# Jensen's inequality

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<math>\phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b)</math> | <math>\phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b)</math> | ||

- | whenever <math>\,0 \leq t \leq 1\,</math> and <math>\,a, b\,</math> are in the domain of <math>\,\phi\,</math>. | + | whenever <math>\,0 \leq t \leq 1\,</math> and <math>\,a\,, b\,</math> are in the domain of <math>\,\phi\,</math>. |

It follows by induction on | It follows by induction on | ||

Line 11: | Line 11: | ||

<br>Jensen's inequality says this: <br>If <math>\,\mu\,</math> is a probability | <br>Jensen's inequality says this: <br>If <math>\,\mu\,</math> is a probability | ||

- | measure on <math>\,X\,</math>, <br><math>\,f\,</math> is a real-valued function on <math>\,X\,</math>, | + | measure on <math>\,X\,</math> , <br><math>\,f\,</math> is a real-valued function on <math>\,X\,</math> , |

<br><math>\,f\,</math> is integrable, and <br><math>\,\phi\,</math> is convex on the range | <br><math>\,f\,</math> is integrable, and <br><math>\,\phi\,</math> is convex on the range | ||

of <math>\,f\,</math> then | of <math>\,f\,</math> then | ||

Line 18: | Line 18: | ||

<br>'''Proof 1:''' By some limiting argument we can assume | <br>'''Proof 1:''' By some limiting argument we can assume | ||

- | that <math>\,f\,</math> is simple ( | + | that <math>\,f\,</math> is simple. (This limiting argument is a missing detail to this proof...) |

- | detail | + | <br>That is, <math>\,X\,</math> is the disjoint union of <math>\,X_1 \,\ldots\, X_n\,</math> |

- | and <math>\,f\,</math> is constant on each <math>\,X_j\,</math>. | + | and <math>\,f\,</math> is constant on each <math>\,X_j\,</math> . |

+ | |||

+ | Say <math>\,t_j=\mu(X_j)\,</math> and <math>\,a_j\,</math> is the value of <math>\,f\,</math> on <math>\,X_j\,</math> . | ||

- | Say <math>\,t_j=\mu(X_j)\,</math> and <math>\,a_j\,</math> is the value of <math>\,f\,</math> on <math>\,X_j\,</math>. | ||

Then (1) and (2) say exactly the same thing. QED. | Then (1) and (2) say exactly the same thing. QED. | ||

- | <br>'''Proof 2:''' The lemma shows | + | <br>'''Proof 2:''' |

- | derivative at every point and | + | |

- | lies above the "tangent" line through any point on the | + | Lemma. If <math>\,a < b,\, \,a' < b',\, \,a \leq a'\,</math> and <math>\,b \leq b'\,</math> then |

- | graph with slope | + | |

+ | <math>\,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond</math> | ||

+ | |||

+ | The lemma shows: | ||

+ | *<math>\,\phi\,</math> has a right-hand derivative at every point, and | ||

+ | *the graph of <math>\,\phi\,</math> lies above the "tangent" line through any point on the graph with slope equal to the right derivative. | ||

+ | |||

+ | Say <math>\,a = \int f d \mu\,</math> | ||

- | + | Let <math>\,m\,</math> be the right derivative of <math>\,\phi\,</math> | |

- | at <math>\,a\,</math>, and let | + | at <math>\,a\,</math> , and let |

<math>\,L(t) = \phi(a) + m(t-a)\,</math> | <math>\,L(t) = \phi(a) + m(t-a)\,</math> | ||

- | The | + | The bullets above say <math>\,\phi(t)\geq L(t)\,</math> for |

- | all <math>\,t\,</math> in the domain of <math>\,\phi\,</math>. So | + | all <math>\,t\,</math> in the domain of <math>\,\phi\,</math> . So |

<math>\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\ | <math>\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\ | ||

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&= \phi(a) + (m \int f) - ma\\ | &= \phi(a) + (m \int f) - ma\\ | ||

&= \phi(a)\\ | &= \phi(a)\\ | ||

- | &= \phi(\int f)\end{array}</math> | + | &= \phi(\int f)\end{array} |

+ | </math> | ||

- | <math>\,-\,</math> | + | <math>\,-\,</math>David C. Ullrich |

## Current revision

By definition is convex if and only if

whenever and are in the domain of .

It follows by induction on that if for then

(1)

Jensen's inequality says this:

If is a probability
measure on ,

is a real-valued function on ,

is integrable, and

is convex on the range
of then

(2)

**Proof 1:** By some limiting argument we can assume
that is simple. (This limiting argument is a missing detail to this proof...)

That is, is the disjoint union of
and is constant on each .

Say and is the value of on .

Then (1) and (2) say exactly the same thing. QED.

**Proof 2:**

Lemma. If and then

The lemma shows:

- has a right-hand derivative at every point, and
- the graph of lies above the "tangent" line through any point on the graph with slope equal to the right derivative.

Say

Let be the right derivative of at , and let

The bullets above say for all in the domain of . So

David C. Ullrich