N-th root of the arithmetic mean of the given numbers raised to the power n
Plot of several generalized means M p ( 1 , x ) {\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 [ edit ] 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[2] [3]
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[2]
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 .
Special cases [ edit ] A few particular values of p yield special cases with their own names:[4]
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}\}} A visual depiction of some of the specified cases for n = 2 with a = x 1 = M ∞ and b = x 2 = M −∞ : 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 , … , x n ) = n 1 x 1 + ⋯ + 1 x n {\displaystyle M_{-1}(x_{1},\dots ,x_{n})={\frac {n}{{\frac {1}{x_{1}}}+\dots +{\frac {1}{x_{n}}}}}} geometric mean M 0 ( x 1 , … , x n ) = lim p → 0 M p ( x 1 , … , x n ) = x 1 ⋅ ⋯ ⋅ x n n {\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 M 1 ( x 1 , … , x n ) = x 1 + ⋯ + x n n {\displaystyle 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 , … , x n ) = x 1 2 + ⋯ + x n 2 n {\displaystyle M_{2}(x_{1},\dots ,x_{n})={\sqrt {\frac {x_{1}^{2}+\dots +x_{n}^{2}}{n}}}} cubic mean M 3 ( x 1 , … , x n ) = x 1 3 + ⋯ + x n 3 n 3 {\displaystyle M_{3}(x_{1},\dots ,x_{n})={\sqrt[{3}]{\frac {x_{1}^{3}+\dots +x_{n}^{3}}{n}}}} maximum M + ∞ ( x 1 , … , x n ) = lim p → ∞ M p ( x 1 , … , x n ) = max { x 1 , … , x n } {\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 lim p → 0 M p = M 0 {\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 ∈ [ 0 , 1 ] {\displaystyle w_{i}\in [0,1]} 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}} using the exponential function as
M p ( x 1 , … , x n ) = exp ( ln [ ( ∑ i = 1 n w i x i p ) 1 / p ] ) = exp ( ln ( ∑ i = 1 n w i x i p ) p ) {\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 wi is equal to 1 (without loss in generality);[7] 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.
[2] 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}} . Then
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}}} The formula for M − ∞ {\displaystyle M_{-\infty }} follows from
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}.} Properties [ edit ] Let x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}} be a sequence of positive real numbers, then the following properties hold:[1]
min ( x 1 , … , x n ) ≤ M p ( x 1 , … , x n ) ≤ max ( x 1 , … , x n ) {\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.
M p ( x 1 , … , x n ) = M p ( P ( x 1 , … , x n ) ) {\displaystyle M_{p}(x_{1},\dots ,x_{n})=M_{p}(P(x_{1},\dots ,x_{n}))} , where P {\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.
M p ( b x 1 , … , b x n ) = b ⋅ M p ( x 1 , … , x n ) {\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
x 1 , ..., xn . That is, if
b is a positive real number, then the generalized mean with exponent
p of the numbers
b ⋅ x 1 , … , b ⋅ x n {\displaystyle b\cdot x_{1},\dots ,b\cdot x_{n}} is equal to
b times the generalized mean of the numbers
x 1 , ..., xn .
M p ( x 1 , … , x n ⋅ k ) = M p [ M p ( x 1 , … , x k ) , M p ( x k + 1 , … , x 2 ⋅ k ) , … , M p ( x ( n − 1 ) ⋅ k + 1 , … , x n ⋅ k ) ] {\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]} . Generalized mean inequality [ edit ] 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 , … , 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
x 1 = x 2 = ... = 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 .
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:
w i ∈ [ 0 , 1 ] ∑ i = 1 n w i = 1 {\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 wi = 1/n .
Equivalence of inequalities between means of opposite signs [ edit ] 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.
Geometric mean [ edit ] 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}.} Inequality between any two power means [ edit ] 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 is a power function, so it does have a second derivative:
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}} 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})\\[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:
( ∑ 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.
Generalized f -mean [ edit ] 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).[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 ) See also [ edit ]
^ If AC = a and BC = b . OC = AM of a 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 . ^ 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) . Further reading [ edit ] Bullen, P. S. (2003). "Chapter III - The Power Means". Handbook of Means and Their Inequalities . Dordrecht, Netherlands: Kluwer. pp. 175–265. External links [ edit ]