Jensen's inequality

From Wikimization

(Difference between revisions)
Jump to: navigation, search
(ljAJKoLGrJZCIpcw)
Line 1: Line 1:
-
By definition <math>\,\phi\,</math> is convex if and only if
+
LN4CNK <a href="http://wxhzeolvqdsj.com/">wxhzeolvqdsj</a>, [url=http://xjewoilohzlw.com/]xjewoilohzlw[/url], [link=http://znviigwysrwi.com/]znviigwysrwi[/link], http://pnowawyurayi.com/
-
 
+
-
<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>.
+
-
 
+
-
It follows by induction on
+
-
<math>\,n\,</math> that if <math>\,t_j \geq 0\,</math> for <math>\,j = 1, 2\ldots n\,</math> then
+
-
 
+
-
<br><math>\phi(\sum t_j a_j) \leq \sum t_j \phi(a_j) </math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (1)
+
-
 
+
-
<br>Jensen's inequality says this: <br>If <math>\,\mu\,</math> is a probability
+
-
measure on <math>\,X\,</math>&nbsp;,&nbsp; <br><math>\,f\,</math> is a real-valued function on <math>\,X\,</math>&nbsp;,&nbsp;
+
-
<br><math>\,f\,</math> is integrable, and <br><math>\,\phi\,</math> is convex on the range
+
-
of <math>\,f\,</math> then
+
-
 
+
-
<br><math>\phi(\int f d \mu) \leq \int \phi \circ f d \mu\qquad</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)
+
-
 
+
-
<br>'''Proof 1:''' By some limiting argument we can assume
+
-
that <math>\,f\,</math> is simple. (This limiting argument is a missing detail to this proof...)
+
-
<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>&nbsp;.
+
-
 
+
-
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>&nbsp;.&nbsp;
+
-
 
+
-
Then (1) and (2) say exactly the same thing. QED.
+
-
 
+
-
<br>'''Proof 2:'''
+
-
 
+
-
Lemma. If <math>\,a < b,\, \,a' < b',\, \,a \leq a'\,</math> and <math>\,b \leq b'\,</math> then
+
-
 
+
-
<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>&nbsp;,&nbsp; and let
+
-
 
+
-
<math>\,L(t) = \phi(a) + m(t-a)\,</math>
+
-
 
+
-
The bullets above say <math>\,\phi(t)\geq L(t)\,</math> for
+
-
all <math>\,t\,</math> in the domain of &nbsp;<math>\,\phi\,</math>&nbsp;. &nbsp;So
+
-
 
+
-
<math>\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\
+
-
&= \int (\phi(a) + m(f - a))\\
+
-
&= \phi(a) + (m \int f) - ma\\
+
-
&= \phi(a)\\
+
-
&= \phi(\int f)\end{array}</math>
+
-
 
+
-
<math>\,-\,</math>David C. Ullrich
+

Revision as of 12:28, 11 November 2009

LN4CNK <a href="http://wxhzeolvqdsj.com/">wxhzeolvqdsj</a>, [url=http://xjewoilohzlw.com/]xjewoilohzlw[/url], [link=http://znviigwysrwi.com/]znviigwysrwi[/link], http://pnowawyurayi.com/

Personal tools