Is $E(E(Y|Z)|X)^2 \le E(E(Y|Z)^2|X))$ correct? (Jensen'sinequality)

Question

I know that Jensen's inequality states that: $\varphi(E(V)) \le E(\varphi(V))$, where $E$ stand for expected value, $V$ for a random variable and $\varphi$ must be a convex function.

Let $X$, $Y$ and $Z$ be random variables. If we have $\varphi = E()^2$, and $V = E(Y|Z)$, can I say that $E(E(Y|Z)|X)^2 \le E(E(Y|Z)^2|X))$. Mi concern is about the conditioning on $X$. Does it have implication when applying the Jensen's inequality? Can I directly do that?

Thanks in advance.

Answer 1

Conditional form of Jensen's inequality says $\phi (E(X |\mathcal G)) \leq E(\phi(X)|\mathcal G))$ for any convex function $\phi$ and any r.v. $X$ such that $X$ anf $\phi (X)$ have finite expectation.

There is an easy proof of this based on the fact that $\phi(x)$ can be written as $\sup_i (a_ix+b_i)$ for some constants $a_i,b_i$.

Answer 2

So in the end you want to know if $\mathbb E[V|X]^2\leq \mathbb E[V^2|X]$, this is implied by the conditional version of Jensen inequality. You can find such a statement here