Jensen's inequalityJensensInequality2001-10-15 22:48:172006-09-13 13:40:49TheoremConvexConcaveInequality\usepackage{graphicx}
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\newcommand{\Expect}{\operatorname{\mathbbmss{E}}}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 $\mu_{k}\geq0$ one has:
$$f\left(\frac{\sum_{k=1}^{n}\mu_{k}x_{k}}{\sum_{k}^{n}\mu_{k}}\right)\leq\frac{\sum_{k=1}^{n}\mu_{k}f\left(x_{k}\right)}{\sum_{k}^{n}\mu_{k}}.$$
A common situation occurs when $\mu_1+\mu_2+\cdots+\mu_n=1$; in this case, the inequality simplifies to:
$$f\left(\sum_{k=1}^n \mu_k x_k\right)\leq \sum_{k=1}^n \mu_k f(x_k)$$
where $0\le \mu_k\le 1$.
If $f$ is a concave function, the inequality is reversed.
\medskip
\textbf{Example:}\\
$f(x)=x^2$ is a convex function on $[0,10]$.
Then
$$(0.2\cdot4+ 0.5\cdot3+0.3\cdot7)^2 \leq 0.2(4^2) + 0.5(3^2)+0.3(7^2).$$
\bigskip
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.