Expected value of exponential transformation of poisson

Let X be a poisson distributed variable with parameter $\lambda$. I would like to calculate the expectation of $E[e^{-ax}]$, where $a$ is a constant. I use the Taylor series approximation as follows, although I get a really ugly result. Is there a better way to solve it so that the answer is more convenient?

$$E[e^{-ax}] = \sum_{i=0}^\infty e^{-ai}\frac{e^{-\lambda}\lambda^i}{i!} = e^{-\lambda}\sum_{i=0}^\infty \frac{ (\lambda/e^a)^i }{i!}$$

which is the Taylor series for $e^{\lambda/e^a}$. Thus,

$$E[e^{-ax}] = e^{-\lambda} e^{\lambda/e^a}$$

Is this correct? or is there a better / more correct way to do it?

• I see nothing wrong with that other than a bunch of typos. (And disagree that it's real ugly. In fact it's one of the most beautiful MGFs out there.) – spaceisdarkgreen May 15 '18 at 23:33
• Thank you! all typos are fixed – dleal May 15 '18 at 23:43
• you still missed one (or two if you count multiplicity). – spaceisdarkgreen May 15 '18 at 23:44
• (In the sum it should be $i!,$ not $\lambda !$) – spaceisdarkgreen May 16 '18 at 2:07