$\require{mathtools}
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The mutual information measures the dependence between $\BX,\BY$ by comparing their distribution to what it would have been if $\BX$ and $\BY$ had been independent, using information divergence. Formally, the mutual information is the divergence between the joint distribution $P_{\BX,\BY}$ and the product of the marginals $P_\BX$ and $P_\BY$:
\[
\al{
\I[\BX;\BY]
&\ce \D\ppa{P_{\BX,\BY}}{P_\BX P_\BY}\\
&= \E_{(x,y) \sim (\BX,\BY)}\b{\frac{\Pr[\BX=x\and\BY=y]}{\Pr[\BX=x]\Pr[\BY=y]}}\\
%&= \E_{(x,y) \sim (\BX,\BY)}\b{\frac{\Pr[(\BX,\BY) = (x,y)]}{\Pr[(\BX,\BY') = (x,y)]}}\\
&= \H[\BX] + \H[\BY] - \H[\BX,\BY]\tag{1}\\
%&= \H[\BX] - \H[\BX \mid \BY]\\
&= \H[\BY] - \H[\BY \mid \BX].\tag{2}
}
\]
Equivalently, let $\BX'$ be distributed identically to $\BX$ and $\BY'$ be distributed identically to $\BY$, but with $\BX',\BY'$ independent. Then
\[
\I[\BX;\BY] = \D\bpa{\BX,\BY}{\BX',\BY'}.
\]
Since the divergence is nonnegative, $(1)$ shows the subadditivity of entropy and $(2)$ shows that conditioning only decreases entropy.
Properties
We also have
\[
\al{
\I[\BX;\BY]
&= \E_{(x,y) \sim (\BX,\BY)}\b{\frac{\Pr\bco{\BY=y}{\BX=x}}{\Pr[\BY=y]}}\\
&= \E_{x \sim \BX}\b{\D\bpa{\at{\BY}{\BX=x}}{\BY}}.
}
\]
In other words, the mutual information measures how much (on average) the distribution of $\BY$ changes when it gets conditioned on $\BX$. This expression is convenient if we want to lower bound the mutual information because the term $\D\bpa{\at{\BY}{\BX=x}}{\BY}$ is always nonnegative, and if $\BY$ is binary we can use asymptotic estimates for the binary divergence to compute it.
Random bits
If $\BX,\BY$ are uniform random bits with correlation $\rho$, then using the properties of entropy deficit (and assuming WLOG that $\BX,\BY \in \pmo$), their mutual information is
\[
\al{
\I[\BX;\BY]
&= \E_{x \sim \BX}\b{\D\bpa{\at{\BY}{\BX=x}}{\BY}}\\
&= \E_{x \sim \BX}\b{\D\b{\at{\BY}{\BX=x}}}\tag{$\BY$ uniform}\\
&= \E_{x \sim \BX}\b{\Theta\p{\E\b{\BY|\BX =x}^2}}\tag{$\BY \in \pmo$}\\
&= \E_{x \sim \BX}\b{\Theta\p{(\rho x)^2}}\\
&= \Theta\p{\rho^2}\\
}
\]
If $\BX,\BY$ are biased bits with $\Pr[\BX=1] = \Pr[\BY=1] = p$ (with $p<1/2$), then their mutual information is
- up to $\H[\BX] = H(p) = \Theta\p{p \log\frac1p}$ if they’re positively correlated;
- only up to $\Theta(p^2)$ if they’re negatively correlated (note that their correlation cannot be lower than $-\frac{p}{1-p}$).
Additivity
Mutual information itself is not generally sub- or superadditive. For example, if $\BX_1, \ldots, \BX_n,\BY$ are all the same random coin flip, we have
\[
\I[\BX;\BY] = 1 \qquad \sum\I[\BX_i;\BY] = n,
\]
but if $\BX_1, \ldots, \BX_n$ are independent coin flips and $\BY \ce \BX_1 \xor \cdots \xor \BX_n$ is their parity, then
\[
\I[\BX;\BY] = 1 \qquad \sum\I[\BX_i;\BY] = 0.
