We know from the law of iterated expectations that $$E[E[X|Y]] = E[X]$$ However, does the same hold true for the square of a conditional expectation? I.e. is the following expression true, $$E[E[X|Y]^2] = E[X]^2$$


No the proposed relation does not hold which is clear from the special case when $X$ and $Y$ are identical, the l.h.s. being then $E(X^2)$.

With $\text{Var}[X \vert Y] := E[X^2 \vert Y ] - E[X \vert Y ]^2$ one can use the following relation $$ \text{Var}(X) = E\{ \text{Var}[X \vert Y] \} + \text{Var}\{E[X \vert Y]\} $$
which is named law of the total variance. By rearranging we find
$$ E\{E[X \vert Y]^2\} = E(X)^2 + \text{Var}\{E[X \vert Y]\}. $$

  • $\begingroup$ I'm having a little trouble deriving the formula for $Var[X|Y] = E[X^2|Y] - E[X|Y]^2$. Particularly, when expanding the squared expression $E[(Y-E(Y|X))^2|X]$, it is required to simplify the term $E[E[Y|X]^2|X]$ to $E[Y|X]^2$. What is the justification for $E[Y|X]^2$ being constant in $X$? $\endgroup$ – shem Mar 21 at 12:41
  • $\begingroup$ Well $E[Y \vert X]^2$ is a function of $X$ so conditioning on $X$ does not change it. It is often useful to regard the conditional expectation as a projection. $\endgroup$ – Yves Mar 21 at 18:41

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