Index: The Book of Statistical ProofsGeneral Theorems ▷ Information theory ▷ Shannon entropy ▷ Gibbs' inequality

Theorem: Let $X$ be a discrete random variable and consider two probability distributions with probability mass functions $p(x)$ and $q(x)$. Then, Gibbs’ inequality states that the entropy of $X$ according to $P$ is smaller than or equal to the cross-entropy of $P$ and $Q$:

$\label{eq:Gibbs-ineq} - \sum_{x \in \mathcal{X}} p(x) \, \log_b p(x) \leq - \sum_{x \in \mathcal{X}} p(x) \, \log_b q(x) \; .$

Proof: Without loss of generality, we will use the natural logarithm, because a change in the base of the logarithm only implies multiplication by a constant:

$\label{eq:log-ln} \log_b a = \frac{\ln a}{\ln b} \; .$

Let $I$ be the of all $x$ for which $p(x)$ is non-zero. Then, proving \eqref{eq:Gibbs-ineq} requires to show that

$\label{eq:Gibbs-ineq-s1} \sum_{x \in I} p(x) \, \ln \frac{p(x)}{q(x)} \geq 0 \; .$

Because $\ln x \leq x - 1$, i.e. $-\ln x \geq 1 - x$, for all $x > 0$, with equality only if $x = 1$, we can say about the left-hand side that

$\label{eq:Gibbs-ineq-s2} \begin{split} \sum_{x \in I} p(x) \, \ln \frac{p(x)}{q(x)} &\geq \sum_{x \in I} p(x) \left( 1 - \frac{p(x)}{q(x)} \right) \\ &= \sum_{x \in I} p(x) - \sum_{x \in I} q(x) \; . \end{split}$

Finally, since $p(x)$ and $q(x)$ are probability mass functions, we have

$\label{eq:p-q-pmf} \begin{split} 0 \leq p(x) \leq 1, \quad \sum_{x \in I} p(x) &= 1 \quad \text{and} \\ 0 \leq q(x) \leq 1, \quad \sum_{x \in I} q(x) &\leq 1 \; , \end{split}$

such that it follows from \eqref{eq:Gibbs-ineq-s2} that

$\label{eq:Gibbs-ineq-s3} \begin{split} \sum_{x \in I} p(x) \, \ln \frac{p(x)}{q(x)} &\geq \sum_{x \in I} p(x) - \sum_{x \in I} q(x) \\ &= 1 - \sum_{x \in I} q(x) \geq 0 \; . \end{split}$
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Metadata: ID: P164 | shortcut: gibbs-ineq | author: JoramSoch | date: 2020-09-09, 02:18.