Jensen's inequality

From Wikimization

(Difference between revisions)
Jump to: navigation, search
Line 20: Line 20:
that <math>\,f\,</math> is simple. (This limiting argument is a missing detail to this proof...)
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>
<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>&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>.
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>.
Line 42: Line 42:
<math>\,L(t) = \phi(a) + m(t-a)\,</math>
<math>\,L(t) = \phi(a) + m(t-a)\,</math>
-
The comment above says that <math>\,\phi(t) \geq L(t)\,</math> for
+
The bullets above say <math>\,\phi(t)\geq L(t)\,</math> for
-
all <math>\,t\,</math> in the domain of <math>\,\phi\,</math>. So
+
all <math>\,t\,</math> in the domain of <math>\,\phi\,</math>&nbsp;. &nbsp;So
<math>\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\
<math>\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\

Revision as of 11:59, 26 July 2008

By definition LaTeX: \,\phi\, is convex if and only if

LaTeX: \phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b)

whenever LaTeX: \,0 \leq t \leq 1\, and LaTeX: \,a\,, b\, are in the domain of LaTeX: \,\phi\,.

It follows by induction on LaTeX: \,n\, that if LaTeX: \,t_j \geq 0\, for LaTeX: \,j = 1, 2\ldots n\, then


LaTeX: \phi(\sum t_j a_j) \leq \sum t_j \phi(a_j)           (1)


Jensen's inequality says this:
If LaTeX: \,\mu\, is a probability measure on LaTeX: \,X\,,
LaTeX: \,f\, is a real-valued function on LaTeX: \,X\,,
LaTeX: \,f\, is integrable, and
LaTeX: \,\phi\, is convex on the range of LaTeX: \,f\, then


LaTeX: \phi(\int f d \mu) \leq \int \phi \circ f d \mu\qquad          (2)


Proof 1: By some limiting argument we can assume that LaTeX: \,f\, is simple. (This limiting argument is a missing detail to this proof...)
That is, LaTeX: \,X\, is the disjoint union of LaTeX: \,X_1 \,\ldots\, X_n\, and LaTeX: \,f\, is constant on each LaTeX: \,X_j\, .

Say LaTeX: \,t_j=\mu(X_j)\, and LaTeX: \,a_j\, is the value of LaTeX: \,f\, on LaTeX: \,X_j\,. Then (1) and (2) say exactly the same thing. QED.


Proof 2:

Lemma. If LaTeX: \,a < b,\, \,a' < b',\, \,a \leq a'\, and LaTeX: \,b \leq b'\, then

LaTeX: \,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond

The lemma shows:

  • LaTeX: \,\phi\, has a right-hand derivative at every point, and
  • the graph of LaTeX: \,\phi\, lies above the "tangent" line through any point on the graph with slope equal to the right derivative.

Say LaTeX: \,a = \int f d \mu\,

Let LaTeX: \,m\, be the right derivative of LaTeX: \,\phi\, at LaTeX: \,a\,, and let

LaTeX: \,L(t) = \phi(a) + m(t-a)\,

The bullets above say LaTeX: \,\phi(t)\geq L(t)\, for all LaTeX: \,t\, in the domain of LaTeX: \,\phi\, .  So

LaTeX: \begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\ </p> <pre> &= \int (\phi(a) + m(f - a))\\ &= \phi(a) + (m \int f) - ma\\ &= \phi(a)\\ &= \phi(\int f)\end{array}

LaTeX: \,-\,D. Ullrich

Retrieved from "http://www.convexoptimization.com/wikimization/index.php/Jensen%27s_inequality"
Personal tools