Calculating Expectation of "exponential of a return time" for random walk on an unbiased graph.

Question

Consider an unbiased random walk starting at the vertex marked $0$ on the graph shown in image : Graph - Unbiased random walk and the following transition probabilities:
Transition probabilities

$~~~$i.e. $X_0 = 0.$ where $X_n$ is the vertex position after n time steps.

Calculate $\operatorname{E} (e^{s\tau})$ , $s$ is a constant, $0 < s < 1$ and $\tau$ is the first return time to $0$, $\tau = min\{n > 0; X_n=0 \}$

Answer 1

Let's introduce the variable $Y_n$: $$ Y_n = \begin{cases} \begin{align*} 0 \quad &\text{if}\; X_n = 0 \\ 1\quad &\text{if}\; X_n = 1,2,3 \\ 2 \quad &\text{if}\; X_n = 4 \end{align*} \end{cases} $$ It's easy to see that $\tau = \min\{n > 0 : Y_n = 0\}$. Draw the graph with probabilites of $Y_n$. Clearly, $\tau$ can only be even and we have $$P(\tau = 2k) = \left(\frac{1}{2}\right)^{k}.$$ This leads to $$\mathbb{E}[e^{s\tau}] = \sum_{k=1}^{\infty} P(\tau = 2k) e^{2sk} = \sum_{k=1}^{\infty} \left(\frac{e^{2s}}{2}\right)^{k}. $$ Which converges iff $s < \ln(2)/2$, in which case you obtain

$$\mathbb{E}[e^{s\tau}] =\frac{2}{2-e^{2s}} -1 = \frac{e^{2s}}{2-e^{2s}}$$