Gibbs' inequality

In information theory, Gibbs' inequality is a statement about the information 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 ${\displaystyle P=\{p_{1},\ldots ,p_{n}\}}$ and ${\displaystyle Q=\{q_{1},\ldots ,q_{n}\}}$ are discrete probability distributions. Then

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

with equality if and only if ${\displaystyle p_{i}=q_{i}}$ for ${\displaystyle i=1,\dots n}$.^[1]: 68 Put in words, the information entropy of a distribution ${\displaystyle P}$ is less than or equal to its cross entropy with any other distribution ${\displaystyle Q}$.

The difference between the two quantities is the Kullback–Leibler divergence or relative entropy, so the inequality can also be written:^[2]: 34

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

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

For simplicity, we prove the statement using the natural logarithm, denoted by ln, since

${\displaystyle \log _{b}a={\frac {\ln a}{\ln b}},}$

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

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

${\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}=-\sum _{i\in I}q_{i}+1\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 ${\displaystyle I}$, we have:

${\displaystyle -\sum _{i\in I}p_{i}\ln {\frac {q_{i}}{p_{i}}}\geq 0}$,

or equivalently

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

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

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

For equality to hold, we require

1. ${\displaystyle {\frac {q_{i}}{p_{i}}}=1}$ for all ${\displaystyle i\in I}$ so that the equality ${\displaystyle \ln {\frac {q_{i}}{p_{i}}}={\frac {q_{i}}{p_{i}}}-1}$ holds,
2. and ${\displaystyle \sum _{i\in I}q_{i}=1}$ which means ${\displaystyle q_{i}=0}$ if ${\displaystyle i\notin I}$, that is, ${\displaystyle q_{i}=0}$ if ${\displaystyle p_{i}=0}$.

This can happen if and only if ${\displaystyle p_{i}=q_{i}}$ for ${\displaystyle i=1,\ldots ,n}$.

Alternative proofs

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. Below we give a proof based on Jensen's inequality:

Because log is a concave function, we have that:

${\displaystyle \sum _{i}p_{i}\log {\frac {q_{i}}{p_{i}}}\leq \log \sum _{i}p_{i}{\frac {q_{i}}{p_{i}}}=\log \sum _{i}q_{i}\leq 0}$

Where the first inequality is due to Jensen's inequality, and the last equality is due to the same reason given in the above proof.

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

${\displaystyle {\frac {q_{1}}{p_{1}}}={\frac {q_{2}}{p_{2}}}=\cdots ={\frac {q_{n}}{p_{n}}}}$

and

${\displaystyle \sum _{i}q_{i}=1}$

Suppose that this ratio is ${\displaystyle \sigma }$, then we have that

${\displaystyle 1=\sum _{i}q_{i}=\sum _{i}\sigma p_{i}=\sigma }$

Where we use the fact that ${\displaystyle p,q}$ are probability distributions. Therefore, the equality happens when ${\displaystyle p=q}$.

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.

See also

• Information entropy
• Bregman divergence
• Log sum inequality

References