A combinatorics problem and the probability interpretation

Question

For a gaussian vector variable $w\sim N(0,I_{n\times n})$, the moments of square norm are $\mathbb{E} \|w\|^{2 r} = \prod_{t=0}^{r-1} (n + 2 t)$.

Based on Isserlis' theorem, $\mathbb{E} \|w\|^{2 r}$ can also be evaluated as $$\sum_{\pi\in \mathcal{P}([r]), |\pi|\leq n}\frac{n!}{(n-|\pi|)!}\prod_{p\in\pi}(2 |p|-1)!!$$ where $\mathcal{P}([r])$ means all partitions on set $[r]=\{1,\dots,r\}$, $\pi$ is a partition, $p$ is one block in a partition, $|\pi|$ and $|p|$ are number of blocks and number of elements in a block.

Now consider a variant of the above problem. $$\sum_{\pi\in \mathcal{P}([r]), |\pi|\leq n}\frac{n!}{(n-|\pi|)!}\prod_{p\in\pi}\frac{1}{2}~(2 |p|-1)!!$$ The above formula only differs from moments of square norm of gaussian vector variable with a factor $\frac{1}{2}$. Is there a similar finite product solution and probability interpretation for the above formula?

Answer 1

Fix $n$. Let $$ G(x) = \sum_{i=0}^n \frac{n!}{(n-i)!}\frac{x^i}{i!} = (1+x)^n. $$ Let $$ F(x) = \sum_{j\geq 1}\frac 12 (2j-1)!!\frac{x^j}{j!} = \frac{1}{2\sqrt{1-2x}}-\frac 12. $$ By the Compositional Formula (Theorem 5.1.4 of Enumerative Combinatorics, vol. 2), the number you want is $r!$ times the coefficient of $x^r$ in $$ G(F(x)) = \frac{1}{2^n}\left( 1+\frac{1}{\sqrt{1-2x}}\right)^n. $$ You can expand this by the binomial theorem and then expand each term into a power series to get a formula for your number as a sum with $n$ terms.