Problem H$2\mathrm{.}1[15$ points$]$ Consider an infinite DTMC with transition probabilities

$p_{ij}=e^{-}{\displaystyle \underset{n=0}{\overset{j}{}}}(\frac{i}{n})p^n{q}^{i-n}\frac{{}^{j-n}}{(j-n)!},i,j0$

where we recall that $(\frac{i}{n})=0$ for $n>i,$ and we assume that $p+q=1$ with ${}_{i}i{}_i={\displaystyle \underset{j=0}{\overset{}{}}}{}_j{p}_{ji}P(z)={\displaystyle \underset{i=0}{\overset{}{}}}{}_i{z}^{i}zP(z)P(1+p(z-1))P(z)n^{th}P(z)=e^{(z-1)(1+p+p^2+\text{d}\text{o}\text{t}\text{s}+p^{n-1})}P(1+p^n(z-1))P(z){}_{i}0.$

$[2pt]Is$ this chain irreducible, periodic? Explain.

$[6pts]We$ want $to$ ultimately obtain $an$ expression for the probability ${}_{i}of$ being $in$ state $i.$

Assuming the existence $of$ a stationary distribution, $we$ know that

${}_i={\displaystyle \underset{j=0}{\overset{}{}}}{}_j{p}_{ji}$

Using this expression and denoting $asP(z)={\displaystyle \underset{i=0}{\overset{}{}}}{}_i{z}^i,$ the $z-$transform $of$ the stationary probabilities,

find a relation between $P(z)$ and $P(1+p(z-1)).$

Hint: It's mostly a question $of$ using $Eq.(1)inP(z)$ and some algebraic manipulations, including standard

permutations $of$ summations.

$[4pts]$ Recursively applying the relationship you just derived, show that the $n^{th}$ iteration $of$ that recursion

gives

$P(z)=e^{(z-1)(1+p+p^2+\text{d}\text{o}\text{t}\text{s}+p^{n-1})}P(1+p^n(z-1))$

Hint: Prove $itby$ induction.

$[3pts]$ Find $an$ expression for $P(z)$ from the result $of$ the previous question and use $itto$ identify the

resulting distribution and obtain $an$ expression for ${}_i.$