Jensen's inequality
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<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 |
Revision as of 13:04, 26 July 2008
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