Proof of Conditional Expectation for Jointly Gaussian Random Variables is Affine

Fix a Gaussian representation of $(X,U_1,\dots,U_r)$: $m$ (zero or a natural number), independent standard normal $Z_1,\dots,Z_m$, real numbers $\mu,\nu_1,\dots,\nu_r$, and, when $m\ge1$, coefficient vectors $x,u_1,\dots,u_r\in\mathbb{R}^{m}$ with

writing $v\cdot Z=\sum_{j=1}^{m}v_jZ_j$ (an empty sum, equal to $0$, when $m=0$).

Step 1: Choice of the coefficients. If $m=0$ or every $u_k$ is the zero vector, set $\beta_k=0$ for $1\le k\le r$ and $\beta_0=\mu$. Otherwise apply Gram-Schmidt orthonormalization to $u_1,\dots,u_r$, obtaining $s\ge1$ and an orthonormal family $e_1,\dots,e_s$ in $\mathbb{R}^{m}$ with the expansion property $u_k=\sum_{u=1}^{s}(u_k\cdot e_u)\,e_u$ and the span property $e_u=\sum_{k=1}^{r}\gamma_{uk}\,u_k$ for real numbers $\gamma_{uk}$; set, with the dot product,

In every case define $Y=\beta_0+\sum_{k=1}^{r}\beta_kU_k$. Then $Y$ is measurable with respect to $\sigma(U_1,\dots,U_r)$: the generators $U_k$ are measurable by Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, constants are measurable with respect to every $\sigma$-algebra, and sums and scalar multiples preserve measurability by the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product. Moreover $(U_1,\dots,U_r)$ is a Gaussian random vector (subfamily clause of Affine Transformations of Gaussian Random Vectors are Gaussian), so $(Y)$ is a Gaussian random vector by Affine Transformations of Gaussian Random Vectors are Gaussian, and $Y$ is square-integrable by Claim 1 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector.

Step 2: The residual and its representation. The tuple $(X-Y,U_1,\dots,U_r)$ is a Gaussian random vector, being an affine transformation of $(X,U_1,\dots,U_r)$ by Affine Transformations of Gaussian Random Vectors are Gaussian (with $X-Y=-\beta_0+X-\sum_k\beta_kU_k$). In the main case of Step 1, the span identity gives, as an identity of vectors in $\mathbb{R}^{m}$,

set $w=x-\sum_{u=1}^{s}c_u\,e_u$ (and $w=x$ in the case that every $u_k$ is zero with $m\ge1$). Off the union of the $r+1$ defining events' complements, which has probability $0$ by the countable additivity and monotonicity of the measure $P$, substituting the representation into $X-Y$ and rearranging finite sums gives

using the definition of $\beta_0$. Hence $\bigl(m,(0,\nu_1,\dots,\nu_r),(w,u_1,\dots,u_r),(Z_j)\bigr)$ is a Gaussian representation of $(X-Y,U_1,\dots,U_r)$; for $m=0$ this reads $P(X-Y=0)=1$ with all coefficient data empty.

Step 3: Moments of the residual. In the main case, orthonormality gives $w\cdot e_u=x\cdot e_u-c_u=0$ for every $u$, and then the expansion property gives

the same conclusion is trivial when every $u_k$ is zero or $m=0$. By Claim 2 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector applied to the representation of Step 2, $\mathbb{E}[X-Y]=0$ and, with the covariance, $\operatorname{Cov}(X-Y,U_k)=w\cdot u_k=0$ for every $k$. Since $(X-Y)$ is a Gaussian random vector (subfamily clause of Affine Transformations of Gaussian Random Vectors are Gaussian), this proves parts 2 and 3.

Step 4: Independence of the residual. Parts 2 and 3 verify the hypotheses of Uncorrelated Jointly Gaussian Blocks are Independent for the Gaussian random vector $(X-Y,U_1,\dots,U_r)$ with the blocks $(X-Y)$ and $(U_1,\dots,U_r)$; hence $\sigma(X-Y)$ and $\sigma(U_1,\dots,U_r)$ are independent, proving part 4.

Step 5: The averaging property. Write $\mathcal{G}=\sigma(U_1,\dots,U_r)$ and let $A\in\mathcal{G}$. The random variable $X$ is square-integrable by Claim 1 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector, so $X-Y$ is square-integrable, hence integrable, by Square-Integrable Random Variables and the Mean-Square Inner Product. The indicator $\mathbf{1}_{A}$ satisfies $\sigma(\mathbf{1}_{A})\subseteq\mathcal{G}$, since its preimages of Borel sets are among $\emptyset$, $A$, $\Omega\setminus A$, $\Omega$, all in $\mathcal{G}$. By part 4, every event of $\sigma(X-Y)$ multiplies with every event of $\mathcal{G}$, hence with every event of $\sigma(\mathbf{1}_{A})$; thus $X-Y$ and $\mathbf{1}_{A}$ are independent random variables (using the identification of independence of random variables with independence of their generated $\sigma$-algebras recorded in Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras). Both are integrable ($\mathbf{1}_{A}$ is bounded), so by Expectation of a Product of Independent Random Variables and Step 3,

and the linearity of expectation from Linearity and Monotonicity of the Lebesgue Integral (the products $X\mathbf{1}_{A}$ and $Y\mathbf{1}_{A}$ being integrable by Square-Integrable Random Variables and the Mean-Square Inner Product) gives $\mathbb{E}[X\mathbf{1}_{A}]=\mathbb{E}[Y\mathbf{1}_{A}]$. Together with the measurability and square-integrability of $Y$ from Step 1, $Y$ satisfies conditions (i)-(iii) of Conditional Expectation of a Square-Integrable Random Variable, so $Y$ is a conditional expectation of $X$ given $\mathcal{G}$. The final clause of part 1 is the uniqueness assertion of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables combined with the notational convention of Conditional Expectation of a Square-Integrable Random Variable. $\blacksquare$