Jensen's inequality
From Wikimization
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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 | ||
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<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>. | ||
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<math>\,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond</math> | <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 | ||
Revision as of 12:54, 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 comment above says that for
all
in the domain of
. So
D. Ullrich