Is this true: $\lim_{\lambda \rightarrow 0}E\left[ e^{-\lambda X} \right] = P\left(X < \infty \right)$?

Question

Is it true that $\lim_{\lambda \rightarrow 0} E\left[ e^{-\lambda X} \right] = P\left(X < \infty \right)$ ?

I've seen this in a few places online but I can't seem to be able to find a proof online / be able to prove it on my own.

Intuitively it seems to make sense, since if $X=\infty$, no matter how small we take $\lambda$ to be, $e^{-\lambda X} = 0$.

Anyways if any of you out there can point me to the right place or let me know how to show this (if it is indeed true) let me know!

(I saw it somewhere in the comments on this page.)

(Edit: forgot to write $\lim_{\lambda \rightarrow 0}$)

Answer 1

Not super rigorous, but it makes sense:

$e^{-\lambda x} = 1+O(\lambda)$ for any fixed $x$ as $\lambda \to 0$

$E[e^{-\lambda X}]=E[e^{-\lambda X}|X<\infty]P(X<\infty)+E[e^{-\lambda X}|X=\infty]P(X=\infty)$

$=E[1+O(\lambda)|X<\infty]P(X<\infty)=(1+O(\lambda))P(X<\infty)\to P(X<\infty)$

Answer 2

If $X$ is a discrete random variable with $P(X=1)=1$, then $$\mathbb{E}e^{-\lambda X} = e^{-\lambda},$$ which is not equal to $P(X<\infty)=1$, as long as $\lambda \neq 0$. Hence, it's not true in general.