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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.

## DefinitionEdit

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

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

## PropertiesEdit

• 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 Edit

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 $x_1 = x_2 = \dots = x_n$. 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 Edit

• $\lim_{p\to-\infty} M_p(x_1,\dots,x_n) = \min \{x_1,\dots,x_n\}$ - 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 inequalityEdit

### Equivalence of inequalities between means of opposite signsEdit

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 meanEdit

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:

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

since the inequality holds for any q, however small, and, as will be shown later, the expressions on the left and right approximate the geometric mean better as q approaches 0, the limit of the power mean for q approaching 0 is the geometric mean:

$\lim_{q\rightarrow 0}\sqrt[q]{\sum_{i=1}^nw_ix_i^{q}}=\prod_{i=1}^nx_i^{w_i}$

### Inequality between any two power meansEdit

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 maximumEdit

Minimum and maximum are assumed to be the power means with exponents of $-/+\infty$. Thus for any q:

$\min (x_1,x_2,\ldots ,x_n)\leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}\leq \max (x_1,x_2,\ldots ,x_n)$

For maximum the proof is as follows: Assume WLoG that the sequence xi is nonincreasing and no weight is zero.

Then the inequality is equivalent to:

$\sqrt[q]{\sum_{i=1}^nw_ix_i^q}\leq x_1$

After raising both sides to the power of q we get (depending on the sign of q) one of the inequalities:

$\sum_{i=1}^nw_ix_i^q\leq {\color{red} \geq} x_1^q$

≤ for q>0, ≥ for q<0.

After subtracting $w_1x_1$ from the both sides we get:

$\sum_{i=2}^nw_ix_i^q\leq {\color{red} \geq} (1-w_1)x_1^q$

After dividing by $(1-w_1)$:

$\sum_{i=2}^n\frac{w_i}{(1-w_1)}x_i^q\leq {\color{red} \geq} x_1^q$

1 - w1 is nonzero, thus:

$\sum_{i=2}^n\frac{w_i}{(1-w_1)}=1$

Substacting x1q leaves:

$\sum_{i=2}^n\frac{w_i}{(1-w_1)}(x_i^q-x_1^q)\leq {\color{red} \geq} 0$

which is obvious, since x1 is greater or equal to any xi, and thus:

$x_i^q-x_1^q\leq {\color{red} \geq} 0$

For minimum the proof is almost the same, only instead of x1, w1 we use xn, wn, QED.

## Generalized $f$-mean Edit

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 Edit

### Signal processing Edit

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)