Chain of birth and dead. Consider a Markov chain $(X_n)_{n \gt 0}$ with state space $E=\left \{ 0,1,2,\ldots \right \} $, such that, starting from some state $x$, it will go to $x+1,x$ or $x-1$ to the next trial with probabilities $p_x,r_x,q_x=1 \; \forall x \in E$ and with $q_0=0$
(a) Find the transition matrix of this chain
(b) If $X_0 \sim \text{Poisson}(\lambda)$, find the distribution of $X_1$
(c) Find $\mathbb{P}(X_2=2) \text{ if } \lambda =2$
It was easy to find (a) The transition matrix is given by:
$$\begin{pmatrix}
r_0 & p_0 & 0 & 0 & 0 & \cdots \\
q_1 & r_1 & p_1 & 0 & 0 & \cdots \\
0 & q_2 & r_2 & p_2 & 0 & \cdots \\
0 & 0 & q_3 & r_3 & p_3 & \cdots \\
0 & 0 & 0 & q_4 & r_4 & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots
\end{pmatrix}$$
But for (b) I don´t know how having the distribution of $X_0$ will help me to find $X_1$
Best Answer
If $x\ge 1$ then \begin{align} & \Pr(X_1 = x) \\[8pt] = {} & \Pr\Big((X_0=x-1\ \&\ X_1=x) \text{ or } (X_0=x\ \&\ X_1 = x) \text{ or }(X_0=x+1\ \&\ X_1=x)\Big) \\[8pt] = {} & \frac{\lambda^{x-1} e^{-\lambda}}{(x-1)!} \cdot p_{x-1} + \frac{\lambda^x e^{-\lambda}}{x!} \cdot r_x + \frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!} \cdot q_{x+1} \\[8pt] = {} & \frac{\lambda^{x-1} e^{-\lambda}}{(x-1)!} \left( p_{x-1} + \frac \lambda x\cdot r_x + \frac{\lambda^2}{x(x+1)} \cdot q_{x+1} \right). \end{align}
However, since you have given no information about the way in which $p_x,r_x,q_x$ depend on $x,$ I am inclined to doubt that more can be said.