Suppose $S_n:= \sum^n_{i}X_i$ be asymmetric simple random walk. with $P(X_{1}=+1)=p>1/2$, $P(X_{1}=-1)=q$, $p+q=1$, and $S_{0}=0$. Define $T_1 := \inf\{n: S_n = 1\}$. And we know that for $\theta > 0$, $e^{\theta}E\varphi(\theta)^{-T_1}=1$, where $$\varphi(\theta)= pe^{\theta} +qe^{-\theta}.$$
$\textbf{Question}$: Find the probability $$P(T_1=2k+1), k\ge 0.$$
I am really stuck on this question, could you please give me some details about how to solve this question? Thank you!
Let $T_{x} = \inf\{n \geq 0 : S_{n} = x\}$ for $x \in \mathbb{Z}$. Now you can show that for all $x,y \in \mathbb{Z}$ \begin{equation} \mathbb{P}(T_{y} = n | S_{0}=x) = \mathbb{P}(T_{y-x} = n | S_0 = 0),\qquad n \in \mathbb{N}. \end{equation} So we can assume without loss of generality that $S_{0} = 0$. Now define the interarrival times $\xi_{n} := T_{n} - T_{n-1}$ for $n \geq 1$.
It can be shown that the sequence $\{\xi_{n}\}_{n \in \mathbb{N}}$ is i.i.d.
Because of this you can compute the probability generating function of $T_{x}$ and $T_{-x}$ if you can determine the probability generating function of $T_{1}$ and $T_{-1}$, respectively since for $\varphi(s) = \mathbb{E}(s^{T_{1}} | S_{0} = 0)$ we get \begin{equation} \mathbb{E}(s^{T_{x}} | S_{0} = 0) = (\varphi(s))^{x}. \end{equation}
Now with first step analysis we get
\begin{equation} \varphi(s) = \mathbb{E}(s^{T_{1}} | S_{0} = 0) = \frac{1 - \sqrt{1 - 4pqs^{2}}}{2qs}. \end{equation} Similarly \begin{equation} \psi(s) = \mathbb{E}(s^{T_{-1}} | S_{0} = 0) = \frac{1 - \sqrt{1 - 4pqs^{2}}}{2ps}. \end{equation} So for all $x \in \mathbb{Z}$ we get \begin{equation*} \mathbb{E}(s^{T_{x}}|S_{0} = 0)= \begin{cases} \left(\varphi(s)\right)^{x}, & \mbox{if $x > 0$ } \\ \left(\psi(s)\right)^{|x|}, & \mbox{if $x < 0$}. \end{cases} \end{equation*}
Now $\varphi^{(2k+1)}(0) = (2k+1)! \cdot \mathbb{P}(T_{1} = 2k + 1)$