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I asked this over on cross validated, but thought it might also get an answer here:

The law of the conditional Gaussian distribution (the mean and covariance) are frequently mentioned to extend to the separable Hilbert spaced valued case, i.e., for $(X,Y)$, $$ \mu_{X|Y=y} = \mu_X - C_{XY}C_{Y}^{-1}(\mu_Y - y) $$ and $$ C_{X|Y=y} = C_{X} - C_{XY}C_Y^{-1}C_{YX} $$

I was trying to trace a proof for this in the separable Hilbert space case, and all the papers I found tended to point to Linear Estimators and Measurable Linear Transformations on a Hilbert Space by A. Mandelbaum (1984). Digging through that paper, there's one a part of the proof I'm stumbling on: enter image description here

I'm struggling with that first equality in (3.6), where the conditional expectation becomes the summation. Thanks for any help. Alternatively, if someone has a reference to another (better?) proof, let me know.

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$\newcommand\Si\Sigma\newcommand\X{\mathbf X}$If $Y,X_1,\dots,X_n$ are jointly normal zero-mean (real-valued) random variables, then $$E(Y|X_1,\dots,X_n)=\Si_{12}\Si_{22}^{-1}\X,$$ where $\X:=[X_1,\dots,X_n]^\top$, $\Si_{22}:=Cov\,\X$ (the covariance matrix of $\X$), and $\Si_{12}:=Cov(Y,\X)=[Cov(Y,X_1),\dots,Cov(Y,X_n)]=[EYX_1,\dots,EYX_n]$. If $X_1,\dots,X_n$ are independent, then $\Si_{22}$ is the diagonal matrix with diagonal entries $EX_1^2,\dots,EX_n^2$.

Applying these observations to $Y=(\theta,h)$ and $X_i=(X,e_i)$ for $i=1,\dots,n$, we get the equality in question.

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  • $\begingroup$ This seems to assume that you know $(Y, X_1,\ldots, X_n)$ are jointly Gaussian; is that obvious? $\endgroup$ Jan 23 at 1:05
  • $\begingroup$ @user2379888 : In Theorem 2 of Mandelbaum's paper, $\theta$ and $X$ are assumed to be jointly Gaussian (and zero-mean), which implies that $Y,X_1,\dots,X_n$ are jointly normal. Such an implication follows, in particular, from the third sentence of Section 3.3 of the paper. $\endgroup$ Jan 23 at 3:41

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