Given {$Y_n,n\ge0$} i.i.d., $P(Y_n=1)=p\ge0$, $P(Y_n=-1)=q=1-p\ge0$. Let $X_0=Y_0=0$ ,$X_n=\sum_{k=0}^n Y_k$, $n\ge0$, $T_{0k}\triangleq\min \{n:n>0,X_0=0,X_n=k\}$.
Now, please use probability generating functions to find the distribution of $T_{01}$.
Here is what I have tried:
$$ \begin{align*} g_{T_{01}}(s)&=\mathbb{E}(s^{T_{01}}| X_0=0) \\ &=\sum_{n=1}^{\infty}s^nP(T_{01}=n | X_0=0) \\ &=\sum_{n=1}^{\infty}s^nP(X_n=1,X_{n-1}\neq1,\dots,X_1\neq1|X_0=0) \\ &=\sum_{n=1}^{\infty}s^nP(\sum_{k=1}^nY_k=1,\dots,Y_1\neq1|Y_0=0) \\ &=sp+\sum_{n=2}^{\infty}s^nP(\sum_{k=2}^nY_k=2,\dots,Y_2\neq2,Y_1=-1|Y_0=0) \\ &=sp+\sum_{n=1}^{\infty}s^{n+1}P(\sum_{k=1}^nY_k=2,\dots,Y_1\neq2,|Y_0=-1) q\\ \end{align*} $$
Use infinite summation, I tried to construct the equation like: $g(s)=f(s,g(s))$ but failed.
Thanks for @Benjamin Wang's tips on the Catalan number!
Rather than using infinite summation, it's easier to use convolution properties of the PGF.
We have the formula: $p_{01}^{(n)}=\sum_{l=1}^n f_{01}^{(l)}p_{11}^{(n-l)}=f_{01}^{(n)}*p_{11}^{(n)}$,where $p_{01}^{(n)}=P(X_n=1|X_0=0)$, $f_{01}^{(l)}=P(T_{01}=l|X_0=0)$, "$*$" denotes the convolution of two functions.
The PGF of $\{n:n\ge0,X_n=1|X_0=1\}$ is defined as $\mathscr{G}(p_{01}^{(n)})=\sum_{n=0}^\infty s^np_{01}^{(n)}$, and other PGFs can also be defined similarly. Then we have the equation: $\mathscr{G}(p_{01}^{(n)})=\mathscr{G}(f_{01}^{(l)})\mathscr{G}(p_{11}^{(n)})$. Therefore, the PGF of $T_{01}$=$\mathscr{G}(p_{01}^{(n)})/\mathscr{G}(p_{11}^{(n)})$.
$$ \begin{align*} \mathscr{G}(p_{11}^{(n)})&=\sum_{n=0}^\infty s^nP(X_n=1|X_0=1) \\ &=\sum_{n=0}^\infty s^nP(\sum_{k=1}^n Y_k=0) =\sum_{m=0}^\infty s^{2m}P(\sum_{k=1}^{2m} Y_k=0) & \text{if n is odd, P(·)=0} \\ &=\sum_{m=0}^\infty s^{2m}\binom{2m}{m}p^mq^m =\sum_{m=0}^\infty \binom{2m}{m}z^m & z=s^2pq<\frac{1}{4} \\ \end{align*} $$
From the wiki https://en.m.wikipedia.org/wiki/Catalan_number, we can find that the Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}$, and its generating function $c(x)=\sum_{n=0}^\infty C_nx^n=\frac{1-\sqrt{1-4x}}{2}$. Therefore,
$$ \mathscr{G}(p_{11}^{(n)})=\sum_{n=0}^\infty (m+1)C_mz^m=(zc(z))^\prime=\frac{1}{\sqrt{1-4z}} $$
Likewise,
$$ \begin{align*} \mathscr{G}(p_{01}^{(n)})&=\sum_{n=0}^\infty s^nP(X_n=1|X_0=0) \\ &=\sum_{n=0}^\infty s^nP(\sum_{k=1}^n Y_k=1) =\sum_{m=0}^\infty s^{2m+1}P(\sum_{k=1}^{2m+1} Y_k=1) & \text{if n is even, P(·)=0} \\ &=\sum_{m=0}^\infty s^{2m+1}\binom{2m+1}{m}p^{m+1}q^m =sp\sum_{m=0}^\infty \binom{2m+1}{m}z^m & z =s^2pq<\frac{1}{4} \\ &=sp\sum_{m=0}^\infty ((2m+2)-1)C_mz^m=sp(2(zc(z))^\prime-c(z)) \\ &=sp\frac{1-\sqrt{1-4z}}{2z\sqrt{1-4z}} \end{align*} $$
Now, we can get the PGF of $T_{01}$:
$$ \mathscr{G}(f_{01}^{(n)})=\mathscr{G}(p_{01}^{(n)})/\mathscr{G}(p_{11}^{(n)})=sp\frac{1-\sqrt{1-4z}}{2z}=spc(z) $$
Surprisingly, the PGF of $T_{01}$ is just like the generating function of Catalan number. Hence, the distribution of $T_{01}$:
$$ P(T_{01}=n|X_0=0)= \left\{ \begin{align*} &0 & n=2m \\ &C_mp^{m+1}q^m & n=2m+1 \\ \end{align*} \right. $$
Since the the coefficient of the distribution of $T_{01}$ is equal to Catalan number, I wonder if there is any way or view to get the distribution directly through the Catalan number. I would be grateful if anyone could provide some ideas.