\]
On the other hand, whenever the $\BX_i$’s are independent like in the second example, then it is superadditive:
\[
\I[\BX;\BY] \ge \sum\I[\BX_i;\BY].
\]
Indeed,
\[
\al{
\I[\BX;\BY]
&= \H[\BX] - \H[\BX \mid \BY]\\
&= \p{\sum\H[\BX_i]} - \H[\BX \mid \BY]\tag{independence}\\
&\ge \p{\sum\H[\BX_i] - \H[\BX_i \mid \BY]}\tag{subadditivity}\\
&\ge \sum\I[\BX_i;\BY].
}
\]
This is an example of a reverse entropic inequality.
Lower and upper bounds
#to-write
- two main bounds from
murphy--probabilistic-ml-advanced.pdf
- and entropy as a special case?
Similarly, the conditional mutual information $\I\bco{\BX; \BY}{\BZ}$ measures the average (over $z \sim \BZ$) divergence between the conditional random variables $\at{(\BX,\BY)}{\BZ=z}$ and an independent copy $\at{\BX'}{\BZ=z},\at{\BY'}{\BZ=z}$:
\[
\al{
\I\bco{\BX;\BY}{\BZ}
&\ce \E_{z \sim \BZ}\b{\D\bpa{\at{(\BX,\BY)}{\BZ=z}}{\at{\BX'}{\BZ=z},\at{\BY'}{\BZ=z}}}\\
&= \E\b{\log\frac{\Pr\bco{\BX=x \and \BY=y}{\BZ = z}}{\Pr\bco{\BX=x}{\BZ=z}\Pr\bco{\BY=y}{\BZ=z}}}\\
%&= \E\b{\log\frac{\Pr\b{\BX=x \and \BY=y \and \BZ = z}\Pr[\BZ=z]}{\Pr\b{\BX=x \and \BZ=z}\Pr\b{\BY=y \and \BZ=z}}}\\
&= \H\bco\BX\BZ + \H\bco\BY\BZ - \H\bco{\BX,\BY}{\BZ}\\
&= \H\bco\BY\BZ - \H\bco\BY{\BX,\BZ}.\tag{3}
}
\]
#to-write
- also define this one using distributions instead of random variables?
- or just rather, just define it based on the non-conditional version
Chain rule
By telescoping, we get
\[
\al{
\I[\BX_1, \ldots, \BX_n ; \BY]
&= \H[\BY] - \H\bco{\BY}{\BX_1, \ldots, \BX_n}\tag{by (2)}\\
&= \sum \H\bco{\BY}{\BX_1, \ldots, \BX_{i-1}} - \H\bco{\BY}{\BX_1, \ldots, \BX_i}\\
&= \sum \I\bco{\BX_i; \BY}{\BX_1, \ldots, \BX_{i-1}}.\tag{by (3)}
}
\]
#to-write
- argue that the base case (1 variable) is under-defined: it could be taken as either $-\H[\BX]$ or $\D[\BX]$, given that the entropy of the reference distribution gets cancelled out from then on?
- in this choice of sign, $\I[\BX;\BY;\BZ] = \I\bco{\BX;\BY}{\BZ} - \I\b{\BX;\BY}$ (or denoted with $\D$?) makes sense (?) because
- it’s positive when $\BX,\BY$ have a “shared effect”, e.g. $\BZ = \BX\xor\BY$, which leads us to be able to predict them better together than expected
- (but otoh if you think about it like shared entropy, then it’s clearly negative: everything was looking completely independent so far and now you have a correlation reducing your entropy)
- it’s negative when $\BX,\BY$ have a “shared cause”, e.g. $\BX = \BZ$ and $\BY = \comp{\BZ}$, which leads us to be somewhat disappointed in our ability to predict them together given that the pairwise dependences were so strong!
- (and if you think about like shared entropy, then it’s positive (??))
- I think this choice of sign makes more sense because it’s about revealing information about things that looked more random, it’s about being able to predict better than expected!
- if you want $\I\b{\BX;\BE} = \E\b{\I\bco{\BX}{\BE}}$, that would force the prior to be $[\BX]$ itself?