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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.


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}. $


  • 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} $


$ \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)

See alsoEdit

External linksEdit

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