Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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, $$E(2^X)=\sum\limits_{n\geqslant0}\mathrm e^{-x}\frac{x^n}{n!}2^n=\mathrm e^{-x}\sum\limits_{n\geqslant0}\frac{(2x)^n}{n!}=\mathrm e^{-x}\cdot\mathrm e^{2x}=\mathrm e^x.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.