Generalized mean

In mathematics, generalised means are a family of functions for aggregating sets of numbers, that include as special cases the arithmetic, geometric, and harmonic means. The generalised mean is also known as power mean or Hölder mean (named after Otto Hölder).

Definition

If p is a non-zero real number, we can define the generalised mean or power mean with exponent p of the positive real numbers $x_{1},\dots ,x_{n}$ as:

$M_{p}(x_{1},\dots ,x_{n})=\left({\frac {1}{n}}\sum _{i=1}^{n}x_{i}^{p}\right)^{\frac {1}{p}}$

Note the relationship to the p-norm. For p = 0 we assume that it's equal to the geometric mean (which is, in fact, the limit of means with exponents approaching zero, as proved below for the general case):

$M_{0}(x_{1},\dots ,x_{n})={\sqrt[{n}]{\prod _{i=1}^{n}x_{i}}}$

Furthermore, for a sequence of positive weights w_i with sum $\sum w_{i}=1$ we define the weighted power mean as:

${\begin{aligned}M_{p}(x_{1},\dots ,x_{n})&=\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{\frac {1}{p}}\\M_{0}(x_{1},\dots ,x_{n})&=\prod _{i=1}^{n}x_{i}^{w_{i}}\end{aligned}}$

The unweighted means correspond to setting all w_i = 1/n. For exponents equal to positive or negative infinity the means are maximum and minimum, respectively, regardless of weights (and they are actually the limit points for exponents approaching the respective extremes, as proved below):

${\begin{aligned}M_{\infty }(x_{1},\dots ,x_{n})&=\max(x_{1},\dots ,x_{n})\\M_{-\infty }(x_{1},\dots ,x_{n})&=\min(x_{1},\dots ,x_{n})\end{aligned}}$

Proof of $\textstyle \lim _{p\to 0}M_{p}=M_{0}$ (geometric mean)

We can rewrite the definition of Mp using the exponential function $M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left[\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\right]}\right)}=\exp {\left({\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}\right)}$ In the limit p → 0, we can apply L'Hôpital's rule to the exponential component, $\lim _{p\to 0}{\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}=\lim _{p\to 0}{\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}=\sum _{i=1}^{n}w_{i}\ln {x_{i}}=\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}$ By the continuity of the exponential function, we can substitute back into the above relation to obtain $\lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}\right)}=\prod _{i=1}^{n}x_{i}^{w_{i}}=M_{0}(x_{1},\dots ,x_{n})$ as desired.

Proof of $\textstyle \lim _{p\to \infty }M_{p}=M_{\infty }$ and $\textstyle \lim _{p\to -\infty }M_{p}=M_{-\infty }$

Assume (possibly after relabeling and combining terms together) that $x_{1}\geq \dots \geq x_{n}$. Then $\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\lim _{p\to \infty }\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}=x_{1}\lim _{p\to \infty }\left(\sum _{i=1}^{n}w_{i}\left({\frac {x_{i}}{x_{1}}}\right)^{p}\right)^{1/p}=x_{1}=M_{\infty }(x_{1},\dots ,x_{n}).$ The formula for $M_{-\infty }$ follows from $M_{-\infty }(x_{1},\dots ,x_{n})={\frac {1}{M_{\infty }(1/x_{1},\dots ,1/x_{n})}}.$

Properties

• Like most means, the generalized mean is a homogeneous function of its arguments x₁, ..., x_n. That is, if b is a positive real number, then the generalized mean with exponent p of the numbers $b\cdot x_{1},\dots ,b\cdot x_{n}$ is equal to b times the generalized mean of the numbers x₁, …, x_n.
• Like the quasi-arithmetic means, the computation of the mean can be split into computations of equal sized sub-blocks.

$M_{p}(x_{1},\dots ,x_{n\cdot k})=M_{p}(M_{p}(x_{1},\dots ,x_{k}),M_{p}(x_{k+1},\dots ,x_{2\cdot k}),\dots ,M_{p}(x_{(n-1)\cdot k+1},\dots ,x_{n\cdot k}))$

Generalized mean inequality

In general,

if p < q, then $M_{p}(x_{1},\dots ,x_{n})\leq M_{q}(x_{1},\dots ,x_{n})$

and the two means are equal if and only if x₁ = x₂ = ... = x_n.

The inequality is true for real values of p and q, as well as positive and negative infinity values.

It follows from the fact that, for all real p,

${\frac {\partial }{\partial p}}M_{p}(x_{1},\dots ,x_{n})\geq 0$

which can be proved using Jensen's inequality.

In particular, for p in {−1, 0, 1}, the generalized mean inequality implies the Pythagorean means inequality as well as the inequality of arithmetic and geometric means.

Special cases

A visual depiction of some of the specified cases for n=2.

