This articleneeds additional citations forverification. Please helpimprove this article byadding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Generalized mean" – news ·newspapers ·books ·scholar ·JSTOR(June 2020) (Learn how and when to remove this message)

Inmathematics,generalized means (orpower mean orHölder mean fromOtto Hölder)^[1] are a family of functions for aggregating sets of numbers. These include as special cases thePythagorean means (arithmetic,geometric, andharmonicmeans).

Ifp is a non-zeroreal number, and${\displaystyle x_1,\dots ,x_n}$ arepositive real numbers, then thegeneralized mean orpower mean with exponentp of these positive real numbers is^[2][3]

$${\displaystyle M_p(x_1,\dots ,x_n)={\left(\frac{1}{n}\sum _{i=1}^n{x}_i^p\right)}^{1/p}.}$$

(Seep-norm). Forp = 0 we set it equal to the geometric mean (which is the limit of means with exponents approaching zero, as proved below):

$${\displaystyle M_0(x_1,\dots ,x_n)={\left(\prod _{i=1}^n{x}_i\right)}^{1/n}.}$$

Furthermore, for asequence of positive weightsw_i we define theweighted power mean as^[2]$${\displaystyle M_p(x_1,\dots ,x_n)={\left(\frac{\sum _{i=1}^n{w}_i{x}_i^p}{\sum _{i=1}^n{w}_i}\right)}^{1/p}}$$and whenp = 0, it is equal to theweighted geometric mean:

$${\displaystyle M_0(x_1,\dots ,x_n)={\left(\prod _{i=1}^n{x}_i^{w_i}\right)}^{1/\sum _{i=1}^n{w}_i}.}$$

The unweighted means correspond to setting allw_i = 1.

For some values of${\displaystyle p}$, the mean${\displaystyle M_p(x_1,\dots ,x_n)}$ corresponds to a well known mean.

Name	Exponent	Value
Minimum	${\displaystyle p=-\mathrm{\infty }}$	${\displaystyle min\{x_1,\dots ,x_n\}}$
Harmonic mean	${\displaystyle p=-1}$	${\displaystyle \frac{n}{\frac{1}{x_1}+\cdots +\frac{1}{x_n}}}$
Geometric mean	${\displaystyle p=0}$	${\displaystyle \sqrt[n]{x_1\dots x_n}}$
Arithmetic mean	${\displaystyle p=1}$	${\displaystyle \frac{x_1+\cdots +x_n}{n}}$
Root mean square	${\displaystyle p=2}$	${\displaystyle \sqrt{\frac{x_1^2+\cdots +x_n^2}{n}}}$
Cubic mean	${\displaystyle p=3}$	${\displaystyle \sqrt[3]{\frac{x_1^3+\cdots +x_n^3}{n}}}$
Maximum	${\displaystyle p=+\mathrm{\infty }}$	${\displaystyle max\{x_1,\dots ,x_n\}}$

Proof of$\underset{p\to 0}{lim}{M}_p=M_0$ (geometric mean)

For the purpose of the proof, we will assume without loss of generality that$${\displaystyle w_i\in [0,1]}$$and$${\displaystyle \sum _{i=1}^n{w}_i=1.}$$

We can rewrite the definition of${\displaystyle M_p}$ using the exponential function as

$${\displaystyle M_p(x_1,\dots ,x_n)=\exp \left(\ln \left[{\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}\right]\right)=\exp \left(\frac{\ln \left(\sum _{i=1}^n{w}_i{x}_i^p\right)}{p}\right)}$$

In the limitp → 0, we can applyL'Hôpital's rule to the argument of the exponential function. We assume that${\displaystyle p\in \mathbb{R}}$ butp ≠ 0, and that the sum ofw_i is equal to 1 (without loss in generality);^[4] Differentiating the numerator and denominator with respect top, we have

By the continuity of the exponential function, we can substitute back into the above relation to obtain$${\displaystyle \underset{p\to 0}{lim}{M}_p(x_1,\dots ,x_n)=\exp \left(\ln \left(\prod _{i=1}^n{x}_i^{w_i}\right)\right)=\prod _{i=1}^n{x}_i^{w_i}=M_0(x_1,\dots ,x_n)}$$as desired.^[2]

