Suppose that $X$ and $Y$ are random variables such that $E(Y¦X) =X$ and $E(Y^2¦X)=X^2$; also, $Y$ is in $L^2(\Omega,\mathcal{A},\mathbb{P})$. I need to show that $Y=X$ almost surely.
I know the definition of conditional expectation as a projection, I. E. $E(X¦Y) =W$ is the unique $W$ in $L^2(\Omega,\sigma(X),\mathbb{P})$ satisfying $E(WZ) =E(YZ) $ for all $Z$ in that same $L^2$.
My intuition is that we get the result by applying this definition and making appropriate choices for $Z$. I've tried, but I got stuck.
I am also worried that I don't know where the $L^2$ should come in. Why not $L^3$ or $L^1$, instead? Whats this have to do with it?
The reason for choosing $L^2$ instead of $L^1$ or $L^3$ is its Hilbert space structure. Also, $E[YZ]$ would not necessarily be well-defined if we merely assume $Z\in L^1$ (the Cauchy-Schwarz inequality is used to prove well-definedness when both are in $L^2$).
As for your question, you obtain by the tower rule $$ E\left((Y-X)^2\right) =E\left(Y^2-2XY+X^2\right) =E\left(E(Y^2\mid X)-2XE(Y\mid X)+X^2\right) =0, $$ so $(Y-X)^2=0$ $\mathbb P$-almost surely.