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


1 Answer 1


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.

  • $\begingroup$ Thanks, guess I was barking up the wrong tree. Forgive me if this is a stupid question, can you explain why $e^{-k}P[Y\leq k] \leq E[e^{-Y}]$ ? $\endgroup$
    – Oliver
    Apr 29, 2020 at 19:48
  • $\begingroup$ 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 $e^{-Y} \geq e^{-k}$. $\endgroup$ Apr 29, 2020 at 20:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .