proving that $Var(E[X|G]) \le Var(X)$ before having defined conditional variance

Question

In my class I am tasked to prove that given any $X,Y$ r.v. $Var(E[X|Y]) \le Var(X)$ with only the properties of the conditional expected value.

I looked up the proof for the total variance law but I don't want to use it since we have not yet defined conditional variance.

my attempt to prove this goes as follows:

$$0 \le Var(X) = E(X^2) - E(X)^2 = E(E(X^2|Y)) - E(E(X|Y))^2 = Var(E(X|Y)) + E(E(X^2|Y)) - E(E(X|Y)^2) $$

so this reduces to prove that $E(E(X^2|Y)) \ge E(E(X|Y)^2)$.

To prove this it seemed simple at first, call $G$ the sigma algebra on which $Y$ is defined

then $$E[E[X^2 | G]1_{G_1}] = E[X^21_{G_1}] \ge E[X1_{G_1}]^2 = E[E[X | G]1_{G_1}]^2 \quad \forall{G_{1}} \in G$$

But this is not what I want even if it's close. How could I adjust my reasoning?

Any other paths to the solution are welcome.

Answer 1

You need the conditional form of Jenson's inequality:

$$f(E[X|Y])\leq E[f(X)|Y],$$

whenever $f$ is convex. You can find a proof here.