Prove Y = X given $Y = E[X|\mathscr{G}] $ and $EY^2 = EX^2$

Question

Prove Y = X, given $Y = E[X|\mathscr{G}] $ and $EY^2 = EX^2$

Attempt:

Suppose $Y = E[X|\mathscr{G}] $. Then $E[X|\mathscr{G}] $ is $\mathscr{G}$-measureable. For every A $\in \mathscr{G}$:

$\int_A Y dP = \int_A E[X|\mathscr{G}] dP = \int_A X dP $

Since Y is $\mathscr{G}$-measureable and $\int_A Y dP = \int_A X dP $, then Y = X.

Any help would be much appreciated.

Answer 1

Since your functions are in $L^2$ with the norm $‖Z‖^2 := E(Z^2)$, the conditional expectation is orthogonal projection onto $L^2(\mathcal{G})$. That is, $Y$ is orthogonal to $X-Y$.

By orthogonality (i.e. Pythagoras's theorem), $‖X‖^2 = ‖X-Y‖^2 + ‖Y‖^2$. Thus $$‖ Y - X‖^2 = ‖X‖^2 - ‖Y‖^2 = 0$$ So $Y=X$.

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The essential part of the above is orthogonality: instead of appealing to Hilbert space theory, we can prove $EY(X-Y) = 0$ directly: \begin{align} \newcommand{\G}{\mathcal{G}} E(YX) = E(E[X|\G] \ X) &= E\left(E\big(\color{blue}{E[X|\G]} \ X\left|\G\big)\right.\right) && \text{(Tower Law i.e. $EZ = EE[Z|\G]$)} \\ &= E\left(\color{blue}{E[X|\G]}E\big( X\left|\G\big)\right.\right) && \text{($\color{blue}{E[X|\G]}$ is $\G$-measurable)}\\ & = E(E[X|\G]^2)\\ & =:E(Y^2) && \text{(Definition of $Y$)} \end{align} Therefore $$E(X-Y)^2 = EX^2 + EY^2 - 2EXY = EX^2 - EY^2 - 2EY(X-Y) = EX^2 - EY^2 = 0$$ So $Y=X$ again.