TITLE: The theorem inverse to Isserlis' theorem
QUESTION [6 upvotes]: In probability theory there exists the Isserlis' theorem, which states that if $X=(X_1,X_2,\ldots,X_{2n})$ is a zero mean multivariate normal random vector, then an expectation value of a product $X_1X_2\ldots X_{2n}$ can be calculated as
$$\mathbb{E}\left(X_1X_2\ldots X_{2n}\right)=\sum{\mathbb{E}(X_{i_1}X_{j_1})\mathbb{E}(X_{i_2}X_{j_2})\ldots \mathbb{E}(X_{i_n}X_{j_n})}$$ where sum is for all possible pairings of the set $\left\{1,2,\ldots,2n\right\}$. 
It's fairly easy to prove this, but in my textbook (non-math) there is a statement I cannot proof: if there is a matrix $D_{mn}$ and all of its eigenvalues are positive and every expectation value for any set of variables from $\left\{X_1,X_2,\ldots,X_{2n}\right\}$ can be calculated by pairing chosen set of variables as $(X_{i_s},X_{i_r})$, taking for each pair corresponding $D_{i_s r_s}$ and multipliing them and then summing productions for all possible pairings, then the vector $X=(X_1,X_2,\ldots,X_{2n})$ is a zero mean multivariate normal random vector with covariance matrix $D$.

REPLY [4 votes]: Let $Y = (Y_1, \ldots, Y_{2n})$ be jointly normal with mean $0$ and covariance matrix $D$.  
The Taylor expansion of the moment generating function for $X$ 
is
$$ M(z) = \mathbb E\left[ e^{z \cdot X}\right] = \sum_{j=0}^\infty \frac{\mathbb E \left[ (z \cdot X)^j\right]}{j!} $$
which can be computed using sums of expected values of products of the $X_i$, and this will be the same as for $Y$.  In particular, this series converges absolutely for all $z \in \mathbb C$, and $X$ has the same moment generating function as $Y$.  By the uniqueness theorem for (multivariate) moment generating functions, $X$ has the same joint distribution as $Y$.