This example is taken from HERE ,page 424:
What is the expected value $X(v)$ of this series: $X(v)=E(\Sigma_{t=0}^{\infty}\delta^t p_t(v))$ where $p_t(v)\in[0,1]$ with $\delta\in(0,1)$? Are all data needed for the expected value given?
I.e. what is the formula for such an expected value of a discounted series of probabilities?
Suppose $X$ has probability mass function $\{p_n\}$, i.e. $p_n\geqslant0$ for all $n$ and $\sum_{n=0}^\infty p_n = 1$. Let $\delta\in(0,1)$. Then the generating function of $X$ is $$\mathbb E\left[\delta^X\right] = \sum_{n=0}^\infty p_n\delta^n. $$ For example, if $X\sim\operatorname{Pois}(\lambda)$ then $$\mathbb E\left[\delta^X\right] = \sum_{n=0}^\infty \frac{e^{-\lambda}\lambda^n}{n!}\delta^n=e^{-\lambda}\sum_{n=0}^\infty \frac{(\lambda\delta)^n}{n!} = e^{-\lambda(1-\delta)}. $$