# Conditional variance definition

This question relates to this one: Understanding a Substep of the Proof for the Law of Total Variance

It is written there, that we should be able to define conditional variance as $$Var[X|\mathcal{F}]=E[X^2|\mathcal{F}]-(E[X|\mathcal{F}])^2$$. But in order to pass to this definition from usual one, i.e. $$Var[X|\mathcal{F}]=E[(X-E[X|\mathcal{F}])^2|\mathcal{F}]$$ I need to know that $$E[XE[X|\mathcal{F}]|\mathcal{F}]=(E[X|\mathcal{F}])^2.$$ Seems plausible, but how I can see that?

$$E(XY|\mathcal F)=YE(X|\mathcal F)$$ whenever $$Y$$ is measurable w.r.t. $$\mathcal F$$ and $$E|XY| <\infty$$. [Here $$X$$ and $$Y$$ have finite second moments so $$E|XY| \leq \sqrt {EX^{2}} \sqrt {EY^{2}} <\infty$$]. Take $$Y=E(X|\mathcal F)$$ in this formula.
• Shouldn't $Y$ be bounded? – Igor Sikora Aug 22 '19 at 12:30
• It is enough if $E|XY| <\infty$. $X$ and $Y$ have finite second moments: $EY^{2} \leq EX^{2}$ by conditional Jensen's in equality. – Kavi Rama Murthy Aug 22 '19 at 12:34