Mutual information and moment generating functions
The moment generating function $M_X$ is a property of a random variable $X$. It's defined by the expected value of $e^{tX}$ (where $t$ is the argument).

Since the exponential function $e^x = \sum_0^\infty \frac{x^n}{n!}$ contains all natural powers of its argument as a summand, the expected value of a sum is the sum of the expected values ($\mathbb{E}(\sum_i X_i)=\sum_i\mathbb{E}(X_i)$) and the expected value of a natural power of $X$ ($\mathbb{E}(X^n)$) is called it's $n$-th moment, the $n$-th moment is present in the $n$-th summand:

$$M_X(t)=\mathbb{E}(e^{tX})=\sum_{i=0}^\infty \frac{t^i\mathbb{E}(X^i)}{i!} \quad .$$

If you now consider the $k$-times derivative of $M_X$:

$$M_X^{(k)}(t)=\mathbb{E}(e^{tX})=\sum_{i=0}^\infty \frac{\mathbb{E}(X^{i+k})}{i!} \quad ,$$

and use $0$ as an argument, you get
$$M_X^{(k)}(0)=\mathbb{E}(X^k)\quad,$$

so the $k$-th moment was generated.



Now look at the mutual information:

$$I(X,Y) = \sum_{(x,y)}P(X=x,Y=y)\log\left(\frac{P(X=x,Y=y)}{P(X=x)\cdot P(Y=y)}\right) = \mathbb{E}(\mathrm{PMI}(X,Y)), $$

which is the expected value of the pointwise mutual information (it's likely that they actually deal with the continuous case where $I$ and $\mathrm{PMI}$ are defined using integrals and densities, respectively). So mutual information does not have a moment (or moment generating function), but it is the first moment of a random variable, so:

$$I(X,Y) = M_{\mathrm{PMI}(X,Y)}'(0)\quad.$$