Trying to solve a problem and got stuck trying to express this
$E[2^{N(t)-N(s)}], (t>s)$
Where $N(t)$ is a Poisson process with unit rate, i.e. I'm trying to find $E[2^X]$ where $X$ has expected value $t-s$. I would have thought the answer would be $E[2^X] = 2^{t-s}$, but the answer is actually $e^{t-s}$. Can anyone explain why this is so?
Note first that, by convexity, $E(2^X)\ne2^{E(X)}$ for every non constant random variable $X$.
Furthermore, if the distribution of $X$ is Poisson with parameter $x$, then, by definition, $$E(2^X)=\sum\limits_{n\geqslant0}\mathrm e^{-x}\frac{x^n}{n!}2^n=\mathrm e^{-x}\sum\limits_{n\geqslant0}\frac{(2x)^n}{n!}=\mathrm e^{-x}\cdot\mathrm e^{2x}=\mathrm e^x.$$