Inequality for nondegenerate random variable

Question

Given X nondegenerate random variable with finite expected value and with $X\geq a\gt0$ show that $$E(\sqrt{X^2-a^2}) \lt \sqrt{E^2(X)-a^2}$$ ($E^2(X)$ = $[E(X)]^2$)

My attempt:

By cauchy-schwarz inequality: $E(\sqrt{X^2-a^2}) \leq \sqrt{E(X^2-a^2)} = \sqrt{E(X^2) -E(a^2)} = \sqrt{E(X^2) - a^2}$

However, I am unable to relate this to the expression I am trying to prove. I'm not sure where to use that X is greater than or equal to a. Any help would be greatly appreciated. Thanks

Answer 1

The approach you've started unfortunately won't work, and as far as I can tell, can't be repaired. The issue is that your Cauchy-Schwarz bound is too generous.

Consider the concrete example where $X$ is $2$ or $3$ with equal probability and $a = 1$. Then \begin{align*} \mathbb E[\sqrt{X^2 - a^2}] &= \frac{1}{2} \sqrt 8 + \frac 1 2 \sqrt 3 \approx 2.28 \\ \sqrt{(\mathbb EX)^2 - a^2)} &= \sqrt{(5/2)^2 - 1} \approx 2.29 \\ \text{but, } \qquad \sqrt{\mathbb E(X^2 - a^2)} &= \sqrt{\frac 1 2 \cdot 8 + \frac 1 2 \cdot 3} \approx 2.35 \end{align*} indicating that while the overall inequality is true (for this particular example), your use of Cauchy-Schwarz overshoots the desired upper bound. Instead, as @WimC suggested, you might consider as an alternate approach applying Jensen's inequality for concave functions.

As for why it matters that $X \geq a$: without that condition, the left-hand side of the original expression may not even be defined. We could technically relax the condition $X \geq a$ a bit to $|X| \geq a$, but the point is that $X \geq a$ is sufficient.

EDIT: As @NCh pointed out in another answer, the condition $X \geq a$ is actually needed in the execution of Jensen's inequality, so it can't be relaxed as I described. (I hadn't thought through the execution to quite that level; apologies for the mistake.)

Answer 2

Look at function $f(x)=\sqrt{x^2-a^2}$. It is concave on $x\geq a$ since $$ \frac{\partial^2}{\partial x^2}f(x)= - \frac{a^2}{(x^2-a^2)^{3/2}}<0. $$ So by Jensen inequality $$\mathbb E\left(\sqrt{X^2-a^2}\right)=\mathbb E(f(X))\leq f(\mathbb E(X))=\sqrt{\bigl(\mathbb E(X)\bigr)^2-a^2}.$$ And the strict inequality follows from the fact that $f(x)$ is strictly concave and $X$ is nondegenerate.

Note that $f(x)$ is not concave on $|x|\geq a$, so if we replace condition $X\geq a$ by $|X|\geq a$ or $X^2\geq a^2$ then inequality fails. Any symmetric r.v. with zero mean can be a counterexample then.