Suppose $Q(D)$ is a Markov chain with state space $E= \{0,1,...\}$. Further the transition matrix of $Q(D)$ is given by: $$P_D=\begin{pmatrix} \delta_0 & \delta_1 & \delta_2 & \delta_3 & ...\\ \delta_0 & \delta_1 & \delta_2 & \delta_3 & ... \\ 0 & \delta_0 & \delta_1 & \delta_2 & ...\\ 0 & 0 & \delta_0 & \delta_1 & ...\\ &&&&... \end{pmatrix}$$ where $$ \delta_j=\mathbb{E} \left[\frac{(\lambda S)^j}{j!} e^{-\lambda S} \right] $$ and $S$ is a random variable with $\mathbb{E}[S]=\frac{1}{\lambda}.$
Then in my book it is written, that it is not dificult to see the following claim. But that is not so obvious for me. Can somebody help me out?
Claim: $\{ y_j : j \geq 1 \}$, where $y_j$ is given by $y_j=j$ for $j \geq 1$, solves the following inequalities: $$y_j \geq \sum_{j=1}^{\infty} p_{ij} y_j , \text{ for } i \geq 1.$$
If I take for example $i=1$, then I get: $$\sum_{j=1}^{\infty} p_{1j} y_j=\sum_{j=1}^{\infty} p_{1j} j=p_{11} 1+p_{12} 2+p_{13} 3+...=\delta_1 + 2 \delta_2 +3 \delta_3+...$$
So I have to prove that this less or equal 1, but why? If I would get that step, then I think I would get if for the particular case. ( By the way, since it is a transition matrix, we also know that $\sum_j \delta_j$=1.)
Can anybody help me? I appreciate any idea, help or advice!
Thank you so far!
$$\delta_1 + 2 \delta_2 +3 \delta_3+\cdots=\sum_{j\geqslant1}j\,\delta_j=\sum_{j\geqslant1}j\,E\left[\frac{(\lambda S)^j}{j!} e^{-\lambda S} \right]=\sum_{i\geqslant0}E\left[\frac{(\lambda S)^{i+1}}{i!} e^{-\lambda S} \right]=E\left[\lambda S\right]$$