# Show that a random variable Y is infinite

I have the following question where I am required to show that a random variable is almost surely infinite:

Let $$Y$$ be a non-negative random variable, given that $$E[e^{-Y}]=0,$$ show that $$Y$$ is almost surely infinite.

I know that "$$Y$$ is almost surely infinite" is equivalent to showing that $$P[Y=\infty] = 1.$$

I have attempted to use the Markov Inequality $$P[e^{-Y}\geq\epsilon] \leq \frac{E[|e^{-Y}|^r]}{\epsilon^r},$$ we can pick $$r=1$$ and we know $$e^{-Y}>0$$ giving $$P[e^{-Y}\geq\epsilon] \leq \frac{E[e^{-Y}]}{\epsilon} = 0 \implies P[e^{-Y}\geq\epsilon] = 0.$$ I think that the next step involves showing that $$P[Y>k] = 1,$$ however I am unsure how to do so rigorously, and where I would go from here.

I would really appreciate any help steering me in the right direction. Thanks

Without using Markov's inequality, you can proceed by writing $$\mathbb{P}[Y= \infty] = \mathbb{P}(\bigcap_{k=1}^\infty \{Y > k\}) = \lim_{k \to \infty} \mathbb{P}(Y > k)$$
This implies that if $$\mathbb{P}(Y = \infty) < 1$$ then there exists a $$k$$ such that $$\mathbb{P}(Y>k) < 1$$. Equivalently, for this $$k$$, $$\mathbb{P}(Y \leq k) > 0$$. But then, since $$e^{-Y} \geq 0$$ and $$x \mapsto e^{-x}$$ is decreasing, we can write $$0 < e^{-k} \mathbb{P}(Y \leq k) \leq \mathbb{E}[e^{-Y}] = 0$$ which is a contradiction.
• Thanks, guess I was barking up the wrong tree. Forgive me if this is a stupid question, can you explain why $e^{-k}P[Y\leq k] \leq E[e^{-Y}]$ ? Apr 29, 2020 at 19:48
• Sure! You can write $$E[e^{-Y}] = E[e^{-Y}1_{\{Y \leq k\}}] + E[e^{-Y}1_{\{Y > k\}}]$$ The second term is non-negative and the first term satisfies $$E[e^{-Y} 1_{\{Y \leq k\}}] \geq e^{-k} P[Y \leq k]$$ since if $Y \leq k$ then $e^{-Y} \geq e^{-k}$. Apr 29, 2020 at 20:53