Generalized mean

400px|thumb|right|Plot of several generalized means <math>M_p(1, x)</math>

In mathematics, generalized means (or power mean or Hölder mean from Otto Hölder)

<math display=block>M_p(x_1,\dots,x_n) = \left( \frac{1}{n} \sum_{i=1}^n x_i^p \right)^ .</math>

(See -norm). For we set it equal to the geometric mean (which is the limit of means with exponents approaching zero, as proved below):

<math display="block">M_0(x_1, \dots, x_n) = \left(\prod_{i=1}^n x_i\right)^{1/n} .</math>

Furthermore, for a sequence of positive weights we define the weighted power mean as differentiating the numerator and denominator with respect to , we have

<math display=block>\begin{align}

\lim_{p \to 0} \frac{\ln{\left(\sum_{i=1}^n w_ix_{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{align}</math>

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

<math display=block>\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)</math>

as desired.

Properties

Let <math>x_1, \dots, x_n</math> be a sequence of positive real numbers, then the following properties hold:

<math>\min(x_1, \dots, x_n) \le M_p(x_1, \dots, x_n) \le \max(x_1, \dots, x_n)</math>.<!--

-->

<math>M_p(x_1, \dots, x_n) = M_p(P(x_1, \dots, x_n))</math>, where <math>P</math> is a permutation operator.<!--

-->

<math>M_p(b x_1, \dots, b x_n) = b \cdot M_p(x_1, \dots, x_n)</math>.<!--

-->

<math>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]</math>.<!--

-->

Generalized mean inequality

In general, if , then

and the two means are equal if and only if .

The inequality is true for real values of and , as well as positive and negative infinity values.

It follows from the fact that, for all real ,

<math display=block>\frac{\partial}{\partial p}M_p(x_1, \dots, x_n) \geq 0</math>

which can be proved using Jensen's inequality.

In particular, for in , 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:

<math display="block">\begin{align}

w_i \in [0, 1] \\

\sum_{i=1}^nw_i = 1

\end{align}</math>

The proof for unweighted power means can be easily obtained by substituting .

Equivalence of inequalities between means of opposite signs

Suppose an average between power means with exponents and holds:

<math display="block">\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}</math>

applying this, then:

<math display="block">\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}</math>

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

<math display="block">\left(\sum_{i=1}^nw_ix_i^{-p}\right)^{-1/p}

= \left(\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^p\right)^{1/p}

\leq \left(\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^q\right)^{1/q}

= \left(\sum_{i=1}^nw_ix_i^{-q}\right)^{-1/q}</math>

We get the inequality for means with exponents and , 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 and non-negative weights summing to 1, the following inequality holds:

<math display="block">\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}.</math>

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

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

Taking -th powers of the yields

<math display=block>\begin{align}

&\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{align}</math>

Thus, we are done for the inequality with positive ; the case for negatives is identical but for the swapped signs in the last step:

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

<math display=block>\prod_{i=1}^n x_i^{w_i} \geq \left(\sum_{i=1}^n w_i x_i^{-q}\right)^{-1/q}.</math>

Inequality between any two power means

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

<math display="block">\left(\sum_{i=1}^n w_i x_i^p\right)^{1/p} \leq \left(\sum_{i=1}^nw_ix_i^q\right)^{1/q}</math>

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

<math display="block">\left(\sum_{i=1}^nw_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}</math>

The proof for positive and is as follows: Define the following function: <math>f(x)=x^{\frac{q}{p</math>. is a power function, so it does have a second derivative:

<math display="block">f(x) = \left(\frac{q}{p} \right) \left( \frac{q}{p}-1 \right)x^{\frac{q}{p}-2}</math>

which is strictly positive within the domain of , since , so we know is convex.

Using this, and the Jensen's inequality we get:

<math display="block">\begin{align}

f \left( \sum_{i=1}^nw_ix_i^p \right) &\leq \sum_{i=1}^nw_if(x_i^p) \\[3pt]

\left(\sum_{i=1}^n w_i x_i^p\right)^{q/p} &\leq \sum_{i=1}^nw_ix_i^q

\end{align}</math>

after raising both side to the power of (an increasing function, since is positive) we get the inequality which was to be proven:

<math display="block">\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}</math>

Using the previously shown equivalence we can prove the inequality for negative and by replacing them with and , respectively.

Generalized f-mean

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

<math display=block> M_f(x_1,\dots,x_n) = f^{-1} \left({\frac{1}{n}\cdot\sum_{i=1}^n{f(x_i)\right) </math>

This covers the geometric mean without using a limit with . The power mean is obtained for . Properties of these means are studied in de Carvalho (2016).