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 12:04, 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