Jensen's inequality
From Wikimization
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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 that <math>\,\phi\,</math> has a right-hand | + | <br>'''Proof 2:''' |
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+ | Lemma. If <math>\,a < b,\, \,a' < b',\, \,a \leq a'\,</math> and <math>\,b \leq b'\,</math> then | ||
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+ | <math>\,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond</math> | ||
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+ | The lemma shows that <math>\,\phi\,</math> has a right-hand | ||
derivative at every point and that the graph of <math>\,\phi\,</math> | derivative at every point and that the graph of <math>\,\phi\,</math> | ||
lies above the "tangent" line through any point on the | lies above the "tangent" line through any point on the |
Revision as of 12:36, 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 the missing
detail).
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 that has a right-hand
derivative at every point and that the graph of
lies above the "tangent" line through any point on the
graph with slope = the right derivative.
Say , let the right derivative of at , and let
The comment above says that for all in the domain of . So
D. Ullrich