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In mathematics, generalized means (or power mean or Hölder mean from Otto Hölder) are a family of functions for aggregating sets of numbers. These include as special cases the Pythagorean means (arithmetic, geometric, and harmonic means).
If p is a non-zero real number, and x 1 , … , x n {\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
M p ( x 1 , … , x n ) = ( 1 n ∑ i = 1 n x i p ) 1 / p . {\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):
M 0 ( x 1 , … , x n ) = ( ∏ i = 1 n x i ) 1 / n . {\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 wi we define the weighted power mean as
M p ( x 1 , … , x n ) = ( ∑ i = 1 n w i x i p ∑ i = 1 n w i ) 1 / p {\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: M 0 ( x 1 , … , x n ) = ( ∏ i = 1 n x i w i ) 1 / ∑ i = 1 n w i . {\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 wi = 1/n.
A few particular values of p yield special cases with their own names:
minimum M − ∞ ( x 1 , … , x n ) = lim p → − ∞ M p ( x 1 , … , x n ) = min { x 1 , … , x n } {\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}\}}For the purpose of the proof, we will assume without loss of generality that
w i ∈ {\displaystyle w_{i}\in } and ∑ i = 1 n w i = 1. {\displaystyle \sum _{i=1}^{n}w_{i}=1.}We can rewrite the definition of M p {\displaystyle M_{p}}
M p ( x 1 , … , x n ) = exp ( ln ) = exp ( ln ( ∑ i = 1 n w i x i p ) p ) {\displaystyle M_{p}(x_{1},\dots ,x_{n})=\exp {\left(\ln {\left}\right)}=\exp {\left({\frac {\ln {\left(\sum _{i=1}^{n}w_{i}x_{i}^{p}\right)}}{p}}\right)}} using the exponential function asIn the limit p → 0, we can apply L'Hôpital's rule to the argument of the exponential function. We assume that p ∈ R {\displaystyle p\in \mathbb {R} } but p ≠ 0, and that the sum of wi is equal to 1 (without loss in generality); Differentiating the numerator and denominator with respect to p, we have
lim p → 0 ln ( ∑ i = 1 n w i x i p ) p = lim p → 0 ∑ i = 1 n w i x i p ln x i ∑ j = 1 n w j x j p 1 = lim p → 0 ∑ i = 1 n w i x i p ln x i ∑ j = 1 n w j x j p = ∑ i = 1 n w i ln x i ∑ j = 1 n w j = ∑ i = 1 n w i ln x i = ln ( ∏ i = 1 n x i w i ) {\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
lim p → 0 M p ( x 1 , … , x n ) = exp ( ln ( ∏ i = 1 n x i w i ) ) = ∏ i = 1 n x i w i = M 0 ( x 1 , … , x n ) {\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. Proof of lim p → ∞ M p = M ∞ {\textstyle \lim _{p\to \infty }M_{p}=M_{\infty }} and lim p → − ∞ M p = M − ∞ {\textstyle \lim _{p\to -\infty }M_{p}=M_{-\infty }}Assume (possibly after relabeling and combining terms together) that x 1 ≥ ⋯ ≥ x n {\displaystyle x_{1}\geq \dots \geq x_{n}}
lim p → ∞ M p ( x 1 , … , x n ) = lim p → ∞ ( ∑ i = 1 n w i x i p ) 1 / p = x 1 lim p → ∞ ( ∑ i = 1 n w i ( x i x 1 ) p ) 1 / p = x 1 = M ∞ ( x 1 , … , x n ) . {\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}}} . ThenThe formula for M − ∞ {\displaystyle M_{-\infty }}
M − ∞ ( x 1 , … , x n ) = 1 M ∞ ( 1 / x 1 , … , 1 / x n ) = x n . {\displaystyle M_{-\infty }(x_{1},\dots ,x_{n})={\frac {1}{M_{\infty }(1/x_{1},\dots ,1/x_{n})}}=x_{n}.} follows fromLet x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}}
be a sequence of positive real numbers, then the following properties hold:In general, if p < q, then
M p ( x 1 , … , x n ) ≤ M q ( x 1 , … , x n ) {\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 = ... = xn.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,
∂ ∂ p M p ( x 1 , … , x n ) ≥ 0 {\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.
