Jensen's inequality

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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

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

$LaTeX: \,-\,$ David C. Ullrich

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

@@ Line 1: / Line 1: @@
-I'm D. Ullrich. I did not post this. I didn't
-write it either - it's wikimization's garbled
-transcription of something I posted elsewhere,
-posted here without my permission.
 By definition <math>\,\phi\,</math> is convex if and only if
 <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
@@ Line 16: / Line 11: @@
 <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>&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
@@ Line 23: / Line 18: @@
 <br>'''Proof 1:''' By some limiting argument we can assume
-that <math>\,f\,</math> is simple (this limiting argument is the missing
+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>
+<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>&nbsp;.&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>.
 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:'''
-derivative at every point and that the graph of <math>\,\phi\,</math>
-lies above the "tangent" line through any point on the
+Lemma. If <math>\,a < b,\, \,a' < b',\, \,a \leq a'\,</math> and <math>\,b \leq b'\,</math> then
-graph with slope = the right derivative.
+<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>
-Say <math>\,a = \int f d \mu\,</math>, let <math>\,m =\,</math> the right derivative of <math>\,\phi\,</math>
+Let <math>\,m\,</math> be the right derivative of <math>\,\phi\,</math>
-at <math>\,a\,</math>, and let
+at <math>\,a\,</math>&nbsp;,&nbsp; and let
 <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 &nbsp;<math>\,\phi\,</math>&nbsp;. &nbsp;So
 <math>\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\
@@ Line 47: / Line 50: @@
          &= \phi(a) + (m \int f) - ma\\
          &= \phi(a)\\
-         &= \phi(\int f)\end{array}</math>
+         &= \phi(\int f)\end{array}
+</math>
-<math>\,-\,</math>D. Ullrich
+<math>\,-\,</math>David C. Ullrich

Jensen's inequality

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