Proof of$\underset{p\to \mathrm{\infty }}{lim}{M}_p=M_{\mathrm{\infty }}$ and$\underset{p\to -\mathrm{\infty }}{lim}{M}_p=M_{-\mathrm{\infty }}$

Assume (possibly after relabeling and combining terms together) that${\displaystyle x_1\ge \cdots \ge x_n}$. Then

The formula for${\displaystyle M_{-\mathrm{\infty }}}$ follows from$${\displaystyle M_{-\mathrm{\infty }}(x_1,\dots ,x_n)=\frac{1}{M_{\mathrm{\infty }}(1/x_1,\dots ,1/x_n)}=x_n.}$$

Let${\displaystyle x_1,\dots ,x_n}$ be a sequence of positive real numbers, then the following properties hold:^[1]

1. ${\displaystyle min(x_1,\dots ,x_n)\le M_p(x_1,\dots ,x_n)\le max(x_1,\dots ,x_n)}$.
   Each generalized mean always lies between the smallest and largest of thex values.
2. ${\displaystyle M_p(x_1,\dots ,x_n)=M_p(P(x_1,\dots ,x_n))}$, where${\displaystyle P}$ is a permutation operator.
   Each generalized mean is a symmetric function of its arguments; permuting the arguments of a generalized mean does not change its value.
3. ${\displaystyle M_p(b{x}_1,\dots ,b{x}_n)=b\cdot M_p(x_1,\dots ,x_n)}$.
   Like mostmeans, the generalized mean is ahomogeneous function of its argumentsx₁, ...,x_n. That is, ifb is a positive real number, then the generalized mean with exponentp of the numbers${\displaystyle b\cdot x_1,\dots ,b\cdot x_n}$ is equal tob times the generalized mean of the numbersx₁, ...,x_n.
4. ${\displaystyle M_p(x_1,\dots ,x_{n\cdot k})=M_p\left[M_p(x_1,\dots ,x_k),M_p(x_{k+1},\dots ,x_{2\cdot k}),\dots ,M_p(x_{(n-1)\cdot k+1},\dots ,x_{n\cdot k})\right]}$.
   Like thequasi-arithmetic means, the computation of the mean can be split into computations of equal sized sub-blocks. This enables use of adivide and conquer algorithm to calculate the means, when desirable.

In general, ifp <q, then$${\displaystyle M_p(x_1,\dots ,x_n)\le M_q(x_1,\dots ,x_n)}$$and the two means are equal if and only ifx₁ =x₂ = ... =x_n.

The inequality is true for real values ofp andq, as well as positive and negative infinity values.

It follows from the fact that, for all realp,$${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }p}{M}_p(x_1,\dots ,x_n)\ge 0}$$which can be proved usingJensen's inequality.

In particular, forp in{−1, 0, 1}, the generalized mean inequality implies thePythagorean means inequality as well as theinequality of arithmetic and geometric means.

We will prove the weighted power mean inequality. For the purpose of the proof we will assume the followingwithout loss of generality:$${\displaystyle \begin{array}{r}{w}_i\in [0,1]\\ \sum _{i=1}^n{w}_i=1\end{array}}$$

The proof for unweighted power means can be easily obtained by substitutingw_i = 1/n.

Suppose an average between power means with exponentsp andq holds:$${\displaystyle {\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}\ge {\left(\sum _{i=1}^n{w}_i{x}_i^q\right)}^{1/q}}$$applying this, then:$${\displaystyle {\left(\sum _{i=1}^n\frac{w_i}{x_i^p}\right)}^{1/p}\ge {\left(\sum _{i=1}^n\frac{w_i}{x_i^q}\right)}^{1/q}}$$

We raise both sides to the power of −1 (strictly decreasing function in positive reals):$${\displaystyle {\left(\sum _{i=1}^n{w}_i{x}_i^{-p}\right)}^{-1/p}={\left(\frac{1}{\sum _{i=1}^n{w}_i\frac{1}{x_i^p}}\right)}^{1/p}\le {\left(\frac{1}{\sum _{i=1}^n{w}_i\frac{1}{x_i^q}}\right)}^{1/q}={\left(\sum _{i=1}^n{w}_i{x}_i^{-q}\right)}^{-1/q}}$$

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.

