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It is well known that $E[X|X+Y]$ is Gaussian if both $X$ and $Y$ are, and the result can be derived using standard density arguments. However, how can one prove it by only resulting to optimization arguments in order to argue that $$ \min_{Z \in L^2(\sigma(X+Y))}E[(Z-X)^2] = \min_{Z \in N}E[(Z-X)^2], $$ where $N$ is the affine subspace of $L^2(\sigma(X+Y))$, spanned by Gaussian random-variables?

Intuition/Sketch: Here is what my trail of thought goes like:

• Since the space of Gaussian random-variables is closed under addition, scalar action, a linear subspace of $L^2(X+Y)$. Moreover, since the limit of a sequence of Gaussians in Gaussian, then $N$ is a closed linear subspace of the Hilbert space $L^2(X+Y)$.
• Therefore, the projection $$ P_N:x \mapsto \operatorname{argmin}_{w \in N}E[(w-x)]^2, $$ is well-defined and single-valued.
• Therefore $L^2(X+Y)\cong N \oplus N^{\perp}$, withthe projection on to the first coordinate, given $P_N$,
• The Triangle-inequality then implies that if $Z \in L^2(X+Y)$, then it's first two moments are well-defined and $$ E[(Z-X)^2]\leq E[(P_N(Z)-X)^2] + E[(P_{N^{\perp}}(Z))^2] , $$ with equality holding if and only if $Z \in N$.
• Hence, if $X$ is Gaussian, then so must the minimzer of $E[(\cdot-X)^2]$ be.

It is well known that $E[X|X+Y]$ is Gaussian if both $X$ and $Y$ are, and the result can be derived using standard density arguments. However, how can one prove it by only resulting to optimization arguments in order to argue that $$ \min_{Z \in L^2(\sigma(X+Y))}E[(Z-X)^2] = \min_{Z \in N}E[(Z-X)^2], $$ where $N$ is the affine subspace of $L^2(\sigma(X+Y))$, spanned by Gaussian random-variables?

Intuition/Sketch: Here is what my trail of thought goes like:

• Since the space of Gaussian random-variables is closed under addition, scalar action, a linear subspace of $L^2(X+Y)$. Moreover, since the limit of a sequence of Gaussians in Gaussian, then $N$ is a closed linear subspace of the Hilbert space $L^2(X+Y)$.
• Therefore, the projection $$ P_N:x \mapsto \operatorname{argmin}_{w \in N}E[(w-x)]^2, $$ is well-defined and single-valued.
• Therefore $L^2(X+Y)\cong N \oplus N^{\perp}$, withthe projection on to the first coordinate, given $P_N$,
• The Triangle-inequality then implies that if $Z \in L^2(X+Y)$, then it's first two moments are well-defined and $$ E[(Z-X)^2]\leq E[(P_N(Z)-X)^2] + E[(P_{N^{\perp}}(Z))^2] , $$ with equality holding if and only if $Z \in N$.
• Hence, if $X$ is Gaussian, then so must the minimzer of $E[(\cdot-X)^2]$ be.

However, this argument doesn't really use the properties of $N$, so it feels like something is missing...

It is well known that $E[X|X+Y]$ is Gaussian if both $X$ and $Y$ are, and the result can be derived using standard density arguments. However, how can one prove it by only resulting to optimization arguments in order to argue that $$ \min_{Z \in L^2(\sigma(X+Y))}E[(Z-X)^2] = \min_{Z \in N}E[(Z-X)^2], $$ where $N$ is the affine subspace of $L^2(\sigma(X+Y))$, spanned by Gaussian random-variables?

Intuition/Sketch: Can we argue by the minimal number of moments? That is if $Z$ is not Gaussian, then there is a Gaussian random-variable with the first $2$ moments equal, and the expectation of the other absolute moments only add to the expression; thus they must be zero?

It is well known that $E[X|X+Y]$ is Gaussian if both $X$ and $Y$ are, and the result can be derived using standard density arguments. However, how can one prove it by only resulting to optimization arguments in order to argue that $$ \min_{Z \in L^2(\sigma(X+Y))}E[(Z-X)^2] = \min_{Z \in N}E[(Z-X)^2], $$ where $N$ is the affine subspace of $L^2(\sigma(X+Y))$, spanned by Gaussian random-variables?

Intuition/Sketch: Here is what my trail of thought goes like:

• Since the space of Gaussian random-variables is closed under addition, scalar action, a linear subspace of $L^2(X+Y)$. Moreover, since the limit of a sequence of Gaussians in Gaussian, then $N$ is a closed linear subspace of the Hilbert space $L^2(X+Y)$.
• Therefore, the projection $$ P_N:x \mapsto \operatorname{argmin}_{w \in N}E[(w-x)]^2, $$ is well-defined and single-valued.
• Therefore $L^2(X+Y)\cong N \oplus N^{\perp}$, withthe projection on to the first coordinate, given $P_N$,
• The Triangle-inequality then implies that if $Z \in L^2(X+Y)$, then it's first two moments are well-defined and $$ E[(Z-X)^2]\leq E[(P_N(Z)-X)^2] + E[(P_{N^{\perp}}(Z))^2] , $$ with equality holding if and only if $Z \in N$.
• Hence, if $X$ is Gaussian, then so must the minimzer of $E[(\cdot-X)^2]$ be.

It is well known that $E[X|X+Y]$ is Gaussian if both $X$ and $Y$ are, and the result can be derived using standard density arguments. However, how can one prove it by only resulting to optimization arguments in order to argue that $$ \min_{Z \in L^2(\sigma(X+Y))}E[(Z-X)^2] = \min_{Z \in N}E[(Z-X)^2], $$ where $N$ is the affine subspace of $L^2(\sigma(X+Y))$, spanned by Gaussian random-variables?

It is well known that $E[X|X+Y]$ is Gaussian if both $X$ and $Y$ are, and the result can be derived using standard density arguments. However, how can one prove it by only resulting to optimization arguments in order to argue that $$ \min_{Z \in L^2(\sigma(X+Y))}E[(Z-X)^2] = \min_{Z \in N}E[(Z-X)^2], $$ where $N$ is the affine subspace of $L^2(\sigma(X+Y))$, spanned by Gaussian random-variables?

Intuition/Sketch: Can we argue by the minimal number of moments? That is if $Z$ is not Gaussian, then there is a Gaussian random-variable with the first $2$ moments equal, and the expectation of the other absolute moments only add to the expression; thus they must be zero?