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).
Definition 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.
Special cases 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.
Properties Let x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}} be a sequence of positive real numbers, then the following properties hold:
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})} .
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.
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})} .
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
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 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 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 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 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
The power mean could be generalized further to the generalized f-mean:
M f