# Generalized mean

In mathematics, generalized means (or power mean or Hölder mean from Otto Hölder)[1] are a family of functions for aggregating sets of numbers. These include as special cases the Pythagorean means (arithmetic, geometric, and harmonic means).

## Definition

If p is a non-zero real number, and ${\displaystyle x_{1},\dots ,x_{n}}$ are positive real numbers, then the generalized mean or power mean with exponent p of these positive real numbers is[2][3]

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

(See p-norm). For p = 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 a sequence of positive weights wi we define the weighted 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 when p = 0, it is equal to the weighted 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 all wi = 1/n.

## Special cases

A few particular values of p yield special cases with their own names:[4]

minimum
${\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})=\lim _{p\to -\infty }M_{p}(x_{1},\dots ,x_{n})=\min\{x_{1},\dots ,x_{n}\}}$
harmonic mean
${\displaystyle M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}}$
geometric mean ${\displaystyle M_{0}(x_{1},\dots ,x_{n})=\lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})={\sqrt[{n}]{x_{1}\cdot \dots \cdot x_{n}}}}$
arithmetic mean
${\displaystyle M_{1}(x_{1},\dots ,x_{n})={\frac {x_{1}+\dots +x_{n}}{n}}}$
root mean square
${\displaystyle M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}}$
cubic mean
${\displaystyle M_{3}(x_{1},\dots ,x_{n})={\sqrt[{3}]{\frac {x_{1}^{3}+\dots +x_{n}^{3}}{n}}}}$
maximum
${\displaystyle M_{+\infty }(x_{1},\dots ,x_{n})=\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\max\{x_{1},\dots ,x_{n}\}}$
Proof of ${\textstyle \lim _{p\to 0}M_{p}=M_{0}}$ (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 limit p → 0, we can apply L'Hôpital's rule to the argument of the exponential function. We assume that ${\displaystyle p\in \mathbb {R} }$ but p ≠ 0, and that the sum of wi is equal to 1 (without loss in generality);[7] Differentiating the numerator and denominator with respect to p, we have

{\displaystyle {\begin{aligned}\lim _{p\to 0}{\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}&=\lim _{p\to 0}{\frac {\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{j=1}^{n}w_{j}x_{j}^{p}}}{1}}\\&=\lim _{p\to 0}{\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{j=1}^{n}w_{j}x_{j}^{p}}}\\&={\frac {\sum _{i=1}^{n}w_{i}\ln {x_{i}}}{\sum _{j=1}^{n}w_{j}}}\\&=\sum _{i=1}^{n}w_{i}\ln {x_{i}}\\&=\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}\end{aligned}}}

By the continuity of the exponential function, we can substitute back into the above relation to obtain

${\displaystyle \lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}\right)}=\prod _{i=1}^{n}x_{i}^{w_{i}}=M_{0}(x_{1},\dots ,x_{n})}$
as desired.[2]

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 ${\displaystyle x_{1}\geq \dots \geq x_{n}}$. Then

{\displaystyle {\begin{aligned}\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}).\end{aligned}}}

The formula for ${\displaystyle M_{-\infty }}$ follows from

${\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})={\frac {1}{M_{\infty }(1/x_{1},\dots ,1/x_{n})}}=x_{n}.}$

## Properties

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

1. ${\displaystyle \min(x_{1},\dots ,x_{n})\leq M_{p}(x_{1},\dots ,x_{n})\leq \max(x_{1},\dots ,x_{n})}$.
Each generalized mean always lies between the smallest and largest of the x 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. ${\displaystyle M_{p}(bx_{1},\dots ,bx_{n})=b\cdot M_{p}(x_{1},\dots ,x_{n})}$.
Like most means, the generalized mean is a homogeneous function of its arguments x1, ..., xn. That is, if b is a positive real number, then the generalized mean with exponent p of the numbers ${\displaystyle b\cdot x_{1},\dots ,b\cdot x_{n}}$ is equal to b times the generalized mean of the numbers x1, ..., xn.
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 the quasi-arithmetic means, the computation of the mean can be split into computations of equal sized sub-blocks. This enables use of a divide and conquer algorithm to calculate the means, when desirable.

### Generalized mean inequality

In general, if p < q, then

${\displaystyle M_{p}(x_{1},\dots ,x_{n})\leq M_{q}(x_{1},\dots ,x_{n})}$
and the two means are equal if and only if x1 = x2 = ... = xn.

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,

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

## Proof of the weighted inequality

We will prove the weighted power mean inequality. For the purpose of the proof we will assume the following without loss of generality:

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

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

### Equivalence of inequalities between means of opposite signs

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

${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$
applying this, then:
${\displaystyle \left(\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{p}}}\right)^{1/p}\geq \left(\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{q}}}\right)^{1/q}}$

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

${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{-p}\right)^{-1/p}=\left({\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{p}}}}}\right)^{1/p}\leq \left({\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{q}}}}}\right)^{1/q}=\left(\sum _{i=1}^{n}w_{i}x_{i}^{-q}\right)^{-1/q}}$

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.

### Geometric mean

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

${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{-q}\right)^{-1/q}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.}$

The proof follows from Jensen's inequality, making use of the fact the logarithm is concave:

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

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

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

Taking q-th powers of the xi yields

{\displaystyle {\begin{aligned}&\prod _{i=1}^{n}x_{i}^{q{\cdot }w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}\\&\prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.\end{aligned}}}

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

${\displaystyle \prod _{i=1}^{n}x_{i}^{-q{\cdot }w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}^{-q}.}$

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

${\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.}$

### Inequality between any two power means

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

${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$
if p is negative, and q is positive, the inequality is equivalent to the one proved above:
${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$

The proof for positive p and q 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 a second 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 of f, since q > p, so we know f is convex.

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

{\displaystyle {\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})\\[3pt]\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{q/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:

${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$

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

## Generalized f-mean

The power mean could be generalized further to the generalized f-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 with f(x) = log(x). The power mean is obtained for f(x) = xp. Properties of these means are studied in de Carvalho (2016).[3]

## 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 one 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)


## Notes

1. ^ If AC = a and BC = b. OC = AM of a and b, and radius r = QO = OG.
Using Pythagoras' theorem, QC² = QO² + OC² ∴ QC = √QO² + OC² = QM.
Using Pythagoras' theorem, OC² = OG² + GC² ∴ GC = √OC² − OG² = GM.
Using similar triangles, HC/GC = GC/OC ∴ HC = GC²/OC = HM.

## References

1. ^ a b Sýkora, Stanislav (2009). "Mathematical means and averages: basic properties". Stan's Library. III. Castano Primo, Italy: Stan's Library. doi:10.3247/SL3Math09.001.
2. ^ a b c P. S. Bullen: Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177
3. ^ a b de Carvalho, Miguel (2016). "Mean, what do you Mean?". The American Statistician. 70 (3): 764‒776. doi:10.1080/00031305.2016.1148632. hdl:20.500.11820/fd7a8991-69a4-4fe5-876f-abcd2957a88c.
4. ^ (retrieved 2019-08-17)
5. ^ Thompson, Sylvanus P. (1965). Calculus Made Easy. Macmillan International Higher Education. p. 185. ISBN 9781349004874. Retrieved 5 July 2020.
6. ^ Jones, Alan R. (2018). Probability, Statistics and Other Frightening Stuff. Routledge. p. 48. ISBN 9781351661386. Retrieved 5 July 2020.
7. ^ Handbook of Means and Their Inequalities (Mathematics and Its Applications).