Finite Gaussian normalization and Wick moments

Let $Z\sim N(0,1)$. For an integer $k\ge 0$, choose an integer $r$ with $2r\ge k$. On $\mathrm{\mid }x\mathrm{\mid }\ge 1$, [F7] gives $e^{x^2\mathrm{/}4}\ge (x^2\mathrm{/}4{)}^r\mathrm{/}r!$, so $\mathrm{\mid }x{\mathrm{\mid }}^k{e}^{-x^2\mathrm{/}2}\le 4^{r}r!e^{-x^2\mathrm{/}4}$. The right side is integrable by [F6], and the integrand is bounded on $[-1,1]$. Since [F2] represents any scalar $N(0,v)$ as $\sqrt{v}Z$ with $\sqrt{v}$ supplied by [F8], every scalar normal has finite absolute moments of every integer order, including the degenerate $v=0$ case.

For fixed $k\ge 1$, define $T_k(u_1,\dots ,u_k):=\mathbb{E}{\prod }_{j=1}^k(u_j^{\mathsf{T}}Q)$. Each projection has a finite $k$th absolute moment by step 1.1 and [F1]. Applying [F5] successively shows the product is integrable. Linearity of each scalar projection and of the integral makes $T_k$ a symmetric $k$-linear form. The law exists under the stated Choice hypothesis [A1]. Since every polynomial is a finite linear combination of coordinate monomials, the triangle inequality also gives every polynomial a finite absolute moment.

On the diagonal, $T_k(u,\dots ,u)=\mathbb{E}[(u^{\mathsf{T}}Q{)}^k]$. Put $v=u^{\mathsf{T}}Cu\ge 0$. By [F1], $u^{\mathsf{T}}Q\sim N(0,v)$, so [F3] gives its characteristic function $e^{-v{t}^2\mathrm{/}2}$. At zero, [F4] gives ${\phi }^{(k)}(0)=i^k\mathbb{E}[(u^{\mathsf{T}}Q{)}^k]$; extracting the $t^k$ coefficient of $e^{-v{t}^2\mathrm{/}2}$ and dividing by $i^k$ yields zero for odd $k$, and for $k=2m$ gives $T_{2m}(u,\dots ,u)=\frac{(2m)!}{2^{m}m!}{v}^m=(2m-1)!!(u^{\mathsf{T}}Cu{)}^m\mathrm{.}$ The identity also holds for $v=0$, when the projected law is the point mass at zero.

For even $k=2m$, define $W_{2m}(u_1,\dots ,u_{2m})$ by the pairing sum in the Statement; for odd $k$, define $W_k=0$. This is a symmetric $k$-linear form. On the diagonal, every pairing contributes $(u^{\mathsf{T}}Cu{)}^m$, and there are $(2m-1)!!$ pairings: pair index $1$ with one of $2m-1$ indices and repeat on the remaining $2m-2$. Hence $W_k(u,\dots ,u)=T_k(u,\dots ,u)$ by step 3.1, also for odd $k$.

For $k\ge 1$, two symmetric $k$-linear forms agreeing on every diagonal agree everywhere. Indeed, for their difference $A$, expand ${\sum }_{\epsilon \in \{-1,1{\}}^k}\left({\prod }_{j=1}^k{\epsilon }_j\right)A\text{\hspace{0.17em}⁣}({\sum }_i{\epsilon }_i{u}_i,\dots ,{\sum }_i{\epsilon }_i{u}_i)\mathrm{.}$ In the multilinear expansion, the sum over signs cancels every term unless each label $1,\dots ,k$ occurs an odd number of times among the $k$ slots. Thus every surviving term uses each label once; symmetry makes all $k!$ such terms equal to $A(u_1,\dots ,u_k)$, and the sign sum is $2^{k}k!A(u_1,\dots ,u_k)$. Since the diagonal of $A=T_k-W_k$ vanishes by step 4.1, this sum is zero and $T_k=W_k$. For $k=0$, the empty product has expectation one and $\mathrm{Pair}(0)$ contains just the empty pairing, whose empty product is one, so the formula also holds.