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Generalized mean - Wikipedia

From Wikipedia, the free encyclopedia

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).

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.

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

A visual depiction of some of the specified cases for n = 2 {\displaystyle n=2} .  

Harmonic mean:

M − 1 ( a , b ) {\displaystyle M_{-1}(a,b)}

.

 

Geometric mean:

M 0 ( a , b ) {\displaystyle M_{0}(a,b)}

.

 

Arithmetic mean:

M 1 ( a , b ) {\displaystyle M_{1}(a,b)}

.

 

Quadratic mean:

M 2 ( a , b ) {\displaystyle M_{2}(a,b)}

.

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 {\displaystyle p\in \mathbb {R} } but p ≠ 0, and that the sum of wi is equal to 1 (without loss in generality);[4] 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}.}

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

  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.

  2. 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.

  3. 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

    x1, ..., 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

    x1, ..., xn

    .

  4. 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 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.

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.

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]

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)
  1. ^ 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.

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