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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?

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$E(XY|\mathcal F)=YE(X|\mathcal F)$ 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.

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    $\begingroup$ Shouldn't $Y$ be bounded? $\endgroup$ – Igor Sikora Aug 22 '19 at 12:30
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Aug 22 '19 at 12:34
  • $\begingroup$ Ok, I think I can see that - the proof should go by classic complication of variable, right? (Characteristic function of a set, then simple function, monotone convergence...). But why do we need the finiteness condition? $\endgroup$ – Igor Sikora Aug 22 '19 at 12:42
  • $\begingroup$ Is it because we need both variables to have expectation and variance? $\endgroup$ – Igor Sikora Aug 22 '19 at 12:43
  • $\begingroup$ In fact, suing your approach we can write down version of montone convergence theorem, convergence theorem etc for conditional expectations. $\endgroup$ – Kavi Rama Murthy Aug 22 '19 at 12:44

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