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Jensen inequality, unexplained distribution, very confusing problem

  1. Jan 29, 2012 #1
    Hi everyone
    I don't know if I can find someone here to help me understand this issue, but I'll try

    the jensen inequality can be found here http://en.wikipedia.org/wiki/Jensen%27s_inequality

    I have the following discrete random variable [itex]X[/itex] with the following pmf:

    x 0 1 2 3
    pr(x) 0.110521 0.359341 0.389447 0.140691

    and the summation is (1).

    defining the function [itex]Y=\frac{1}{X}[/itex]

    Then finding the [itex]\bar{X}=\sum_{i=0}^{3}{iPr(i)=1.560308}[/itex]

    Now calculating [itex]E[Y]=E[\frac{1}{X}]=\sum_{i=1}^{3}{\frac{1}{i}Pr(i)}=0.6009615[/itex]

    According to Jensen inequality what should happen:

    [itex]E[\frac{1}{X}] \geq \frac{1}{E[X]}[/itex]
    but what I got is the opposite case!!!!!!!!:cry:

    Now also I have to mention the function [itex]Y=\frac{1}{X}[/itex] is convex given that the second derivative is strictly larger than zero.

    Do I misunderstand the Jensen inequality? or is there something wrong with the convixity assumption?
    Any idea?
  2. jcsd
  3. Jan 30, 2012 #2
    Without having looked at the details of your calculations, you certaintly have a problem with Y = 1/X for X=0
  4. Jan 30, 2012 #3
    Yes you are right to assume it but in the calculation I assumed X>0 when calculating Y.
    Even with this assumption I can't approve the opposite inequality.
    This can be thought like: choosing one message from X messages problem.

    Thanks for reply
  5. Jan 30, 2012 #4
    Well, you cannot simply "remove" X=0 by setting 1/0=0. Then your function is not convex any more, and this is the source of your problem.
  6. Jan 30, 2012 #5
    thanks winterfors
    But why it's not convex
    is there a way to prove that it's concave?
    Maybe you mean it's non differentiable at X=0

    thanks for reply
  7. Jan 30, 2012 #6
    I suggest you look up the definition of a convex/concave function. A function does not have to be differentiable to be convex, for instance is f(x)=abs(x) convex even though is it not differentiable at x=0. Also, that a function is not convex does not imply it is concave, or vice versa.
  8. Jan 30, 2012 #7
    I know dear
    that's why I asked about why setting Y=0 when X=0 caused the problem
    You said the function is not convex any more, here I didn't get it, does it mean now it's concave?
    the second derivative is positive since X>0
    Anyway thanks for reply I'll do more research on that
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