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# 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 $\mu_{{k}}\geq 0$ 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 $0\leq\mu_{k}\leq 1$.

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

Example:

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

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

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

$f\left(\frac{1}{n}\sum_{{k=1}}^{n}x_{k}\right)\leq\frac{1}{n}\sum_{{k=1}}^{n}f% (x_{k})$ |

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 $X$:

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

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

## Mathematics Subject Classification

81Q30*no label found*26D15

*no label found*39B62

*no label found*18-00

*no label found*

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## Attached Articles

## Corrections

broken by yark ✓

More general form by Andrea Ambrosio ✓

notation by Mathprof ✓