For anyq > 0 and non-negative weights summing to 1, the following inequality holds:$${\displaystyle {\left(\sum _{i=1}^n{w}_i{x}_i^{-q}\right)}^{-1/q}\le \prod _{i=1}^n{x}_i^{w_i}\le {\left(\sum _{i=1}^n{w}_i{x}_i^q\right)}^{1/q}.}$$

The proof follows fromJensen's inequality, making use of the fact thelogarithm is concave:$${\displaystyle \log \prod _{i=1}^n{x}_i^{w_i}=\sum _{i=1}^n{w}_i\log x_i\le \log \sum _{i=1}^n{w}_i{x}_i.}$$

By applying theexponential function to both sides and observing that as a strictly increasing function it preserves the sign of the inequality, we get$${\displaystyle \prod _{i=1}^n{x}_i^{w_i}\le \sum _{i=1}^n{w}_i{x}_i.}$$

Takingq-th powers of thex_i yields

Thus, we are done for the inequality with positiveq; the case for negatives is identical but for the swapped signs in the last step:

$${\displaystyle \prod _{i=1}^n{x}_i^{-q\cdot w_i}\le \sum _{i=1}^n{w}_i{x}_i^{-q}.}$$

Of course, taking each side to the power of a negative number-1/q swaps the direction of the inequality.

$${\displaystyle \prod _{i=1}^n{x}_i^{w_i}\ge {\left(\sum _{i=1}^n{w}_i{x}_i^{-q}\right)}^{-1/q}.}$$

We are to prove that for anyp <q the following inequality holds:$${\displaystyle {\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}\le {\left(\sum _{i=1}^n{w}_i{x}_i^q\right)}^{1/q}}$$ifp is negative, andq is positive, the inequality is equivalent to the one proved above:$${\displaystyle {\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}\le \prod _{i=1}^n{x}_i^{w_i}\le {\left(\sum _{i=1}^n{w}_i{x}_i^q\right)}^{1/q}}$$

The proof for positivep andq is as follows: Define the following function:f :R₊ →R₊${\displaystyle f(x)=x^{\frac{q}{p}}}$.f is a power function, so it does have asecond derivative:$${\displaystyle f^{″}(x)=\left(\frac{q}{p}\right)\left(\frac{q}{p}-1\right)x^{\frac{q}{p}-2}}$$which is strictly positive within the domain off, sinceq >p, so we knowf is convex.

Using this, and the Jensen's inequality we get:after raising both side to the power of1/q (an increasing function, since1/q is positive) we get the inequality which was to be proven:

$${\displaystyle {\left(\sum _{i=1}^n{w}_i{x}_i^p\right)}^{1/p}\le {\left(\sum _{i=1}^n{w}_i{x}_i^q\right)}^{1/q}}$$

Using the previously shown equivalence we can prove the inequality for negativep andq by replacing them with−q and−p, respectively.

The power mean could be generalized further to thegeneralizedf-mean:

$${\displaystyle M_f(x_1,\dots ,x_n)=f^{-1}\left(\frac{1}{n}\cdot \sum _{i=1}^{n}f(x_i)\right)}$$

This covers the geometric mean without using a limit withf(x) = log(x). The power mean is obtained forf(x) =x^p. Properties of these means are studied in de Carvalho (2016).^[3]

A power mean serves a non-linearmoving average which is shifted towards small signal values for smallp and emphasizes big signal values for bigp. Given an efficient implementation of amoving arithmetic mean calledsmooth one can implement a moving power mean according to the followingHaskell code.

powerSmooth::Floatinga=>([a]->[a])->a->[a]->[a]powerSmoothsmoothp=map(**recipp).smooth.map(**p)

• For bigp it can serve as anenvelope detector on arectified signal.
• For smallp it can serve as abaseline detector on amass spectrum.

• Bullen, P. S. (2003). "Chapter III - The Power Means".Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer. pp. 175–265.

• Power mean at MathWorld
• Examples of Generalized Mean
• Aproof of the Generalized Mean onPlanetMath

• Means
• Inequalities (mathematics)

• Articles with short description
• Short description matches Wikidata
• Articles needing additional references from June 2020
• All articles needing additional references
• Articles with example Haskell code