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(Top) 1Definition 2Special cases 3Properties 3.1Generalized mean inequality 4Proof of the weighted inequality 4.1Equivalence of inequalities between means of opposite signs 4.2Geometric mean 4.3Inequality between any two power means 5Generalized f-mean 6Applications 6.1Signal processing 7See also 8Notes 9References 10Further reading 11External links

Generalized mean

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Plot of several generalized means

M_{p}(1,x)

Inmathematics, generalized means (orpower meanorHölder 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[edit]

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

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

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):

M_{0}(x_{1},\dots ,x_{n})=\left(\prod _{i=1}^{n}x_{i}\right)^{1/n}.

M_{0}(x_{1},\dots ,x_{n})=\left(\prod _{i=1}^{n}x_{i}\right)^{1/n}.

Furthermore, for a sequence of positive weights $w i$ we define the weighted power meanas^[2]

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}}

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:

M_{0}(x_{1},\dots ,x_{n})=\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)^{1/\sum _{i=1}^{n}w_{i}}.

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 $w i = 1/n$ .

Special cases[edit]

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

minimum: $M_{-\infty }(x_{1},\dots ,x_{n})=\lim _{p\to -\infty }M_{p}(x_{1},\dots ,x_{n})=\min\{x_{1},\dots ,x_{n}\}$
A visual depiction of some of the specified cases for $n = 2$ with $a = x 1 = M \infty$ and $b = x 2 = M -\infty$ : harmonic mean, $H = M -1 (a, b)$ , geometric mean, $G = M 0 (a, b)$ arithmetic mean, $A = M 1 (a, b)$ quadratic mean, $Q = M 2 (a, b)$ harmonic mean: $M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}$
geometric mean $M_{0}(x_{1},\dots ,x_{n})=\lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})={\sqrt[{n}]{x_{1}\cdot \dots \cdot x_{n}}}$
arithmetic mean: $M_{1}(x_{1},\dots ,x_{n})={\frac {x_{1}+\dots +x_{n}}{n}}$
root mean square or quadratic mean^[5]^[6]: $M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}$
cubic mean: $M_{3}(x_{1},\dots ,x_{n})={\sqrt[{3}]{\frac {x_{1}^{3}+\dots +x_{n}^{3}}{n}}}$
maximum: $M_{+\infty }(x_{1},\dots ,x_{n})=\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\max\{x_{1},\dots ,x_{n}\}$

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

w_{i}\in [0,1]

w_{i}\in [0,1]

and

\sum _{i=1}^{n}w_{i}=1.

\sum _{i=1}^{n}w_{i}=1.

We can rewrite the definition of $M_{p}$ using the exponential function as

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)}

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 $p \to 0$ , we can apply L'Hôpital's rule to the argument of the exponential function. We assume that $p\in \mathbb {R}$ but $p \neq 0$ , and that the sum of $w i$ is equal to 1 (without loss in generality);^[7] Differentiating the numerator and denominator with respect to $p$ , we have

{\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}}

{\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

\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})

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

{\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}}

{\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 $M_{-\infty }$ follows from

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

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

Properties[edit]

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

$\min(x_{1},\dots ,x_{n})\leq M_{p}(x_{1},\dots ,x_{n})\leq \max(x_{1},\dots ,x_{n})$ .
Each generalized mean always lies between the smallest and largest of the $x$ values.
$M_{p}(x_{1},\dots ,x_{n})=M_{p}(P(x_{1},\dots ,x_{n}))$ , where $P$ 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.
$M_{p}(bx_{1},\dots ,bx_{n})=b\cdot M_{p}(x_{1},\dots ,x_{n})$ .
Like most means, the generalized mean is a homogeneous function of its arguments $x 1, ..., x n$ . 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 $x 1, ..., x n$ .
$M_{p}(x_{1},\dots ,x_{n\cdot k})=M_{p}\left[M_{p}(x_{1},\dots ,x_{k}),M_{p}(x_{k+1},\dots ,x_{2\cdot k}),\dots ,M_{p}(x_{(n-1)\cdot k+1},\dots ,x_{n\cdot k})\right]$ .
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[edit]

Geometric proof without words that max (a,b) > root mean square (RMS)orquadratic mean (QM) > arithmetic mean (AM) > geometric mean (GM) > harmonic mean (HM) > min (a,b) of two distinct positive numbers a and b ^{[note 1]}

In general, if $p < q$ , then

M_{p}(x_{1},\dots ,x_{n})\leq M_{q}(x_{1},\dots ,x_{n})

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

x 1 = x 2 = ... = x n

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 $p$ ,

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

{\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[edit]

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

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

{\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 $w i = 1/ n$ .

Equivalence of inequalities between means of opposite signs[edit]

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

\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}

\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:

\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}

\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):

\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}

\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[edit]

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

\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}.

\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:

\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}.

\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

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

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

Taking $q$ -th powers of the $x i$ yields

{\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}}

{\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:

\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}^{-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.

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

\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[edit]

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

\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}

\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}

p

is negative, and

q

is positive, the inequality is equivalent to the one proved above:

\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}

\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: $f : R + \to R +$ ${\displaystyle f($ . $f$ is a power function, so it does have a second derivative:

{\displaystyle f''(

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:

{\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}}

{\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:

\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}

\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[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)

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 $f (x) = x p$ . Properties of these means are studied in de Carvalho (2016).^[3]

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 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)

For big $p$ it can serve as an envelope detector on a rectified signal.
For small $p$ it can serve as a baseline detector on a mass spectrum.

Notes[edit]

^ If AC = a and BC = b. OC = AMofa 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[edit]

^ ^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.

^ ^a ^b ^c P. S. Bullen: Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177

^ ^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.

^ Weisstein, Eric W. "Power Mean". MathWorld. (retrieved 2019-08-17)

^ Thompson, Sylvanus P. (1965). Calculus Made Easy. Macmillan International Higher Education. p. 185. ISBN 9781349004874. Retrieved 5 July 2020.^{[permanent dead link]}

^ Jones, Alan R. (2018). Probability, Statistics and Other Frightening Stuff. Routledge. p. 48. ISBN 9781351661386. Retrieved 5 July 2020.

^ Handbook of Means and Their Inequalities (Mathematics and Its Applications).

External links[edit]

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