Suppose $E(2^X)=4$. Prove that $P(X \ge 3) \le {1 \over 2}$.

Question

Suppose $E(2^X)=4$. Prove that $P(X \ge 3) \le {1 \over 2}$.

I thought that this question could be solved by using Jensen's inequality and Markov's inequality like below...

$E(2^X)=4 \ge 2^{E(X)}$ which implying $2 \ge E(X)$

$P(X \ge 3) \le {E(X) \over 3} \le {2 \over 3}$

How I can get $P(X \ge 3) \le {1 \over 2}$ from this?? Thank you.

Answer 1

There's no need for such powerful tools. If $X \ge 3$, then how large must $2^X$ be?

Answer 2

Here is a solution using Markov's inequality (https://en.wikipedia.org/wiki/Markov%27s_inequality).

Because $\mathbb E(X) \leqslant \mathbb (X|X \geqslant 0)$ (see Remark below), we can assume WLOG that $X \geqslant 0$. This is important because Markov's inequality is only valid for a positive random variable.

Set $Y=2^X \iff X=\log_2(Y)$.

Thus we have

$$\tag{1}\mathbb E(Y)=4$$

Then : $P(X \geqslant 3)=P(\log_2(Y)\geqslant\log_2(2^3))=P(Y \geqslant 2^3)$

(because $\log_2$ is an increasing function, therefore preserves order)

By Markov inequality, $P(Y \geqslant a)\leqslant \dfrac{\mathbb E(Y)}{a}$. Taking $a=9$ and using (1) gives

$$P(X \geqslant 3)=P(Y \geqslant 8) \leqslant \dfrac{4}{8}=\dfrac{1}{2},$$

the desired result.

Remark: Proof of the fact that $\mathbb E(X)\leqslant \mathbb E(X|X \geqslant 0)$

$\mathbb E(X)=\underbrace{\mathbb E(X|X<0)P(X<0)}_{\leqslant 0}+\mathbb E(X|X \geqslant 0)\underbrace{P(X \geqslant 0)}_{\leqslant 1}$ whence the result.

Answer 3

Since $2^x$ is $(1)$ positive for any $x\in \mathbb R$ and $(2)$ increasing for $x\ge 3$, we have that \begin{align}4&=\mathbb E\left(2^X\right)=\mathbb E\left(2^X\mid X<3\right)P(X<3)+\mathbb E\left(2^X\mid X\ge 3\right)P(X\ge 3)\\[0.2cm]^{(1)}&\ge \mathbb E \left(2^X\mid X\ge 3\right)P(X\ge 3)\\[0.2cm]^{(2)}&\ge2^3P(X\ge3)\end{align}