Index: The Book of Statistical ProofsGeneral Theorems ▷ Information theory ▷ Discrete mutual information ▷ Relation to marginal and conditional entropy

Theorem: Let $X$ and $Y$ be discrete random variables with the joint probability $p(x,y)$ for $x \in \mathcal{X}$ and $y \in \mathcal{Y}$. Then, the mutual information of $X$ and $Y$ can be expressed as

\[\label{eq:dmi-mce} \begin{split} \mathrm{I}(X,Y) &= \mathrm{H}(X) - \mathrm{H}(X|Y) \\ &= \mathrm{H}(Y) - \mathrm{H}(Y|X) \end{split}\]

where $\mathrm{H}(X)$ and $\mathrm{H}(Y)$ are the marginal entropies of $X$ and $Y$ and $\mathrm{H}(X \vert Y)$ and $\mathrm{H}(Y \vert X)$ are the conditional entropies.

Proof: The mutual information of $X$ and $Y$ is defined as

\[\label{eq:MI} \mathrm{I}(X,Y) = \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} p(x,y) \log \frac{p(x,y)}{p(x)\,p(y)} \; .\]

Separating the logarithm, we have:

\[\label{eq:MI-s1} \mathrm{I}(X,Y) = \sum_x \sum_y p(x,y) \log \frac{p(x,y)}{p(y)} - \sum_x \sum_y p(x,y) \log p(x) \; .\]

Applying the law of conditional probability, i.e. $p(x,y) = p(x \vert y) \, p(y)$, we get:

\[\label{eq:MI-s2} \mathrm{I}(X,Y) = \sum_x \sum_y p(x|y) \, p(y) \log p(x|y) - \sum_x \sum_y p(x,y) \log p(x) \; .\]

Regrouping the variables, we have:

\[\label{eq:MI-s3} \mathrm{I}(X,Y) = \sum_y p(y) \sum_x p(x|y) \log p(x|y) - \sum_x \left( \sum_y p(x,y) \right) \log p(x) \; .\]

Applying the law of marginal probability, i.e. $p(x) = \sum_y p(x,y)$, we get:

\[\label{eq:MI-s4} \mathrm{I}(X,Y) = \sum_y p(y) \sum_x p(x|y) \log p(x|y) - \sum_x p(x) \log p(x) \; .\]

Now considering the definitions of marginal and conditional entropy

\[\label{eq:ME-CE} \begin{split} \mathrm{H}(X) &= - \sum_{x \in \mathcal{X}} p(x) \log p(x) \\ \mathrm{H}(X|Y) &= \sum_{y \in \mathcal{Y}} p(y) \, \mathrm{H}(X|Y=y) \; , \end{split}\]

we can finally show:

\[\label{eq:MI-qed} \begin{split} \mathrm{I}(X,Y) &= - \mathrm{H}(X|Y) + \mathrm{H}(X) \\ &= \mathrm{H}(X) - \mathrm{H}(X|Y) \; . \end{split}\]

The conditioning of $X$ on $Y$ in this proof is without loss of generality. Thus, the proof for the expression using the reverse conditional entropy of $Y$ given $X$ is obtained by simply switching $x$ and $y$ in the derivation.


Metadata: ID: P19 | shortcut: dmi-mce | author: JoramSoch | date: 2020-01-13, 18:20.