I have a trouble proving the below result:
Let $\left(S_{n}\right)_{n \geq 0}$ be a simple random walk defined by $p(1)=p$ and $p(-1)=q$ where $p+q=1$ Let $k$ be a positive integer.
Prove:
$$ P_{0}\left(T_{0}^{\prime}=2 k \mid S_{2 k}=0\right)=\frac{1}{2 k-1} $$ where $T_{0}^{\prime}=\inf \left\{n \geq 1: S_{n}=0\right\}$.
This is what I have done so far:
Using the definition of conditional probability, I found that:
$$P_{0}\left(T_{0}^{\prime}=2 k \mid S_{2 k}=0\right)= \frac{P_{0}(T_{0}^{\prime}=2 k, S_{2 k}=0)}{P_{0}(S_{2 k}=0)}$$
$P_{0}(T_{0}^{\prime}=2 k, S_{2 k}=0)$ means that $S_{2k}$ is the first time it hits zero. Therefore using Catalan numbers, this is the same as $$ P_{0}\left(S_{1}>0, \ldots, S_{2 k-1}>0, S_{2 k}=0\right)=\frac{1}{2 k-1}\left(\begin{array}{c} 2 k-1 \\ k-1 \end{array}\right) p^{k} q^{k} $$
And $P_{0}(S_{2 k})=0$ is $(\begin{array}{c} 2 k \\ k \end{array}) p^{k} q^{k}$. Thus:
$$\frac{P_{0}(T_{0}^{\prime}=2 k, S_{2 k}=0)}{P_{0}(S_{2 k}=0)} = \frac{ \frac{1}{2 k-1}\left(\begin{array}{c} 2 k-1 \\ k-1 \end{array}\right) p^{k} q^{k} }{ \left(\begin{array}{c} 2 k \\ k \end{array}\right) p^{k} q^{k}}= \frac{1}{2 k-1} * \frac{1}{2} $$
Which is wrong as I am suppose to get $$ \frac{1}{2 k-1} $$
I will be so grateful for any help!
I haven't checked all of your math, but I noticed that for $P_0(T'_0 = 2k, S_{2k} = 0)$, you calculated it as $$P_0(S_1 > 0, \ldots, S_{2k+1} > 0, S_{2k} = 0).$$ This is the probability that the $2k$ is the first time the walk returns to 0 AND the first step is going up. You should also account for the case where the first step is going down, which is $$P_0(S_1 < 0, \ldots, S_{2k+1} < 0, S_{2k} = 0).$$ Once you account for this, I believe the $\frac12$ factor should be resolved.