TITLE: Is it possible to find a better bound for this probability?
QUESTION [1 upvotes]: In my homework I've come across the problem of:
Suppose $\mathbb{E}(2^X) = 4$. Prove $\mathbb{P}(X\geq 3) \leq \frac{1}{2}$.
What I did was note that by Jensen's inequality we have that $\mathbb{E}(2^X) \geq 2^{\mathbb{E}(X)} \implies 4 \geq 2 \implies \mathbb{E}(X) \leq 2$.
Then by Markov's inequality I also noted that $3 \mathbb{P}(x \geq 3) \leq 2 \implies \mathbb{P}(X\geq3)\leq \frac{2}{3}$.
But this only shows the inequality for 2/3, and the question asked for 1/2. I don't need a full solution, but do you think it was a typo and 2/3 is the best bound? Or do I still have to find a better method to crunch this down to 1/2?
Thanks

REPLY [2 votes]: We have 
\begin{align}
\mathbb{P}(X \geq 3) &=\mathbb{P}(2^X\geq 2^3)\\
&\leq \frac{ E ( 2^X  ) }{2^3}\\
&=\frac{1}{2}
\end{align}

REPLY [1 votes]: $$4=E\left[2^X\right]\geq P(X<3)\cdot 0+P(X\geq 3)\cdot 2^3=P(X\geq 3)\cdot 2^3\implies P(X\geq 3)\leq\frac{4}{2^3}=\frac{1}{2}$$
Where we used the fact that $2^x\geq 0$ and is increasing.
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