Gaussian entry moments and triangular independence

Fix an integer $k\ge 0$. By [F4], $t^k{e}^{-t\mathrm{/}2}$ is bounded for $t\ge 1$. Since $t^2\ge 2t$ for $t\ge 2$, $t^k{e}^{-t^2\mathrm{/}2}\le (t^k{e}^{-t\mathrm{/}2})e^{-t\mathrm{/}2}\le C_k{e}^{-t\mathrm{/}2}$ there for some finite $C_k$. The same bound holds with $t=\mathrm{\mid }x\mathrm{\mid }$. On each interval $[n,n+1]$, $n\ge 2$, monotonicity and the addition law bound $e^{-x\mathrm{/}2}$ by $e^{-n\mathrm{/}2}$; the resulting geometric series converges. The integrand is continuous, hence bounded and Riemann integrable on every compact interval, including $[-2,2]$, by [F3] and [F9]. Applying [F5] to the successive compact intervals shows that ${\int }_{\mathbb{R}}\mathrm{\mid }x{\mathrm{\mid }}^k\varphi (x)dx<\mathrm{\infty }$. The nonnegative case of [F2] therefore makes $Z^k$ integrable. The same tail estimate also gives $R^k\varphi (R)\to 0$ as $R\to \mathrm{\infty }$.

Put $I_k={\int }_{\mathbb{R}}{x}^k\varphi (x)dx$. By [F2] and step 1.1, $I_k=\mathbb{E}{Z}^k$. For odd $k$ and $R>0$, $f(x)=x^k\varphi (x)$ is continuous, hence bounded and Riemann integrable on $[-R,0]$, by [F3] and [F9]. The reflection $r:[0,R]\to [-R,0]$, $r(t)=-t$, is a decreasing surjection. Its difference quotient is $-1$, including in the one-sided endpoint limits, so its derivative is the Riemann-integrable constant $-1$ and $\mathrm{\mid }{r}^{\mathrm{\prime }}(t)\mathrm{\mid }=1$. Applying [F10] and using $\varphi (-t)=\varphi (t)$ gives ${\int }_{-R}^0{x}^k\varphi (x)dx={\int }_0^R(-t{)}^k\varphi (-t)dt=-{\int }_0^R{t}^k\varphi (t)dt$. Thus the two half-interval integrals cancel. By [F5] the compact-interval Riemann and Lebesgue integrals agree; monotone convergence for the positive and negative parts, together with absolute integrability from step 1.1, gives $I_k=0$. Also $I_0=1$ by [F1].

For $k\ge 2$ and $R>0$, the polynomial rule in [F3] gives $h^{\mathrm{\prime }}(x)=-x$ for $h(x)=-x^2\mathrm{/}2$ and $u^{\mathrm{\prime }}(x)=(k-1)x^{k-2}$ for $u(x)=x^{k-1}$. The chain and scalar-multiple rules in [F3], on the full real inner and outer domains, give ${\varphi }^{\mathrm{\prime }}(x)=-x\varphi (x)$. Thus $v(x)=-\varphi (x)$ has $v^{\mathrm{\prime }}(x)=x\varphi (x)$. These derivatives are continuous by [F3] and [F9], so [F3] applies on $[-R,R]$. This gives ${\int }_{-R}^R{x}^k\varphi (x)dx=[-x^{k-1}\varphi (x){]}_{-R}^R+(k-1){\int }_{-R}^R{x}^{k-2}\varphi (x)dx$. Both boundary terms vanish as $R\to \mathrm{\infty }$ by step 1.1. The compact-interval integrals equal their Lebesgue counterparts by [F5], and absolute integrability permits passage to the whole line, separately for positive and negative parts. Hence $I_k=(k-1)I_{k-2}$. The base case is $I_2=I_0=1$. For $j\ge 1$, the recurrence gives $I_{2j+2}=(2j+1)I_{2j}$, so [F8], applied to the property at $j=m+1$ for $m\ge 0$, gives $I_{2j}=(2j-1)!!$ for every $j\ge 1$. Each factor is a positive integer, so this moment is at least $1$.

Using the pointwise finite matrix convention [F7], group a finite product of entries by its unordered index pairs. The result is ${\prod }_{i\le j}{G}_{ij}^{m_{ij}}$. Each power is integrable by step 1.1. The functions $x\mapsto x^{m_{ij}}$ are continuous polynomials, hence Borel measurable by [F9]; applying [F6] to them gives the stated factorization, including exponent zero. If $n=0$, the only list of entries is empty and the identity reads $\mathbb{E}1=1$; an empty list for any $n$ gives the same identity. For $i<j$, $G_{ij}=G_{ji}$ and this variable has mean zero and second moment one by steps 2.1 and 3.1. If the two copies were independent, [F6] would give their product expectation as zero, contradicting the second moment. This proves the statement.