Generalized mean

Plot of several generalized means

M_{p} (1, x)

In mathematics, generalized means (or power mean or Hölder 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

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

$M_{p} (x_{1}, \dots, x_{n}) = {(\frac{1}{n} \sum_{i = 1}^{n} x_{i}^{p})}^{1 / p} .$

(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):

$M_{0} (x_{1}, \dots, x_{n}) = {(\prod_{i = 1}^{n} x_{i})}^{1 / n} .$

Furthermore, for a sequence of positive weights $w i$ we define the weighted power mean as^[2] $M_{p} (x_{1}, \dots, x_{n}) = {(\frac{\sum_{i = 1}^{n} w_{i} x_{i}^{p}}{\sum_{i = 1}^{n} w_{i}})}^{1 / p}$ and when $p = 0$ , it is equal to the weighted geometric mean:

$M_{0} (x_{1}, \dots, x_{n}) = {(\prod_{i = 1}^{n} x_{i}^{w_{i}})}^{1 / \sum_{i = 1}^{n} w_{i}} .$

The unweighted means correspond to setting all $w i = 1$ .

Special cases

For some values of $p$ , the mean $M_{p} (x_{1}, \dots, x_{n})$ corresponds to a well known mean.

File:Generalized Means.svg

A visual depiction of some of the specified cases for

n = 2

.

Harmonic mean:

M_{- 1} (a, b)

.

Geometric mean:

M_{0} (a, b)

.

Arithmetic mean:

M_{1} (a, b)

.

Quadratic mean:

M_{2} (a, b)

.

Name	Exponent	Value
Minimum	$p = - \infty$	$\min {x_{1}, \dots, x_{n}}$
Harmonic mean	$p = - 1$	$\frac{n}{\frac{1}{x_{1}} + \dots + \frac{1}{x_{n}}}$
Geometric mean	$p = 0$	$\sqrt[n]{x_{1} \dots x_{n}}$
Arithmetic mean	$p = 1$	$\frac{x_{1} + \dots + x_{n}}{n}$
Root mean square	$p = 2$	$\sqrt{\frac{x_{1}^{2} + \dots + x_{n}^{2}}{n}}$
Cubic mean	$p = 3$	$\sqrt[3]{\frac{x_{1}^{3} + \dots + x_{n}^{3}}{n}}$
Maximum	$p = + \infty$	$\max {x_{1}, \dots, x_{n}}$

Proof of $\lim_{p \to 0} M_{p} = M_{0}$ (geometric mean)

For the purpose of the proof, we will assume without loss of generality that $w_{i} \in [0, 1]$ and $\sum_{i = 1}^{n} w_{i} = 1.$

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

$M_{p} (x_{1}, \dots, x_{n}) = \exp (\ln [{(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{1 / p}]) = \exp (\frac{\ln (\sum_{i = 1}^{n} w_{i} x_{i}^{p})}{p})$

In the limit $p \to 0$ , we can apply L'Hôpital's rule to the argument of the exponential function. We assume that $p \in ℝ$ but $p \neq 0$ , and that the sum of $w i$ is equal to 1 (without loss in generality);^[4] Differentiating the numerator and denominator with respect to $p$ , we have $\begin{matrix} \lim_{p \to 0} \frac{\ln (\sum_{i = 1}^{n} w_{i} x_{i}^{p})}{p} & = \lim_{p \to 0} \frac{\frac{\sum_{i = 1}^{n} w_{i} x_{i}^{p} \ln x_{i}}{\sum_{j = 1}^{n} w_{j} x_{j}^{p}}}{1} \\ = \lim_{p \to 0} \frac{\sum_{i = 1}^{n} w_{i} x_{i}^{p} \ln x_{i}}{\sum_{j = 1}^{n} w_{j} x_{j}^{p}} \\ = \frac{\sum_{i = 1}^{n} w_{i} \ln x_{i}}{\sum_{j = 1}^{n} w_{j}} \\ = \sum_{i = 1}^{n} w_{i} \ln x_{i} \\ = \ln (\prod_{i = 1}^{n} x_{i}^{w_{i}}) \end{matrix}$

