Gibbs' inequality

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Josiah Willard Gibbs

In information theory, Gibbs' inequality is a statement about the mathematical entropy of a discrete probability distribution. Several other bounds on the entropy of probability distributions are derived from Gibbs' inequality, including Fano's inequality. It was first presented by J. Willard Gibbs in the 19th century.

Gibbs' inequality

Suppose that

${\displaystyle P=\{p_{1},\ldots ,p_{n}\}}$

is a probability distribution. Then for any other probability distribution

${\displaystyle Q=\{q_{1},\ldots ,q_{n}\}}$

the following inequality between positive quantities (since the pi and qi are positive numbers less than one) holds[1]:68

${\displaystyle -\sum _{i=1}^{n}p_{i}\log _{2}p_{i}\leq -\sum _{i=1}^{n}p_{i}\log _{2}q_{i}}$

with equality if and only if

${\displaystyle p_{i}=q_{i}\,}$

for all i. 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:[2]:34

${\displaystyle D_{\mathrm {KL} }(P\|Q)\equiv \sum _{i=1}^{n}p_{i}\log _{2}{\frac {p_{i}}{q_{i}}}\geq 0.}$

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.

Proof

Since

${\displaystyle \log _{2}a={\frac {\ln a}{\ln 2}}}$

it is sufficient to prove the statement using the natural logarithm (ln). Note that the natural logarithm satisfies

${\displaystyle \ln x\leq x-1}$

for all x > 0 with equality if and only if x=1.

Let ${\displaystyle I}$ denote the set of all ${\displaystyle i}$ for which pi is non-zero. Then

${\displaystyle -\sum _{i\in I}p_{i}\ln {\frac {q_{i}}{p_{i}}}\geq -\sum _{i\in I}p_{i}\left({\frac {q_{i}}{p_{i}}}-1\right)}$
${\displaystyle =-\sum _{i\in I}q_{i}+\sum _{i\in I}p_{i}}$
${\displaystyle \geq 0.}$

So

${\displaystyle -\sum _{i\in I}p_{i}\ln q_{i}\geq -\sum _{i\in I}p_{i}\ln p_{i}}$

and then trivially

${\displaystyle -\sum _{i=1}^{n}p_{i}\ln q_{i}\geq -\sum _{i=1}^{n}p_{i}\ln p_{i}}$

since the right hand side does not grow, but the left hand side may grow or may stay the same.

For equality to hold, we require:

1. ${\displaystyle {\frac {q_{i}}{p_{i}}}=1}$ for all ${\displaystyle i\in I}$ so that the approximation ${\displaystyle \ln {\frac {q_{i}}{p_{i}}}={\frac {q_{i}}{p_{i}}}-1}$ is exact.
2. ${\displaystyle \sum _{i\in I}q_{i}=1}$ so that equality continues to hold between the third and fourth lines of the proof.

This can happen if and only if

${\displaystyle p_{i}=q_{i}}$

for i = 1, ..., n.

Alternative proofs

The result can alternatively be proved using Jensen's inequality or log sum inequality.

Corollary

The entropy of ${\displaystyle P}$ is bounded by:[1]:68

${\displaystyle H(p_{1},\ldots ,p_{n})\leq \log n.}$

The proof is trivial - simply set ${\displaystyle q_{i}=1/n}$ for all i.