# Conditional Expectation.

1. Let $X$ be a random variable in $L^2(\Omega, \Sigma, P)$ and $\mathcal G$ a sub-$\sigma$-algebra of $\Sigma$.
Prove that $E[(X-E[X\mid\mathcal G])^2] \le E[(X-E[X])^2]$.

As conditional expectation can be viewed as a projection of $X$ from $L^2(\Sigma)$ into a subspace $L^2(\mathcal G)$, the norm $E[(\bullet)^2]$ would be smaller. And the inequality is then obvious. However, this is rather intuitive. I have some trouble in giving a rigorous proof. Can anyone help?

Also I am wondering if $E[|X-E[X\mid\mathcal G]|] \le E[|X-E[X]|]$. This seems true from the above reasoning?

1. Let $T$ and $S$ be stopping times and $\mathcal F_T$ and $\mathcal F_S$ be the stopping time $\sigma$-algebra.
Prove that $E[E[Y\mid\mathcal F_S]\mid\mathcal F_T] = E[E[Y\mid\mathcal F_T]\mid\mathcal F_S] = E[Y\mid\mathcal F_{T\land S}]$.

It seems obvious that the smaller stopping time is the one that matters. But still, I am not able to give the proof.

Thanks a lot!

• Three different questions in the same post are something one should avoid... Re the first question, note that the decomposition $$X-E[X]=X-E[X\mid\mathcal G]+E[X\mid\mathcal G]-E[X]$$ together with the definition of $E[X\mid\mathcal G]$ implies that $$E[(X-E[X])^2]=E[(X-E[X\mid\mathcal G])^2]+E[(E[X\mid\mathcal G]-E[X])^2].$$
– Did
Mar 12, 2016 at 14:34
• Got it! It is similar to the proof in Hilbert Space though..... And sure, I would avoid posting so many questions in one post!
– Simo
Mar 12, 2016 at 15:08