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Log sum inequality

Let ${\displaystyle a_{1},\ldots ,a_{n}}$ and ${\displaystyle b_{1},\ldots ,b_{n}}$ be nonnegative numbers. Denote the sum of all ${\displaystyle a_{i}}$s by ${\displaystyle a}$ and the sum of all ${\displaystyle b_{i}}$s by ${\displaystyle b}$. The log sum inequality states that

${\displaystyle \sum _{i=1}^{n}a_{i}\log {\frac {a_{i}}{b_{i}}}\geq a\log {\frac {a}{b}},}$

with equality if and only if ${\displaystyle {\frac {a_{i}}{b_{i}}}}$ are equal for all ${\displaystyle i}$, in other words ${\displaystyle a_{i}=cb_{i}}$ for all ${\displaystyle i}$.^[1]

(Take ${\displaystyle a_{i}\log {\frac {a_{i}}{b_{i}}}}$ to be ${\displaystyle 0}$if${\displaystyle a_{i}=0}$ and ${\displaystyle \infty }$if${\displaystyle a_{i}>0,b_{i}=0}$. These are the limiting values obtained as the relevant number tends to ${\displaystyle 0}$.)^[1]

Proof[edit]

Notice that after setting ${\displaystyle f($x)=x\log x} we have

${\displaystyle {\begin{aligned}\sum _{i=1}^{n}a_{i}\log {\frac {a_{i}}{b_{i}}}&{}=\sum _{i=1}^{n}b_{i}f\left({\frac {a_{i}}{b_{i}}}\right)=b\sum _{i=1}^{n}{\frac {b_{i}}{b}}f\left({\frac {a_{i}}{b_{i}}}\right)\\&{}\geq bf\left(\sum _{i=1}^{n}{\frac {b_{i}}{b}}{\frac {a_{i}}{b_{i}}}\right)=bf\left({\frac {1}{b}}\sum _{i=1}^{n}a_{i}\right)=bf\left({\frac {a}{b}}\right)\\&{}=a\log {\frac {a}{b}},\end{aligned}}}$

where the inequality follows from Jensen's inequality since ${\displaystyle {\frac {b_{i}}{b}}\geq 0}$, ${\displaystyle \sum _{i=1}^{n}{\frac {b_{i}}{b}}=1}$, and ${\displaystyle f}$ is convex.^[1]

Generalizations[edit]

The inequality remains valid for ${\displaystyle n=\infty }$ provided that ${\displaystyle a<\infty }$ and ${\displaystyle b<\infty }$.^{[citation needed]} The proof above holds for any function ${\displaystyle g}$ such that ${\displaystyle f($x)=xg(x)} is convex, such as all continuous non-decreasing functions. Generalizations to non-decreasing functions other than the logarithm is given in Csiszár, 2004.

Another generalization is due to Dannan, Neff and Thiel, who showed that if ${\displaystyle a_{1},a_{2}\cdots a_{n}}$ and ${\displaystyle b_{1},b_{2}\cdots b_{n}}$ are positive real numbers with ${\displaystyle a_{1}+a_{2}\cdots +a_{n}=a}$ and ${\displaystyle b_{1}+b_{2}\cdots +b_{n}=b}$, and ${\displaystyle k\geq 0}$, then ${\displaystyle \sum _{i=1}^{n}a_{i}\log \left({\frac {a_{i}}{b_{i}}}+k\right)\geq a\log \left({\frac {a}{b}}+k\right)}$. ^[2]

Applications[edit]

The log sum inequality can be used to prove inequalities in information theory. Gibbs' inequality states that the Kullback-Leibler divergence is non-negative, and equal to zero precisely if its arguments are equal.^[3] One proof uses the log sum inequality.

Proof[1]

Let ${\displaystyle P=(p_{i})_{i\in \mathbb {N} }}$ and ${\displaystyle Q=(q_{i})_{i\in \mathbb {N} }}$ be pmfs. In the log sum inequality, substitute ${\displaystyle n=\infty }$, ${\displaystyle a_{i}=p_{i}}$ and ${\displaystyle b_{i}=q_{i}}$ to get ${\displaystyle \mathbb {D} _{\mathrm {KL} }(P\|Q)\equiv \sum _{i}p_{i}\log _{2}{\frac {p_{i}}{q_{i}}}\geq 1\log {\frac {1}{1}}=0}$ with equality if and only if ${\displaystyle p_{i}=q_{i}}$ for all i (as both ${\displaystyle P}$ and ${\displaystyle Q}$ sum to 1).

The inequality can also prove convexity of Kullback-Leibler divergence.^[4]

Notes[edit]

References[edit]

• Cover, Thomas M.; Thomas, Joy A. (1991). Elements of Information Theory. Hoboken, New Jersey: Wiley. ISBN 978-0-471-24195-9.
• Csiszár, I.; Shields, P. (2004). "Information Theory and Statistics: A Tutorial" (PDF). Foundations and Trends in Communications and Information Theory. 1 (4): 417–528. doi:10.1561/0100000004. Retrieved 2009-06-14.
• T.S. Han, K. Kobayashi, Mathematics of information and coding. American Mathematical Society, 2001. ISBN 0-8218-0534-7.
• Information Theory course materials, Utah State University [1]. Retrieved on 2009-06-14.
• MacKay, David J.C. (2003). Information Theory, Inference, and Learning Algorithms. Cambridge University Press. ISBN 0-521-64298-1.