# Jensen's inequality

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 Revision as of 12:28, 11 November 2009 (edit) (ljAJKoLGrJZCIpcw)← Previous diff Revision as of 16:48, 11 November 2009 (edit) (undo)m (Reverted edits by 65.79.227.14 (Talk); changed back to last version by Cslaw)Next diff → Line 1: Line 1: - LN4CNK wxhzeolvqdsj, [url=http://xjewoilohzlw.com/]xjewoilohzlw[/url], [link=http://znviigwysrwi.com/]znviigwysrwi[/link], http://pnowawyurayi.com/ + By definition \,\phi\,[/itex] is convex if and only if + + $\phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b) + + whenever [itex]\,0 \leq t \leq 1\, and [itex]\,a\,, b\,$ are in the domain of $\,\phi\,$. + + It follows by induction on + $\,n\, that if [itex]\,t_j \geq 0\, for [itex]\,j = 1, 2\ldots n\,$ then + +
$\phi(\sum t_j a_j) \leq \sum t_j \phi(a_j)$           (1) + +
Jensen's inequality says this:
If $\,\mu\, is a probability + measure on [itex]\,X\, , [itex]\,f\,$ is a real-valued function on $\,X\,$ ,  +
$\,f\,$ is integrable, and
$\,\phi\,$ is convex on the range + of $\,f\,$ then + +
$\phi(\int f d \mu) \leq \int \phi \circ f d \mu\qquad$          (2) + +
'''Proof 1:''' By some limiting argument we can assume + that $\,f\,$ is simple. (This limiting argument is a missing detail to this proof...) +
That is, $\,X\, is the disjoint union of [itex]\,X_1 \,\ldots\, X_n\, + and [itex]\,f\,$ is constant on each $\,X_j\,$ . + + Say $\,t_j=\mu(X_j)\, and [itex]\,a_j\, is the value of [itex]\,f\,$ on $\,X_j\,$ .  + + Then (1) and (2) say exactly the same thing. QED. + +
'''Proof 2:''' + + Lemma. If $\,a < b,\, \,a' < b',\, \,a \leq a'\, and [itex]\,b \leq b'\, then + + [itex]\,(f(a) - f(b)) / (a - b) \leq (f(a') - f(b')) / (a' - b')\quad\diamond$ + + The lemma shows: + *$\,\phi\, has a right-hand derivative at every point, and + *the graph of [itex]\,\phi\, lies above the "tangent" line through any point on the graph with slope equal to the right derivative. + + Say [itex]\,a = \int f d \mu\, + + Let [itex]\,m\,$ be the right derivative of $\,\phi\,$ + at $\,a\,$ ,  and let + + $\,L(t) = \phi(a) + m(t-a)\,$ + + The bullets above say $\,\phi(t)\geq L(t)\,$ for + all $\,t\,$ in the domain of  $\,\phi\,$ .  So + + $\begin{array}{rl}\int \phi \circ f &\geq \int L \circ f\\ + &= \int (\phi(a) + m(f - a))\\ + &= \phi(a) + (m \int f) - ma\\ + &= \phi(a)\\ + &= \phi(\int f)\end{array}$ + + $\,-\,$David C. Ullrich

## Revision as of 16:48, 11 November 2009

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\\

 &= \int (\phi(a) + m(f - a))\\ &= \phi(a) + (m \int f) - ma\\ &= \phi(a)\\ &= \phi(\int f)\end{array}$

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