Jensen’s inequality

If $f$ is a convex function on the interval $[a,b]$, for each ${\left\{x_k\right\}}_{k=1}^n\in [a,b]$ and each ${\left\{{\mu }_k\right\}}_{k=1}^n$ with ${\mu }_k\ge 0$ one has:

$$f\left(\frac{{\sum }_{k=1}^n{\mu }_k{x}_k}{{\sum }_k^n{\mu }_k}\right)\le \frac{{\sum }_{k=1}^n{\mu }_{k}f\left(x_k\right)}{{\sum }_k^n{\mu }_k}.$$

A common situation occurs when ${\mu }_1+{\mu }_2+\mathrm{\cdots }+{\mu }_n=1$; in this case, the inequality simplifies to:

$$f\left(\sum _{k=1}^n{\mu }_k{x}_k\right)\le \sum _{k=1}^n{\mu }_{k}f(x_k)$$

where $0\le {\mu }_k\le 1$.

If $f$ is a concave function, the inequality is reversed.

Example:
$f\mathbf{}\mathrm{(}x\mathrm{)}\mathrm{=}{x}^{\mathrm{2}}$ is a convex function on $[0,10]$. Then

$${(0.2\cdot 4+0.5\cdot 3+0.3\cdot 7)}^2\le 0.2(4^2)+0.5(3^2)+0.3(7^2).$$

A very special case of this inequality is when ${\mu }_k=\frac{1}{n}$ because then

$$f\left(\frac{1}{n}\sum _{k=1}^n{x}_k\right)\le \frac{1}{n}\sum _{k=1}^{n}f(x_k)$$

that is, the value of the function at the mean of the $x_k$ is less or equal than the mean of the values of the function at each $x_k$.

There is another formulation of Jensen’s inequality used in probability:
Let $X$ be some random variable, and let $f(x)$ be a convex function (defined at least on a segment containing the range of $X$). Then the expected value of $f(X)$ is at least the value of $f$ at the mean of $X$:

$$\mathrm{E}[f(X)]\ge f(\mathrm{E}[X]).$$

With this approach, the weights of the first form can be seen as probabilities.

Title	Jensen’s inequality
Canonical name	JensensInequality
Date of creation	2013-03-22 11:46:30
Last modified on	2013-03-22 11:46:30
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	13
Author	Andrea Ambrosio (7332)
Entry type	Theorem
Classification	msc 81Q30
Classification	msc 26D15
Classification	msc 39B62
Classification	msc 18-00
Related topic	ConvexFunction
Related topic	ConcaveFunction
Related topic	ArithmeticGeometricMeansInequality
Related topic	ProofOfGeneralMeansInequality