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!

  • $\begingroup$ 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 $$E[(X-E[X])^2]=E[(X-E[X\mid\mathcal G])^2]+E[(E[X\mid\mathcal G]-E[X])^2].$$ $\endgroup$
    – Did
    Mar 12, 2016 at 14:34
  • $\begingroup$ Got it! It is similar to the proof in Hilbert Space though..... And sure, I would avoid posting so many questions in one post! $\endgroup$
    – Simo
    Mar 12, 2016 at 15:08

1 Answer 1


The key is to utilize to definition of conditional expectation and its properties, which is based on your thorough understanding of the difference between conditional expectation in advanced probability (based on measure theory) and that in elementary probability (based on calculus).

The solution to the first question.

The solution to the second question.


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