In the question we have a birth-death chain on $\{0,1,2,…\}$ whose only non-zero transitions from $i$ are to $i+1$ and $i-1$, with probabilities $p_i$ and $q_i$, respectively. I have that $p_i$ and $q_i$ are positive for all $i\ge1$, and $h_i$ is the probability of reaching $0$ starting from $i$.
If we write $$\gamma_i = \prod_1^{i-1}{\frac {q_i}{p_i}}$$ it can be shown by iteration that $h_i = 1-(1-h_1)\sum_{k=1}^i{\gamma_k}$. I now have a case where $$p_i=\left(\frac {i+1}{i}\right)^2 q_i$$ and I need to show that if $X_0=1$, then $\mathbb{P}(X_n \ge1 \text{ for all } n\ge1) = 6/\pi^2$. I can see clearly that $$\sum_{k=1}^\infty{\gamma_k}=\frac {\pi^2}{6}$$ but I'm not really sure how to progress beyond there. Any help?
Best Answer
I have doubts about your formula for $h_i$. I think the derivation of it should produce:
\begin{align} h_i &= \dfrac{\sum\limits_{k=i}^{\infty} \rho_k}{1 + \sum\limits_{k=1}^{\infty} \rho_k} \qquad\text{ where } \rho_k = \dfrac{q_1\cdots q_k}{p_1\cdots p_k} = \prod_{j=1}^k\left(\dfrac{j}{j+1}\right)^2 = \dfrac{1}{(k+1)^2},\;k\geq 1 \\ \end{align}
A reference for this is here or Karlin and Taylor's Introduction to Stochastic Modeling.
The required probability is:
\begin{align} 1-h_1 &= \dfrac{1}{1 + \sum\limits_{k=1}^{\infty} \rho_k} \\ & \\ &= \dfrac{1}{\sum\limits_{k=1}^{\infty} \dfrac{1}{k^2}} \\ & \\ &= \dfrac{1}{\dfrac{\pi^2}{6}} \\ & \\ &= \dfrac{6}{\pi^2}. \end{align}