Plot of several generalized means ${\displaystyle M_{p}(1,x)}$.

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 ${\displaystyle x_{1},\dots ,x_{n}}$ are positive real numbers, then the generalized mean or power mean with exponent p of these positive real numbers is^[2][3]

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

(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):

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

Furthermore, for a sequence of positive weights w_i we define the weighted power mean as^[2]

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

and when p = 0, it is equal to the weighted geometric mean:

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

The unweighted means correspond to setting all w_i = 1/n.

Special cases

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

minimum

${\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})=\lim _{p\to -\infty }M_{p}(x_{1},\dots ,x_{n})=\min\{x_{1},\dots ,x_{n}\}}$

A visual depiction of some of the specified cases for n = 2 with a = x₁ = M_∞ and b = x₂ = M_−∞:

  harmonic mean, H = M₋₁(a, b),

  geometric mean, G = M₀(a, b)

  arithmetic mean, A = M₁(a, b)

  quadratic mean, Q = M₂(a, b)

harmonic mean

${\displaystyle M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}}$

geometric mean ${\displaystyle M_{0}(x_{1},\dots ,x_{n})=\lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})={\sqrt[{n}]{x_{1}\cdot \dots \cdot x_{n}}}}$

arithmetic mean

${\displaystyle M_{1}(x_{1},\dots ,x_{n})={\frac {x_{1}+\dots +x_{n}}{n}}}$

root mean square
or quadratic mean^[5][6]

${\displaystyle M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}}$

cubic mean

${\displaystyle M_{3}(x_{1},\dots ,x_{n})={\sqrt[{3}]{\frac {x_{1}^{3}+\dots +x_{n}^{3}}{n}}}}$

maximum

${\displaystyle M_{+\infty }(x_{1},\dots ,x_{n})=\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})=\max\{x_{1},\dots ,x_{n}\}}$

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

$${\displaystyle w_{i}\in [0,1]}$$

and

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

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

$${\displaystyle M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left[\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\right]}\right)}=\exp {\left({\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}\right)}}$$

In the limit p → 0, we can apply L'Hôpital's rule to the argument of the exponential function. We assume that p ∈ ℝ but p ≠ 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

$${\displaystyle {\begin{aligned}\lim _{p\to 0}{\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}&=\lim _{p\to 0}{\frac {\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{j=1}^{n}w_{j}x_{j}^{p}}}{1}}\\&=\lim _{p\to 0}{\frac {\sum _{i=1}^{n}w_{i}x_{i}^{p}\ln {x_{i}}}{\sum _{j=1}^{n}w_{j}x_{j}^{p}}}\\&={\frac {\sum _{i=1}^{n}w_{i}\ln {x_{i}}}{\sum _{j=1}^{n}w_{j}}}\\&=\sum _{i=1}^{n}w_{i}\ln {x_{i}}\\&=\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}\end{aligned}}}$$

By the continuity of the exponential function, we can substitute back into the above relation to obtain

$${\displaystyle \lim _{p\to 0}M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left(\prod _{i=1}^{n}x_{i}^{w_{i}}\right)}\right)}=\prod _{i=1}^{n}x_{i}^{w_{i}}=M_{0}(x_{1},\dots ,x_{n})}$$

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

$${\displaystyle {\begin{aligned}\lim _{p\to \infty }M_{p}(x_{1},\dots ,x_{n})&=\lim _{p\to \infty }\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\\&=x_{1}\lim _{p\to \infty }\left(\sum _{i=1}^{n}w_{i}\left({\frac {x_{i}}{x_{1}}}\right)^{p}\right)^{1/p}\\&=x_{1}=M_{\infty }(x_{1},\dots ,x_{n}).\end{aligned}}}$$