By the continuity of the exponential function, we can substitute back into the above relation to obtain $\lim_{p \to 0} M_{p} (x_{1}, \dots, x_{n}) = \exp (\ln (\prod_{i = 1}^{n} x_{i}^{w_{i}})) = \prod_{i = 1}^{n} x_{i}^{w_{i}} = M_{0} (x_{1}, \dots, x_{n})$ as desired.^[2]

Proof of $\lim_{p \to \infty} M_{p} = M_{\infty}$ and $\lim_{p \to - \infty} M_{p} = M_{- \infty}$

Assume (possibly after relabeling and combining terms together) that $x_{1} \geq \dots \geq x_{n}$ . Then

$\begin{matrix} \lim_{p \to \infty} M_{p} (x_{1}, \dots, x_{n}) & = \lim_{p \to \infty} {(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{1 / p} \\ = x_{1} \lim_{p \to \infty} {(\sum_{i = 1}^{n} w_{i} {(\frac{x_{i}}{x_{1}})}^{p})}^{1 / p} \\ = x_{1} = M_{\infty} (x_{1}, \dots, x_{n}) . \end{matrix}$

The formula for $M_{- \infty}$ follows from $M_{- \infty} (x_{1}, \dots, x_{n}) = \frac{1}{M_{\infty} (1 / x_{1}, \dots, 1 / x_{n})} = x_{n} .$

Properties

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.
$M_{p} (b x_{1}, \dots, b x_{n}) = b \cdot M_{p} (x_{1}, \dots, x_{n})$ .
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 $x 1, ..., x n$ .
$M_{p} (x_{1}, \dots, x_{n \cdot k}) = M_{p} [M_{p} (x_{1}, \dots, x_{k}), M_{p} (x_{k + 1}, \dots, x_{2 \cdot k}), \dots, M_{p} (x_{(n - 1) \cdot k + 1}, \dots, x_{n \cdot k})]$ .
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

File:QM AM GM HM inequality visual proof.svg

Geometric proof without words that max (a,b) > root mean square (RMS) or quadratic 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 $M_{p} (x_{1}, \dots, x_{n}) \leq M_{q} (x_{1}, \dots, x_{n})$ 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 $p$ , $\frac{\partial}{\partial p} M_{p} (x_{1}, \dots, x_{n}) \geq 0$ 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

We will prove the weighted power mean inequality. For the purpose of the proof we will assume the following without loss of generality: $\begin{matrix} w_{i} \in [0, 1] \\ \sum_{i = 1}^{n} w_{i} = 1 \end{matrix}$

The proof for unweighted power means can be easily obtained by substituting $w i = 1/ n$ .

Equivalence of inequalities between means of opposite signs

Suppose an average between power means with exponents $p$ and $q$ holds: ${(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{1 / p} \geq {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{1 / q}$ applying this, then: ${(\sum_{i = 1}^{n} \frac{w_{i}}{x_{i}^{p}})}^{1 / p} \geq {(\sum_{i = 1}^{n} \frac{w_{i}}{x_{i}^{q}})}^{1 / q}$

We raise both sides to the power of −1 (strictly decreasing function in positive reals): ${(\sum_{i = 1}^{n} w_{i} x_{i}^{- p})}^{- 1 / p} = {(\frac{1}{\sum_{i = 1}^{n} w_{i} \frac{1}{x_{i}^{p}}})}^{1 / p} \leq {(\frac{1}{\sum_{i = 1}^{n} w_{i} \frac{1}{x_{i}^{q}}})}^{1 / q} = {(\sum_{i = 1}^{n} w_{i} x_{i}^{- q})}^{- 1 / q}$

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

For any $q > 0$ and non-negative weights summing to 1, the following inequality holds: ${(\sum_{i = 1}^{n} w_{i} x_{i}^{- q})}^{- 1 / q} \leq \prod_{i = 1}^{n} x_{i}^{w_{i}} \leq {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{1 / q} .$

The proof follows from Jensen's inequality, making use of the fact the logarithm is concave: $\log \prod_{i = 1}^{n} x_{i}^{w_{i}} = \sum_{i = 1}^{n} w_{i} \log x_{i} \leq \log \sum_{i = 1}^{n} w_{i} x_{i} .$

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 $\prod_{i = 1}^{n} x_{i}^{w_{i}} \leq \sum_{i = 1}^{n} w_{i} x_{i} .$

