Please forgive me if this is not the right Stack Exchange (I also posted it at Cross Validated). Please also forgive me for inventing terms.

For discrete random variables X and Y, the mutual information of X and Y can be defined as follows: $I(X;Y) = \sum_{y \in Y} \sum_{x \in X} p(x,y) \log{ \left( \frac{p(x,y)}{p_1(x)\,p_2(y)} \right) }, \,\!$

I will define the mutual information of a "cell" $x_0$ to be: $CI(x_0,Y) = \sum_{y \in Y} p(x_0,y) \log{ \left( \frac{p(x_0,y)}{p_1(x_0)\,p_2(y)} \right) }, \,\!$

I'm not sure if this quantity goes by another name. Essentially I'm restricting focus to a single state of variable X (and then the full MI can be calculated by summing all the cell MIs).

My question: is it guaranteed that $CI(x_0,Y) \ge 0$? We know $I(X;Y)\ge0$ and we know that the pointwise mutual information can be negative. I feel like CI should be nonnegative and that I might be missing some obvious proof.


By the simple transformation $p(x_0, y) = p(x_0) p(y | x_0)$ your expression merely becomes $p(x_0) KL(p (y|x), p(y))$, where $KL(p,q)$ is the Kullback-Leibler divergence, and is always positive. therefore, your expression is always positive.

  • $\begingroup$ Terrific, thanks! I was fooling around with similar transformations but somehow missed the obvious. $\endgroup$ – Michael McGowan Dec 22 '10 at 17:13
  • $\begingroup$ yes. this particular transformation turns out to be very handy. $\endgroup$ – Suresh Venkat Dec 22 '10 at 17:32

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