The formula for ${\displaystyle M_{-\infty }}$ follows from

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

Properties

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

1. ${\displaystyle \min(x_{1},\dots ,x_{n})\leq M_{p}(x_{1},\dots ,x_{n})\leq \max(x_{1},\dots ,x_{n})}$.
   Each generalized mean always lies between the smallest and largest of the x values.
2. ${\displaystyle M_{p}(x_{1},\dots ,x_{n})=M_{p}(P(x_{1},\dots ,x_{n}))}$, where ${\displaystyle P}$ 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.
3. ${\displaystyle M_{p}(bx_{1},\dots ,bx_{n})=b\cdot M_{p}(x_{1},\dots ,x_{n})}$.
   Like most means, the generalized mean is a homogeneous function of its arguments x₁, ..., x_n. That is, if b is a positive real number, then the generalized mean with exponent p of the numbers ${\displaystyle b\cdot x_{1},\dots ,b\cdot x_{n}}$ is equal to b times the generalized mean of the numbers x₁, ..., x_n.
4. ${\displaystyle M_{p}(x_{1},\dots ,x_{n\cdot k})=M_{p}\left[M_{p}(x_{1},\dots ,x_{k}),M_{p}(x_{k+1},\dots ,x_{2\cdot k}),\dots ,M_{p}(x_{(n-1)\cdot k+1},\dots ,x_{n\cdot k})\right]}$.
   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

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

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

and the two means are equal if and only if x₁ = x₂ = ... = 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,

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

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:

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

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

Equivalence of inequalities between means of opposite signs

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

$${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$$

applying this, then:

$${\displaystyle \left(\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{p}}}\right)^{1/p}\geq \left(\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{q}}}\right)^{1/q}}$$

We raise both sides to the power of −1 (strictly decreasing function in positive reals):

$${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{-p}\right)^{-1/p}=\left({\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{p}}}}}\right)^{1/p}\leq \left({\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{q}}}}}\right)^{1/q}=\left(\sum _{i=1}^{n}w_{i}x_{i}^{-q}\right)^{-1/q}}$$

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

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

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

The proof follows from Jensen's inequality, making use of the fact the logarithm is concave:

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

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

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

Taking q-th powers of the x_i yields

$${\displaystyle {\begin{aligned}&\prod _{i=1}^{n}x_{i}^{q{\cdot }w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}\\&\prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.\end{aligned}}}$$

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:

$${\displaystyle \prod _{i=1}^{n}x_{i}^{-q{\cdot }w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}^{-q}.}$$

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

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

Inequality between any two power means

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

$${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$$

if p is negative, and q is positive, the inequality is equivalent to the one proved above:

$${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$$

The proof for positive p and q is as follows: Define the following function: f : R₊ → R₊ ${\displaystyle f(x)=x^{\frac {q}{p}}}$. f is a power function, so it does have a second derivative:

$${\displaystyle f''(x)=\left({\frac {q}{p}}\right)\left({\frac {q}{p}}-1\right)x^{{\frac {q}{p}}-2}}$$

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:

$${\displaystyle {\begin{aligned}f\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)&\leq \sum _{i=1}^{n}w_{i}f(x_{i}^{p})\\[3pt]\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{q/p}&\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}\end{aligned}}}$$

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:

$${\displaystyle \left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)^{1/p}\leq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}}$$

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

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

$${\displaystyle M_{f}(x_{1},\dots ,x_{n})=f^{-1}\left({{\frac {1}{n}}\cdot \sum _{i=1}^{n}{f(x_{i})}}\right)}$$

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

powerSmooth :: Floating a => ([a] -> [a]) -> a -> [a] -> [a]
powerSmooth smooth p = map (** recip p) . smooth . map (**p)

See also

Notes

References

Further reading

• Bullen, P. S. (2003). "Chapter III - The Power Means". Handbook of Means and Their Inequalities. Dordrecht, Netherlands: Kluwer. pp. 175–265.

External links

• Power mean at MathWorld
• Examples of Generalized Mean
• A proof of the Generalized Mean on PlanetMath