We will prove the weighted power mean inequality. For the purpose of the proof we will assume the following without loss of generality:
w i ∈ ∑ i = 1 n w i = 1 {\displaystyle {\begin{aligned}w_{i}\in \\\sum _{i=1}^{n}w_{i}=1\end{aligned}}}The proof for unweighted power means can be easily obtained by substituting wi = 1/n.
Suppose an average between power means with exponents p and q holds:
( ∑ i = 1 n w i x i p ) 1 / p ≥ ( ∑ i = 1 n w i x i q ) 1 / q {\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: ( ∑ i = 1 n w i x i p ) 1 / p ≥ ( ∑ i = 1 n w i x i q ) 1 / q {\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):
( ∑ i = 1 n w i x i − p ) − 1 / p = ( 1 ∑ i = 1 n w i 1 x i p ) 1 / p ≤ ( 1 ∑ i = 1 n w i 1 x i q ) 1 / q = ( ∑ i = 1 n w i x i − q ) − 1 / q {\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.
For any q > 0 and non-negative weights summing to 1, the following inequality holds:
( ∑ i = 1 n w i x i − q ) − 1 / q ≤ ∏ i = 1 n x i w i ≤ ( ∑ i = 1 n w i x i q ) 1 / q . {\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:
log ∏ i = 1 n x i w i = ∑ i = 1 n w i log x i ≤ log ∑ i = 1 n w i x i . {\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
∏ i = 1 n x i w i ≤ ∑ i = 1 n w i x i . {\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}.}Taking q-th powers of the xi yields
∏ i = 1 n x i q ⋅ w i ≤ ∑ i = 1 n w i x i q ∏ i = 1 n x i w i ≤ ( ∑ i = 1 n w i x i q ) 1 / q . {\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:
∏ i = 1 n x i − q ⋅ w i ≤ ∑ i = 1 n w i x i − q . {\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.
∏ i = 1 n x i w i ≥ ( ∑ i = 1 n w i x i q ) 1 / q . {\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\geq \left(\sum _{i=1}^{n}w_{i}x_{i}^{q}\right)^{1/q}.}We are to prove that for any p < q the following inequality holds:
( ∑ i = 1 n w i x i p ) 1 / p ≤ ( ∑ i = 1 n w i x i q ) 1 / q {\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: ( ∑ i = 1 n w i x i p ) 1 / p ≤ ∏ i = 1 n x i w i ≤ ( ∑ i = 1 n w i x i q ) 1 / q {\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+ f ( x ) = x q p {\displaystyle f(x)=x^{\frac {q}{p}}}
f ″ ( x ) = ( q p ) ( q p − 1 ) x q p − 2 {\displaystyle f''(x)=\left({\frac {q}{p}}\right)\left({\frac {q}{p}}-1\right)x^{{\frac {q}{p}}-2}} . f is a power function, so it does have a second derivative: 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 ( ∑ i = 1 n w i x i p ) ≤ ∑ i = 1 n w i f ( x i p ) ( ∑ i = 1 n w i x i p ) q / p ≤ ∑ i = 1 n w i x i q {\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})\\\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: ( ∑ i = 1 n w i x i p ) 1 / p ≤ ( ∑ i = 1 n w i x i q ) 1 / q {\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.
The power mean could be generalized further to the generalized f-mean:
M f ( x 1 , … , x n ) = f − 1 ( 1 n ⋅ ∑ i = 1 n f ( x i ) ) {\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) = xp. Properties of these means are studied in de Carvalho (2016).
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 -> -> powerSmooth smooth p = map (** recip p) . smooth . map (**p)