Proving the expected value of the square root of X is less than the square root of the expected value of X

Question

How do I show that $E(\sqrt{X}) \leq \sqrt{E(X)}$ for a positive random variable $X$?

I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 \leq E(X^2)E(Y^2)$, but I'm not sure how.

Answer 1

The Cauchy-Schwarz inequality tells us that if $A=\sqrt{X}$ and $B=1$, then $$E[\sqrt{X}]^2=(E[AB])^2\leq E[A^2]E[B^2]=E[X]$$ so $$E[\sqrt{X}]\leq \sqrt{E[X]}$$

Answer 2

$\sqrt{x}$, $x\geq 0$ is a concave function so Jensen's inequality gives us the result without further effort.

Answer 3

Let us assume a variable x and its probability p. Then we calculate (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.