Let $(X,Y)$ be drawn from a 2-dimensional zero-mean multivariate normal distribution. For $j,k$ non-negative integers, compute $E[X^j Y^k]$ using only $E[X^2]$, $E[Y^2]$, and $E[XY]$, using Isserlis' theorem.
I know that if $j+k$ is odd then the result is zero, but I am not sure about how to compute when $j+k$ is even for arbitrary $j$ and $k$. I realize that it is an inherently combinatorial problem, but I can't figure out how to enumerate all the possibilities as required by Isserlis' theorem. I have not worked with Isserlis' theorem in the past, but I'm pretty sure it's the key to computing this type of thing.
I see that a technique is outlined in https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Properties under "Higher moments", but I'm not sure how to generalize to arbitrary $j,k$.
Any help would be very much appreciated. Thank you in advance!
Let $Z_i = X$ for $i = 1 \ldots j$ and $Z_i = Y$ for $i = j+1 \ldots j+k$. For convenience let's assume $j \le k$. Isserlis says $$ \mathbb E[X^j Y^k] = \mathbb E[Z_1 \ldots Z_{i+j}] = \sum_p \prod_{(s,t) \in p} \mathbb E[Z_s Z_t] $$ where the sum is over all perfect matchings $p$ of $[1,\ldots, j+k]$, i.e. all ways of partitioning $[1,\ldots, j+k]$ into unordered pairs. Such a matching might have $r_1$ pairs where both elements are in $A = [1,\ldots,j]$; if so, it must have $r_2 = j - 2 r_1$ pairs where one is in $A$ and the other in $B = [j+1,\ldots,j+k]$, and then the remaining $r_3 = (j+k)/2 - (j - r_1) = r_1 + (k-j)/2$ pairs where both are in $B$. Of course $r_1, r_2, r_3$ must be nonnegative. For such matchings we have a term of $\mathbb E[X^2]^{r_1} \mathbb E[XY]^{r_2} \mathbb E[Y^2]^{r_3}$.
Now you just have to count how many ways there are to choose such a matching. There are ${j \choose 2 r_1}$ ways to choose which $2 r_1$ members of $A$ are matched together, and then $(2r_1 - 1)!! = (2r_1 - 1)(2r_1 - 3) \ldots 1$ ways of pairing them. Similarly there are ${k \choose 2 r_3}$ ways to choose which $2 r_3$ members of $B$ are matched together, and $(2r_3 - 1)!!$ ways of pairing them. That leaves $r_2 = j - 2 r_1$ members of $A$ and $k - 2 r_3$ members of $B$ which must be paired. Of course we need these to be equal, i.e. $r_3 = r_1 + (k-j)/2$, in which case there are $r_2!$ ways to pair those members of $A$ with those members of $B$. So we end up with (for $j \le k$ with $j+k$ even) $$ \mathbb E[X^j Y^k] = \sum_{r_1=0}^{\lfloor j/2\rfloor} {j \choose 2 r_1} (2 r_1 - 1)!! {k \choose 2 r_1 + k-j} (2 r_1 + k-j -1)!!\; (j - 2 r_1)! \;\mathbb E[X^2]^{r_1} \mathbb E[XY]^{j - 2 r_1} \mathbb E[Y^2]^{r_1 + (k-j)/2} $$