1

I'm reading this paper that's talking about the norm of the conditional expectation operator.

He starts by defining $\mathbb{E}[\ \dot \ |\mathcal{F'}]: \ L^2(X,\mathcal{F}) \mapsto L^2(X,\mathcal{F'})$ where $\mathcal{F'}$ is a sub-sigma algebra of $\mathcal{F}$.

Then it says: "if f is also in $L^1(X,\mathcal{F})$, then $sgn(\mathbb{E}[\ f \ |\mathcal{F'}])$ is in $L^2(X,\mathcal{F'})$ and $\langle f - \mathbb{E}[\ f \ |\mathcal{F'}], sgn(\mathbb{E}[\ f \ |\mathcal{F'}]) \rangle=0$."

Why does $sgn(\mathbb{E}[\ f \ |\mathcal{F'}]) \in L^2(X,\mathcal{F'})$ follow from $f \in L^1 \cap L^2$?

(the paper can be found here: http://people.maths.ox.ac.uk/greenbj/papers/conditional-expect.pdf).

user202542
  • 851
  • 2
    This does not follow from that. Note that the sign of any random variable is bounded hence square integrable. – Did Aug 27 '16 at 11:11
  • Ok, thank you! Why is it bounded? – user202542 Aug 27 '16 at 11:40
  • 3
    For every random variable $Z$, $|\mathrm{sgn}(Z)|\leqslant1$ almost surely. Seems pretty bounded to me. – Did Aug 27 '16 at 11:47
  • 1
    Right, and since it's a probability spaces bounded impliessquare integrable? This doesn't work if the space isn't of bounded measure? – user202542 Aug 27 '16 at 12:04
  • For any function $f\in L^2(X)$, it is clear that $x\mapsto |\operatorname{sgn}f(x)|^2$ is the same map as $x\mapsto \mathsf 1_{{|f(x)|>0}}$, so $f\in L^2(X)$ iff ${x:|f(x)|>0}$ has finite measure. – Math1000 Sep 09 '16 at 10:38

0 Answers0