Give upper bound for random variable Problem: Suppose $X$ is a random variable such that $E[2^X] = 4$. Give an upper bound for $P(X ≥ 3)$. Justify your answer.
Attempt: I know the following equations 
$P(X \ge 3) \le \frac{E[X]}{3}$ 
$E[g(X)] = \int_{-\infty}^{\infty}{g(x)f_x(x)dx} = 4$ where $g(x)=2^x$ 
$E[X] = \int_{-\infty}^{\infty}{xf_x(x)dx}$ 
I thought what I could do is take the second equation and solve for $f_x(x)$ and then use that in the third equation to get $E[X]$, but I don't know how to solve for $f_x(x)$. Is there another way to solve this?
If this is the correct path, how can I solve for $f_x(x)$?
 A: Define a new random variable, $Y$, where $Y=2^X$. Solving for $X$, we get $X=\frac{\ln Y}{\ln 2}$. Then 
$$P(X\geq 3)=P(\frac{\ln Y}{\ln 2} \geq 3) = P(\ln Y \geq 3\ln2) = P(Y \geq \exp\{3\ln2\}).$$
Applying your first inequality, which only applies to non-negative random variables ($Y=2^X$ satisfies this), we get 
$$P(Y \geq \exp\{3\ln2\}) \leq \frac{E(Y)}{\exp\{3\ln2\}}.$$
Substituting in our given value of $E(Y)$, we get.
$$P(X\geq 3) \leq \frac{4}{\exp\{3\ln2\}}=\frac{1}{2}.$$
A: @user240935 has already given a direct proof (+1) of the fact that $P(X\geq 3) \leq \frac 12$ so here is an alternative way (proof by contradiction) of looking at the matter.
Suppose that $P(X \geq 3) > \frac 12$, that is, the total probability mass at $3$ or to the right of $3$ is greater than $\frac 12$. Now, $Y=2^X$ is a positive random variable regardless of the distribution of $X$, and in the distribution of $Y = 2^X$, the total probability mass at $8$ or to the right of $8$ is more than $\frac 12$. Now, $E[Y]$ equals the total moment of the probability mass of $Y$ about the origin, and since we know that the mass at $8$ or to the right of $8$ is more than $\frac 12$, we can can conclude that just this mass alone contributes at least $8\times P(Y\geq 3)$ to the total moment -- exactly $8\times P(Y\geq 3)$ if all the probability mass is sitting (as an atom) at $8$ and more if the mass is spread out beyond $8$. But since $P(X \geq 3) > \frac 12$ by assumption, this contribution to the total moment $E[Y]$ is at least $8\times P(X \geq 3) > 4$ and so $E[Y]$ must be even larger, contradicting the known result that $E[Y]=E[2^X] = 4$. So, our assumption that $P(X \geq 3) > \frac 12$. is incorrect and it must be that $P(X \geq 3) \leq \frac 12$.
