5
$\begingroup$

I went to listen to a workshop and someone from the audience asked the presenter how the moments can improve the mutual information. I am learning about MI (Mutual Information) so didn't have enough knowledge to understand what it means. Then, I did some research but I still have some confusion. I am wondering if someone who has more knowledge about this can clarify things for me. Here are my questions:

  • Mutual information is usually calculated by bin functions to estimate the probability of two random variables which can be a case of two vectors $X$ and $Y$. Is the moment generating function another way to estimate probability?

  • If moment generating functions can present the probability of $X$ and $Y$, how do we calculate it?

  • Does a MI have a moment generating function?

  • If MI has a moment generating function, how can we present a MI of $X$ and $Y$ by its moment functions?

$\endgroup$
6
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.