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A generalized mean, also known as power mean or Hölder mean, is an abstraction of the Pythagorean means including arithmetic, geometric, and harmonic means.

1 Definition
2 Properties
- 2.1 Generalized mean inequality
3 Special cases
4 Proof of power means inequality
5 Generalized f-mean
6 Applications
- 6.1 Signal processing
7 See also
8 External links

Definition

If $p$ is a non-zero real number, we can define the generalized mean with exponent $p$ of the positive real numbers $x_1,\dots,x_n$ as

$M_p(x_1,\dots,x_n) = \left( \frac{1}{n} \cdot \sum_{i=1}^n x_{i}^p \right)^{1/p}.$

Properties

Like most means, the generalized mean is a homogeneous function of its arguments $x_1,\dots,x_n$ . 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 $x_1,\dots, x_n$ .
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 $M_p(x_1,\dots,x_n) \le M_q(x_1,\dots,x_n)$ and the two means are equal if and only if $x_1 = x_2 = \cdots = x_n$ . This follows from the fact that

$\forall p\in\mathbb{R}\ \frac{\partial M_p(x_1,\dots,x_n)}{\partial p}\geq 0,$

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.

$\lim_{p\to-\infty} M_p(x_1,\dots,x_n) = \min \{x_1,\dots,x_n\}$ - minimum,
$M_{-1}(x_1,\dots,x_n) = \frac{n}{\frac{1}{x_1}+\dots+\frac{1}{x_n}}$ - harmonic mean,
$\lim_{p\to0} M_p(x_1,\dots,x_n) = \sqrt[n]{x_1\cdot\dots\cdot x_n}$ - 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,
$\lim_{p\to\infty} M_p(x_1,\dots,x_n) = \max \{x_1,\dots,x_n\}$ - maximum.

Proof of power means inequality

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

$w_i\in (0;1]$

$\sum_{i=1}^nw_i=1$

Proof for unweighted power means is easily obtained by substituting $w_i=\frac{1}{n}$ .

Equivalence of inequalities between means of opposite signs

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

$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$

then:

$\sqrt[p]{\sum_{i=1}^n\frac{w_i}{x_i^p}}\leq \sqrt[q]{\sum_{i=1}^n\frac{w_i}{x_i^q}}$

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

$\sqrt[-p]{\sum_{i=1}^nw_ix_i^{-p}}=\sqrt[p]{\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^p}}}\geq \sqrt[q]{\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^q}}}=\sqrt[-q]{\sum_{i=1}^nw_ix_i^{-q}}$

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 the inequality between mean with exponent q and geometric mean can be transformed in the following way:

$\prod_{i=1}^nx_i^{w_i} \leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$

$\sqrt[q]{\sum_{i=1}^nw_ix_i^q}\leq \prod_{i=1}^nx_i^{w_i}$

(the first inequality is to be proven for positive q, and the latter otherwise)

We raise both sides to the power of q:

$\prod_{i=1}^nx_i^{w_i\cdot q} \leq \sum_{i=1}^nw_ix_i^q$

in both cases we get the inequality between weighted arithmetic and geometric means for the sequence $x_i^q$ , which can be proved by Jensen's inequality, making use of the fact the logarithmic function is concave:

$\sum_{i=1}^nw_i\log(x_i) \leq \log(\sum_{i=1}^nw_ix_i)$

$\log(\prod_{i=1}^nx_i^{w_i}) \leq \log(\sum_{i=1}^nw_ix_i)$

By applying (strictly increasing) exp function to both sides we get the inequality:

$\prod_{i=1}^nx_i^{w_i} \leq \sum_{i=1}^nw_ix_i$

Thus for any positive q it is true that:

$\sqrt[-q]{\sum_{i=1}^nw_ix_i^{-q}}\leq \prod_{i=1}^nx_i^{w_i} \leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$

thus we have proved the inequality between geometric mean and any power mean.

Geometric mean as a limit

Furthermore, we can prove that the geometric mean is the limit of power means for exponent approaching zero. Firstly, we will prove the limit:

$\lim_{p\to0} \frac{\log\left(\sum_{i=1}^nw_ix_i^p\right)}{p}=\sum_{i=1}^nw_i\log(x_i)$

It's easy to conclude that the limits of both the numerator and the denominator are both 0, so we can use L'Hôpital's rule:

$\lim_{p\to 0} \frac{\log\left(\sum_{i=1}^nw_ix_i^p\right)}{p}=\lim_{p\to 0}\frac{1}{\sum_{i=1}^nw_ix_i^p}\cdot\left(\sum_{i=1}^nw_ix_i^p\right)'=$

