Hartley function

The Hartley function coincides with the Shannon entropy (as well as with the Rényi entropies of all orders) in the case of a uniform probability distribution. It is a special case of the Rényi entropy since:

${\displaystyle H_{0}(X)={\frac {1}{1-0}}\log \sum _{i=1}^{|X|}p_{i}^{0}=\log |X|.}$

But it can also be viewed as a primitive construction, since, as emphasized by Kolmogorov and Rényi, the Hartley function can be defined without introducing any notions of probability (see Uncertainty and information by George J. Klir, p. 423).

The Hartley function only depends on the number of elements in a set, and hence can be viewed as a function on natural numbers. Rényi showed that the Hartley function in base 2 is the only function mapping natural numbers to real numbers that satisfies

1. ${\displaystyle H(mn)=H(m)+H(n)}$ (additivity)
2. ${\displaystyle H(m)\leq H(m+1)}$ (monotonicity)
3. ${\displaystyle H(2)=1}$ (normalization)

Condition 1 says that the uncertainty of the Cartesian product of two finite sets A and B is the sum of uncertainties of A and B. Condition 2 says that a larger set has larger uncertainty.

We want to show that the Hartley function, log₂(n), is the only function mapping natural numbers to real numbers that satisfies

1. ${\displaystyle H(mn)=H(m)+H(n)\,}$ (additivity)
2. ${\displaystyle H(m)\leq H(m+1)\,}$ (monotonicity)
3. ${\displaystyle H(2)=1\,}$ (normalization)

Let ƒ be a function on positive integers that satisfies the above three properties. From the additive property, we can show that for any integer n and k,

${\displaystyle f(n^{k})=kf(n).\,}$

Let a, b, and t be any positive integers. There is a unique integer s determined by

${\displaystyle a^{s}\leq b^{t}\leq a^{s+1}.\qquad (1)}$

Therefore,

${\displaystyle s\log _{2}a\leq t\log _{2}b\leq (s+1)\log _{2}a\,}$

and

${\displaystyle {\frac {s}{t}}\leq {\frac {\log _{2}b}{\log _{2}a}}\leq {\frac {s+1}{t}}.}$

On the other hand, by monotonicity,

${\displaystyle f(a^{s})\leq f(b^{t})\leq f(a^{s+1}).\,}$

Using equation (1), one gets

${\displaystyle sf(a)\leq tf(b)\leq (s+1)f(a),\,}$

and

${\displaystyle {\frac {s}{t}}\leq {\frac {f(b)}{f(a)}}\leq {\frac {s+1}{t}}.}$

Hence,

${\displaystyle \left\vert {\frac {f(b)}{f(a)}}-{\frac {\log _{2}(b)}{\log _{2}(a)}}\right\vert \leq {\frac {1}{t}}.}$

Since t can be arbitrarily large, the difference on the left hand side of the above inequality must be zero,

${\displaystyle {\frac {f(b)}{f(a)}}={\frac {\log _{2}(b)}{\log _{2}(a)}}.}$

So,

${\displaystyle f(a)=\mu \log _{2}(a)\,}$

for some constant μ, which must be equal to 1 by the normalization property.

• Rényi entropy
• Min-entropy

• This article incorporates material from Hartley function on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
• This article incorporates material from Derivation of Hartley function on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.