Gibbs' inequality

Suppose that $P\; =\; \backslash \{\; p\_1\; ,\; \backslash ldots\; ,\; p\_n\; \backslash \}$ and $Q\; =\; \backslash \{\; q\_1\; ,\; \backslash ldots\; ,\; q\_n\; \backslash \}$

are discrete probability distributions. Then

:$-\; \backslash sum\_\{i=1\}^n\; p\_i\; \backslash log\; p\_i\; \backslash leq\; -\; \backslash sum\_\{i=1\}^n\; p\_i\; \backslash log\; q\_i$

with equality if and only if

$p\_i\; =\; q\_i$

for $i\; =\; 1,\; \backslash dots\; n$.{{cite book|author=Pierre Bremaud|title=An Introduction to Probabilistic Modeling|date=6 December 2012|publisher=Springer Science & Business Media|isbn=978-1-4612-1046-7}}{{rp|68}} Put in words, the information entropy of a distribution $P$ is less than or equal to its cross entropy with any other distribution $Q$.

The difference between the two quantities is the Kullback–Leibler divergence or relative entropy, so the inequality can also be written:{{cite book|author=David J. C. MacKay|title=Information Theory, Inference and Learning Algorithms|date=25 September 2003|publisher=Cambridge University Press|isbn=978-0-521-64298-9}}{{rp|34}}

:$D\_\{\backslash mathrm\{KL\}\}(P\backslash |Q)\; \backslash equiv\; \backslash sum\_\{i=1\}^n\; p\_i\; \backslash log\; \backslash frac\{p\_i\}\{q\_i\}\; \backslash geq\; 0.$

Note that the use of base-2 logarithms is optional, and

allows one to refer to the quantity on each side of the inequality as an

"average surprisal" measured in bits.

For simplicity, we prove the statement using the natural logarithm, denoted by {{math|ln}}, since

:$\backslash log\_b\; a\; =\; \backslash frac\{\; \backslash ln\; a\; \}\{\; \backslash ln\; b\; \},$

so the particular logarithm base {{math|b > 1}} that we choose only scales the relationship by the factor {{math|1 / ln b}}.

Let $I$ denote the set of all $i$ for which p_i is non-zero. Then, since $\backslash ln\; x\; \backslash leq\; x-1$ for all x > 0, with equality if and only if x=1, we have:

:$-\; \backslash sum\_\{i\; \backslash in\; I\}\; p\_i\; \backslash ln\; \backslash frac\{q\_i\}\{p\_i\}\; \backslash geq\; -\; \backslash sum\_\{i\; \backslash in\; I\}\; p\_i\; \backslash left(\; \backslash frac\{q\_i\}\{p\_i\}\; -\; 1\; \backslash right)$$=\; -\; \backslash sum\_\{i\; \backslash in\; I\}\; q\_i\; +\; \backslash sum\_\{i\; \backslash in\; I\}\; p\_i\; =\; -\; \backslash sum\_\{i\; \backslash in\; I\}\; q\_i\; +\; 1\; \backslash geq\; 0$

The last inequality is a consequence of the p_i and q_i being part of a probability distribution. Specifically, the sum of all non-zero values is 1. Some non-zero q_i, however, may have been excluded since the choice of indices is conditioned upon the p_i being non-zero. Therefore, the sum of the q_i may be less than 1.

So far, over the index set $I$, we have:

:$-\; \backslash sum\_\{i\; \backslash in\; I\}\; p\_i\; \backslash ln\; \backslash frac\{q\_i\}\{p\_i\}\; \backslash geq\; 0$,

or equivalently

:$-\; \backslash sum\_\{i\; \backslash in\; I\}\; p\_i\; \backslash ln\; q\_i\; \backslash geq\; -\; \backslash sum\_\{i\; \backslash in\; I\}\; p\_i\; \backslash ln\; p\_i$.

Both sums can be extended to all $i=1,\; \backslash ldots,\; n$, i.e. including $p\_i=0$, by recalling that the expression $p\; \backslash ln\; p$ tends to 0 as $p$ tends to 0, and $(-\backslash ln\; q)$ tends to $\backslash infty$ as $q$ tends to 0. We arrive at

:$-\; \backslash sum\_\{i=1\}^n\; p\_i\; \backslash ln\; q\_i\; \backslash geq\; -\; \backslash sum\_\{i=1\}^n\; p\_i\; \backslash ln\; p\_i$

For equality to hold, we require

1. $\backslash frac\{q\_i\}\{p\_i\}\; =\; 1$ for all $i\; \backslash in\; I$ so that the equality $\backslash ln\; \backslash frac\{q\_i\}\{p\_i\}\; =\; \backslash frac\{q\_i\}\{p\_i\}\; -1$ holds,
2. and $\backslash sum\_\{i\; \backslash in\; I\}\; q\_i\; =\; 1$ which means $q\_i=0$ if $i\backslash notin\; I$, that is, $q\_i=0$ if $p\_i=0$.

This can happen if and only if $p\_i\; =\; q\_i$ for $i\; =\; 1,\; \backslash ldots,\; n$.

The result can alternatively be proved using Jensen's inequality, the log sum inequality, or the fact that the Kullback-Leibler divergence is a form of Bregman divergence.

Because log is a concave function, we have that:

:$\backslash sum\_i\; p\_i\; \backslash log\backslash frac\{q\_i\}\{p\_i\}\; \backslash le\; \backslash log\backslash sum\_i\; p\_i\backslash frac\{q\_i\}\{p\_i\}\; =\; \backslash log\backslash sum\_i\; q\_i\; =\; 0$

where the first inequality is due to Jensen's inequality, and $q$ being a probability distribution implies the last equality.

Furthermore, since $\backslash log$ is strictly concave, by the equality condition of Jensen's inequality we get equality when

:$\backslash frac\{q\_1\}\{p\_1\}\; =\; \backslash frac\{q\_2\}\{p\_2\}\; =\; \backslash cdots\; =\; \backslash frac\{q\_n\}\{p\_n\}$

and

:$\backslash sum\_i\; q\_i\; =\; 1$.

Suppose that this ratio is $\backslash sigma$, then we have that

:$1\; =\; \backslash sum\_i\; q\_i\; =\; \backslash sum\_i\; \backslash sigma\; p\_i\; =\; \backslash sigma$

where we use the fact that $p,\; q$ are probability distributions. Therefore, the equality happens when $p\; =\; q$.

Alternatively, it can be proved by noting that$$q\; -\; p\; -\; p\backslash ln\backslash frac\; qp\; \backslash geq\; 0$$for all $p,\; q\; >\; 0$, with equality holding iff $p=q$. Then, sum over the states, we have$$\backslash sum\_i\; q\_i\; -\; p\_i\; -\; p\_i\backslash ln\backslash frac\{q\_i\}\{p\_i\}\; \backslash geq\; 0$$with equality holding iff $p\; =\; q$.

This is because the KL divergence is the Bregman divergence generated by the function $t\; \backslash mapsto\; \backslash ln\; t$.

The entropy of $P$ is bounded by:{{rp|68}}

:$H(p\_1,\; \backslash ldots\; ,\; p\_n)\; \backslash leq\; \backslash log\; n.$

The proof is trivial – simply set $q\_i\; =\; 1/n$ for all i.

• Information entropy
• Bregman divergence
• Log sum inequality

{{Reflist}}

Category:Information theory

Category:Coding theory

Category:Probabilistic inequalities

Category:Articles containing proofs