Conditional expectation of markov chain

Question

Let$\{X_n:n\geq0\}$ be a Markov chain with $X_{n+1}$, conditionally on $X_n=x$, a Poisson r.v. with parameter $\lambda x$ where $\lambda$ is a positive parameter. $X_n$ can be thought of as the number of birth/death on a given day. Find $E[X_n\,|\,X_0=x]$.

Currently, I have the transition probability as follow $$p_{j,k}= \begin{cases} P(X_{k}=x+y\,|\,X_{j}=x)=\cfrac{(\lambda\,x)^{x+y}}{(x+y)!}e^{-\lambda\,x},\hspace{0.5cm}&\text{if $k=j+1$}\\ P(X_{k}=x-y\,|\,X_{j}=x)=\cfrac{(\lambda\,x)^{x-y}}{(x-y)!}e^{-\lambda\,x},\hspace{0.5cm}&\text{if $k=j-1$}\\ P(X_{k}=x\,|\,X_{j}=x)=\cfrac{(\lambda\,x)^{x}}{x!}e^{-\lambda\,x},\hspace{0.5cm}&\text{if $k=j$}\\ 0,&\text{otherwise} \end{cases}$$

not sure if it's right or not.

Answer 1

I would probably approach it this way:

HINT 1: Due to the splitting property of Poisson process, each "individual" within the population $X_n$ evolves its own "tree of descendents" separately.

HINT 2: That makes this a branching process. The answer to your question, i.e. $E[X_n \mid X_0 = 1]$, is given in that article, but it provides no proof. You can either try to prove it yourself, or at least this should give you some further keyword to search.

Good luck!

Answer 2

Apply the law of total expectation, that is $$E[X_n\,|\,X_0=x]=E[\cdots\,|\,E[X_2\,|\,X_1]\,|\,X_0=x]$$