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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

File:RMS-AM-GM-HM.gif

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)

See alsoEdit

External linksEdit


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