# Let X be a random variable with finite expectation, and suppose that E[e^{-X}] =< 1. Prove that E[X] >= P(X >= 2).

Let X be a random variable with finite expectation, and suppose that $E\left[{e}^{-X}\right]\le 1$. Prove that $E\left[X\right]\ge P\left(X\ge 2\right)$.<br<My first thought was to use Markov's inequality, but I can only use that if X is non-negative, and in any case, the resulting inequality doesn't seem very helpful. Even if I assume that $X\ge -c$ and apply it to $Y=X+c$, I'm left to show $P\left(X\ge 2\right)\ge \frac{c}{c+1}$, which doesn't seem like it's always true.
I'm primarily unsure of how to use the $E\left[{e}^{-X}\right]$ condition; Jensen's inequality tells me that the expectation is non-negative but there's presumably more to it.
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Antwan Wiley
Step 1
I'm not entirely sure how the hypothesis about the expectation comes into play, but here's my solution. I'll simply use the Markov inequality for a non-negative r.v. and $a>0$ in the form
$E\left[X\right]\ge aP\left(X\ge a\right)$
with $a=2$. Remembering that probabilities are non-negative and using Markov we have:
$E\left[X\right]\ge P\left(X\ge 2\right)+P\left(X\ge 2\right)\ge P\left(X\ge 2\right)$
I think I saw the light :D I use the extended form of Markov inequality for monotonically increasing function:
Step 2
Theorem
Let $\phi$ be a monotonically increasing function for $x>0$ and let X be a r.v., $a>0,\phi \left(a\right)>0$. Then
$P\left(X\ge a\right)\le \frac{\mathbb{E}\left[\phi \left(X\right)\right]}{\phi \left(a\right)}$
In our case I'll consider the (somehwat artificial) function $\phi \left(x\right):=x+{e}^{-x}-1$, which we can prove satisfy all the hypothesis with $a=2$. Hence, since $1/\phi \left(a\right)\le 1$ and using the bound for ${e}^{-X}$, we have
$\begin{array}{rl}P\left(X\ge a\right)& \le \frac{\mathbb{E}\left[X+{e}^{-X}-1\right]}{\phi \left(a\right)}\\ & \le \mathbb{E}\left[X\right]+\mathbb{E}\left[{e}^{-X}\right]-1\\ & \le \mathbb{E}\left[X\right]\end{array}$