Let $E[X\vert\mathcal{G}]= Y$
Why does it follow from
$$ \begin{align} E[(X-Y)^2\mid \mathcal{G}]&=E[X^2-2XY+Y^2\mid \mathcal{G}]\\ &=E(Y^2\mid \mathcal{G})-X^2 \end{align} $$
that the tower law implies that $$ E(X-Y)^2=EY^2-EX^2 $$ ?
2026-04-01 18:52:43.1775069563
Why does Tower Law imply that $E[(X-Y)^{2}]=E[X^{2}]-E[Y^{2}]$
93 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in PROBABILITY
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