}(S_{n-1}=b+1)]=\frac{b}{n}\mathbb{P}(S_{n}=b)$
I don't think the Ballot Theorem justifies it. It comes from evaluating the probabilities involved:
\begin{eqnarray*} &&p\cdot P(S_{n-1} = b-1) - q\cdot P(S_{n-1}=b+1) - \dfrac{b}{n}P(S_n=b) \\ && \\ && \qquad\qquad = p\binom{n-1}{\frac{n+b-2}{2}} p^{\frac{n+b-2}{2}}q^{\frac{n-b}{2}} - q\binom{n-1}{\frac{n+b}{2}} p^{\frac{n+b}{2}}q^{\frac{n-b-2}{2}} - \dfrac{b}{n} \binom{n}{\frac{n+b}{2}} p^{\frac{n+b}{2}}q^{\frac{n-b}{2}} \\ && \\ && \qquad\qquad = 0\qquad\qquad\text{by expanding the binomial cofficients and simplifying.} \\ \end{eqnarray*}