$=\frac{1}{\sum_{i=1}^nw_i}\cdot \lim_{p\to 0}\sum_{i=1}^n(w_i\cdot\log(x_i)\cdot x_i^p)=\sum_{i=1}^nw_i\log(x_i)$

Then we make use of the exponential function's continuity:

$\lim_{p \to 0} \sqrt[p]{\sum_{i=1}^nw_ix_i^p}=\lim_{p \to 0} \exp\left(\frac{\log\left(\sum_{i=1}^nw_ix_i^p\right)}{p}\right)=\exp\left(\lim_{p \to 0} \frac{\log\left(\sum_{i=1}^nw_ix_i^p\right)}{p}\right)=\exp\left(\sum_{i=1}^nw_i\log(x_i)\right)=\prod_{i=1}^nx_i^{w_i}$

which was to be proven.

Inequality between any two power means

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

$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$

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

$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq \prod_{i=1}^nx_i^{w_i} \leq\sqrt[q]{\sum_{i=1}^nw_ix_i^q}$

The proof for positive p and q is as follows: Define the following function: $f:{\mathbb R_+}\rightarrow{\mathbb R_+},$ $f(x)=x^{\frac{q}{p}}$ . f is a power function, so it does have a second derivative: $f''(x)=(\frac{q}{p})(\frac{q}{p}-1)x^{\frac{q}{p}-2},$ 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:

$f(\sum_{i=1}^nw_ix_i^p)\leq\sum_{i=1}^nw_if(x_i^p)$

$\sqrt[\frac{p}{q}]{\sum_{i=1}^nw_ix_i^p}\leq\sum_{i=1}^nw_ix_i^q$

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:

$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq\sqrt[q]{\sum_{i=1}^nw_ix_i^q}$

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

Minimum and maximum

Minimum and maximum are the limits of power means at, respectively, $- \infty$ and $+\infty$ . The proof is as follows:

Suppose without loss of generality that x₁ is the largest, while x_n is the smallest of x_i. First, using the squeeze theorem we will prove that:

$\lim_{p \to \infty}\left(\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_i^p}{x_1^p}\right)\right)=0$

It suffices to notice that for positive p the inequalities hold:

$\frac{1}{p}\ln(w_1)=\frac{1}{p}\ln\left(\frac{w_1x_1^p}{x_1^p}\right)\leq\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_i^p}{x_1^p}\right)\leq\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_1^p}{x_1^p}\right)=\ln(1)=0$

Then, making use of the limit:

$\lim_{p \to \infty}\frac{1}{p}\ln\left(\sum_{i=1}^nw_ix_i^p\right)=\lim_{p \to \infty}\frac{1}{p}\ln\left(x_1^p\cdot\frac{\sum_{i=1}^nw_ix_i^p}{x_1^p}\right)=\lim_{p \to \infty}\left(\frac{\ln(x_1^p)}{p}\right)+\lim_{p \to \infty}\left(\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_i^p}{x_1^p}\right)\right)=\ln(x_1)+0=\ln(x_1)$

and finally, we use the fact that the exponential function is continuous:

$\lim_{p \to \infty}\sqrt[p]{\sum_{i=1}^nw_ix_i^p}=\lim_{p \to \infty}\exp\left(\frac{1}{p}\ln\left(\sum_{i=1}^nw_ix_i^p\right)\right)=\exp\left(\lim_{p \to \infty}\frac{1}{p}\ln\left(\sum_{i=1}^nw_ix_i^p\right)\right)=x_1$

Similarly, for negative p:

$\lim_{p \to -\infty}\left(\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_n^p}{x_n^p}\right)\right)=0$

since (for p<0):

$\frac{1}{p}\ln(w_n)=\frac{1}{p}\ln\left(\frac{w_nx_n^p}{x_n^p}\right) \geq\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_i^p}{x_n^p}\right)\geq\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_n^p}{x_n^p}\right)=\ln(1)=0$

Thus:

$\lim_{p \to-\infty}\frac{1}{p}\ln\left(\sum_{i=1}^nw_ix_i^p\right)=\lim_{p \to -\infty}\left(\frac{\ln(x_n^p)}{p}\right)+\lim_{p \to -\infty}\left(\frac{1}{p}\ln\left(\frac{\sum_{i=1}^nw_ix_i^p}{x_n^p}\right)\right)=\ln(x_n)$

and again, by continuity of the exp function:

$\lim_{p \to-\infty}\sqrt[p]{\sum_{i=1}^nw_ix_i^p}=\exp\left(\lim_{p \to -\infty}\frac{1}{p}\ln\left(\sum_{i=1}^nw_ix_i^p\right)\right)=x_n$

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 e.g. the geometric mean without using a limit. The power mean is obtained for $f\left(x\right)=x^p$ .

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.

External links

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Generalized mean

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