# Jensen's inequality

(Difference between revisions)
 Revision as of 12:04, 26 July 2008 (edit)← Previous diff Revision as of 11:28, 11 November 2009 (edit) (undo) (ljAJKoLGrJZCIpcw)Next diff → Line 1: Line 1: - By definition \,\phi\,[/itex] is convex if and only if + LN4CNK wxhzeolvqdsj, [url=http://xjewoilohzlw.com/]xjewoilohzlw[/url], [link=http://znviigwysrwi.com/]znviigwysrwi[/link], http://pnowawyurayi.com/ - + - $\phi(ta + (1-t)b) \leq t \phi(a) + (1-t) \phi(b) + - + - whenever [itex]\,0 \leq t \leq 1\, and [itex]\,a\,, b\,$ 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 +