Taking $q$ -th powers of the $x i$ yields $\begin{matrix} \prod_{i = 1}^{n} x_{i}^{q \cdot w_{i}} \leq \sum_{i = 1}^{n} w_{i} x_{i}^{q} \\ \prod_{i = 1}^{n} x_{i}^{w_{i}} \leq {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{1 / q} . \end{matrix}$

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:

$\prod_{i = 1}^{n} x_{i}^{- q \cdot w_{i}} \leq \sum_{i = 1}^{n} w_{i} x_{i}^{- q} .$

Of course, taking each side to the power of a negative number $-1/ q$ swaps the direction of the inequality.

$\prod_{i = 1}^{n} x_{i}^{w_{i}} \geq {(\sum_{i = 1}^{n} w_{i} x_{i}^{- q})}^{- 1 / q} .$

Inequality between any two power means

We are to prove that for any $p < q$ the following inequality holds: ${(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{1 / p} \leq {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{1 / q}$ if $p$ is negative, and $q$ is positive, the inequality is equivalent to the one proved above: ${(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{1 / p} \leq \prod_{i = 1}^{n} x_{i}^{w_{i}} \leq {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{1 / q}$

The proof for positive $p$ and $q$ is as follows: Define the following function: $f : R + \to R +$ $f (x) = x^{\frac{q}{p}}$ . $f$ 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: $\begin{matrix} f (\sum_{i = 1}^{n} w_{i} x_{i}^{p}) & \leq \sum_{i = 1}^{n} w_{i} f (x_{i}^{p}) \\ {(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{q / p} & \leq \sum_{i = 1}^{n} w_{i} x_{i}^{q} \end{matrix}$ 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:

${(\sum_{i = 1}^{n} w_{i} x_{i}^{p})}^{1 / p} \leq {(\sum_{i = 1}^{n} w_{i} x_{i}^{q})}^{1 / q}$

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

The power mean could be generalized further to the generalized $f$ -mean:

$M_{f} (x_{1}, \dots, x_{n}) = f^{- 1} (\frac{1}{n} \cdot \sum_{i = 1}^{n} f (x_{i}))$

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

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 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

^ If NM = a and PM = b. AM = AM of a and b, and radius r = AQ = AG.
Using Pythagoras' theorem, QM² = AQ² + AM² ∴ QM = √AQ² + AM² = QM.
Using Pythagoras' theorem, AM² = AG² + GM² ∴ GM = √AM² − AG² = GM.
Using similar triangles, ⁠HM/GM⁠ = ⁠GM/AM⁠ ∴ HM = ⁠GM²/AM⁠ = HM.

References

^ ^a ^b Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).
^ ^a ^b ^c P. S. Bullen: Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177
^ ^a ^b Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).
^ Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).

External links

[5] If NM = a and PM = b. AM = AM of a and b, and radius r = AQ = AG.
Using Pythagoras' theorem, QM² = AQ² + AM² ∴ QM = √AQ² + AM² = QM.
Using Pythagoras' theorem, AM² = AG² + GM² ∴ GM = √AM² − AG² = GM.
Using similar triangles, ⁠HM/GM⁠ = ⁠GM/AM⁠ ∴ HM = ⁠GM²/AM⁠ = HM.

[sykora-1] Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).

[Bullen1-2] P. S. Bullen: Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer, 2003, pp. 175-177

[dC2016-3] Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).

[4] Lua error in Module:Citation/CS1/Configuration at line 2172: attempt to index field '?' (a nil value).

[1]

[2]

[3]

[4]

[note 1]

Generalized mean

Contents

Definition

Special cases

Properties

Generalized mean inequality

Proof of the weighted inequality

Equivalence of inequalities between means of opposite signs

Geometric mean

Inequality between any two power means

Generalized f-mean

Applications

Signal processing

See also

Notes

References

Further reading

External links

Navigation menu

Generalized mean

Definition

Special cases

Properties

Generalized mean inequality

Proof of the weighted inequality

Equivalence of inequalities between means of opposite signs

Geometric mean

Inequality between any two power means

Generalized f-mean

Applications

Signal processing

See also

Notes

References

Further reading

External links

Navigation menu

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