$M_{-\infty }(x_{1},\dots ,x_{n})=\lim _{p\to -\infty }M_{p}(x_{1},\dots ,x_{n})=\min\{x_{1},\dots ,x_{n}\}$	minimum
$M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}$	harmonic mean
$M_{0}(x_{1},\dots ,x_{n})=\lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})={\sqrt[{n}]{x_{1}\cdot \dots \cdot x_{n}}}$	geometric mean
$M_{1}(x_{1},\dots ,x_{n})={\frac {x_{1}+\dots +x_{n}}{n}}$	arithmetic mean
$M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}$	quadratic mean, a.k.a. root mean square
$M_{+\infty }(x_{1},\dots ,x_{n})=\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\max\{x_{1},\dots ,x_{n}\}$	maximum

Proof of power means inequality

We will prove weighted power means inequality, for the purpose of the proof we will assume the following without loss of generality:

${\begin{aligned}w_{i}\in [0;1]\\\sum _{i=1}^{n}w_{i}=1\end{aligned}}$

Proof for unweighted power means is easily obtained by substituting w_i = 1/n.

Equivalence of inequalities between means of opposite signs

Suppose an average between power means with exponents p and q holds:

${\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\geq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}$

applying this, then:

${\sqrt[{p}]{\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{p}}}}}\geq {\sqrt[{q}]{\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{q}}}}}$

We raise both sides to the power of −1 (strictly decreasing function in positive reals):

${\sqrt[{-p}]{\sum _{i=1}^{n}w_{i}x_{i}^{-p}}}={\sqrt[{p}]{\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{p}}}}}}\leq {\sqrt[{q}]{\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{q}}}}}}={\sqrt[{-q}]{\sum _{i=1}^{n}w_{i}x_{i}^{-q}}}$

We get the inequality for means with exponents −p and −q, and we can use the same reasoning backwards, thus proving the inequalities to be equivalent, which will be used in some of the later proofs.

Geometric mean

For any q > 0, and non-negative weights summing to 1, the following inequality holds

${\begin{aligned}{\sqrt[{-q}]{\sum _{i=1}^{n}w_{i}x_{i}^{-q}}}&\leq \prod _{i=1}^{n}x_{i}^{w_{i}}&\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}\\\end{aligned}}$

The proof is as follows. From Jensen's inequality, making use of the fact the logarithmic function is concave:

${\begin{aligned}\log \left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)=\sum _{i=1}^{n}w_{i}\log(x_{i})&\leq \log \left(\sum _{i=1}^{n}w_{i}x_{i}\right)\\\end{aligned}}$

By applying the exponential function to both sides and observing that as a strictly increasing function it preserves the sign of the inequality, we get

$\prod _{i=1}^{n}x_{i}^{w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}$

and taking qth powers of the x_i, we are done for the inequality with positive q, and the case for negatives is identical.

Inequality between any two power means

We are to prove that for any p < q the following inequality holds:

${\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}$

if p is negative, and q is positive, the inequality is equivalent to the one proved above:

${\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}$

The proof for positive p and q is as follows: Define the following function: f : R₊ → R₊ $f(x)=x^{\frac {q}{p}}$. f is a power function, so it does have a second derivative:

$f''(x)=\left({\frac {q}{p}}\right)\left({\frac {q}{p}}-1\right)x^{{\frac {q}{p}}-2}$

which is strictly positive within the domain of f, since q > p, so we know f is convex.

Using this, and the Jensen's inequality we get:

${\begin{aligned}f\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)&\leq \sum _{i=1}^{n}w_{i}f(x_{i}^{p})\\{\sqrt[{\frac {p}{q}}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}&\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}\end{aligned}}$

after raising both side to the power of 1/q (an increasing function, since 1/q is positive) we get the inequality which was to be proven:

${\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}$

Using the previously shown equivalence we can prove the inequality for negative p and q by substituting them with, respectively, −q and −p, QED.

Generalized f-mean

Main article: Generalized ƒ-mean

The power mean could be generalized further to the generalized f-mean:

$M_{f}(x_{1},\dots ,x_{n})=f^{-1}\left({{\frac {1}{n}}\cdot \sum _{i=1}^{n}{f(x_{i})}}\right)$

Which covers the geometric mean without using a limit with f(x) = log(x). The power mean is obtained for f(x) = x^p.

Applications

Signal processing

A power mean serves a non-linear moving average which is shifted towards small signal values for small p and emphasizes big signal values for big p. Given an efficient implementation of a moving arithmetic mean called smooth you can implement a moving power mean according to the following Haskell code.

 powerSmooth :: Floating a => ([a] -> [a]) -> a -> [a] -> [a]
 powerSmooth smooth p = map (** recip p) . smooth . map (**p)

• For big p it can serve an envelope detector on a rectified signal.
• For small p it can serve an baseline detector on a mass spectrum.

See also

• Arithmetic mean
• Arithmetic-geometric mean
• Average
• Geometric mean
• Harmonic mean
• Heronian mean
• Inequality of arithmetic and geometric means
• Lehmer mean – also a mean related to powers
• Root mean square

External links

• Power mean at MathWorld
• Examples of Generalized Mean
• A proof of the Generalized Mean on PlanetMath