Expected value of $e^x$ as $X\sim exp(\lambda)$

Question

Let $X\sim exp(\lambda)$

Calculate the Expected value of $$Y=e^X$$

Is there any elegant way to do it?

Answer 1

Observe that $E(Y)=E(e^X)=E(e^{1\cdot X})=m_X(1)$ where $m_X$ is the moment generating function of $X$. So if you know the moment generating function of an $\exp(\lambda)$ distribution, then you're good to go.

Note: for this method you need $\lambda>1$.

Answer 2

What's not-elegant in, ($\lambda > 1$),

$$E(e^X) = \int_0^{\infty} e^x\lambda e^{-\lambda x} dx = \frac {\lambda}{\lambda-1}\int_0^{\infty} (\lambda-1) e^{-(\lambda-1) x} dx = \frac {\lambda}{\lambda-1} $$ ?

Seen (comically) narrowly, Mathematical Statistics is a field were people spend their life finding pairs of integrands and domains over which the integrands integrate to unity... for us mortals to use this repository of knowledge.