Gaussian covariance for positive quadratic forms and a boundary counterexample

We construct $R$ with $K=R^{\mathsf{T}}R$ by induction on $n$. For $n=1$, take $R=\sqrt{K_{11}}>0$. For $n>1$, write $K=\left(\begin{array}{cc}a& b^{\mathsf{T}}\\ b& C\end{array}\right)$; positive definiteness gives $a>0$ and $S:=C-b{b}^{\mathsf{T}}\mathrm{/}a$ positive definite, since $y^{\mathsf{T}}Sy=(-b^{\mathsf{T}}y\mathrm{/}a,y{)}^{\mathsf{T}}K(-b^{\mathsf{T}}y\mathrm{/}a,y)>0$ for $y\mathrm{\ne }0$. By induction $S=R_1^{\mathsf{T}}{R}_1$; then $R=\left(\begin{array}{cc}\sqrt{a}& b^{\mathsf{T}}\mathrm{/}\sqrt{a}\\ 0& R_1\end{array}\right)$ satisfies $R^{\mathsf{T}}R=K$. Its diagonal entries are positive, so $\det R>0$; taking determinants gives $\det K=(\det R{)}^2$ and therefore $\det R=\sqrt{\det K}$.

Put $T(x)=Rx$. This is a $C^1$ diffeomorphism of ${\mathbb{R}}^n$ with $\mathrm{\mid }\det DT\mathrm{\mid }=\det R$. Since $H(x)=\mathrm{\mid }Rx{\mathrm{\mid }}^2\mathrm{/}2$, [F2] gives $Z=(\det R{)}^{-1}{\int }_{{\mathbb{R}}^n}{e}^{-\beta \mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}2}d{\lambda }_n(y)\mathrm{.}$ By [F3], ${\lambda }_n$ is the finite product of $n$ sigma-finite one-dimensional Lebesgue measures; Tonelli and the factorization $e^{-\beta \mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}2}={\prod }_i{e}^{-\beta y_i^2\mathrm{/}2}$ give the integral as $I_0(\beta \mathrm{/}2{)}^n$. By [F1] and [F4], $I_0(\beta \mathrm{/}2)=\sqrt{2\pi \mathrm{/}\beta }$. Hence $Z=(2\pi \mathrm{/}\beta {)}^{n\mathrm{/}2}(\det K{)}^{-1\mathrm{/}2},$ which is finite and positive.

Let $Y$ have the normalized density proportional to $e^{-\beta \mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}2}$ with respect to ${\lambda }_n$. For each $i$, [F3] and Tonelli factor the nonnegative second-moment integral into $I_1(\beta \mathrm{/}2)I_0(\beta \mathrm{/}2{)}^{n-1}$, while the normalizer is $I_0(\beta \mathrm{/}2{)}^n$; [F1] gives $\mathbb{E}[Y_i^2]=I_1(\beta \mathrm{/}2)\mathrm{/}{I}_0(\beta \mathrm{/}2)=1\mathrm{/}\beta$. Also $\int \mathrm{\mid }{y}_i\mathrm{\mid }{e}^{-\beta \mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}2}d{\lambda }_n<\mathrm{\infty }$, since $\mathrm{\mid }{y}_i\mathrm{\mid }\le 1+y_i^2$ and the same Tonelli factorization makes both integrals finite. For $i\mathrm{\ne }j$, $\mathrm{\mid }{y}_i{y}_j\mathrm{\mid }\le (y_i^2+y_j^2)\mathrm{/}2$, so the cross moments are absolutely integrable. Fubini [F3] factors these signed moments; the one-dimensional first and odd moments vanish by [F1]. Thus $\mathbb{E}[Y]=0$ and $\mathbb{E}[Y{Y}^{\mathsf{T}}]={\beta }^{-1}I$.

By [F2], the normalized law of $X$ is the pullback of the law of $Y$ under $x=R^{-1}y$. The absolute integrability established in step 3.1 permits the same change of variables for signed first and second moments by applying [F2] to their positive and negative parts. Hence $\mathbb{E}[X]=0$ and $\mathbb{E}[X{X}^{\mathsf{T}}]=R^{-1}({\beta }^{-1}I)R^{-\mathsf{T}}={\beta }^{-1}(R^{\mathsf{T}}R{)}^{-1}={\beta }^{-1}{K}^{-1}\mathrm{.}$ Since $H(X)=\mathrm{\mid }Y{\mathrm{\mid }}^2\mathrm{/}2$, summing the diagonal second moments gives $\mathbb{E}[H]=n\mathrm{/}(2\beta )$.

For the stated boundary counterexample, the density on $[-L,L]$ is the constant $(2L{)}^{-1}$ because $H=0$ and the normalizing integral is $2L$ by [F3]. Since $H^{\mathrm{\prime }}(x)=0$ everywhere, $\mathbb{E}[x{H}^{\mathrm{\prime }}(x)]=0\mathrm{\ne }{\beta }^{-1}$ for $\beta >0$.