Jensen inequality, unexplained distribution, very confusing problem

AI Thread Summary
The discussion revolves around the application of Jensen's inequality to a discrete random variable X with a defined probability mass function. The user calculates the expected values E[Y] and Y(μ) but finds that E[Y] is less than Y(μ), contradicting Jensen's inequality. The issue arises from the handling of the function Y = 1/X, particularly at X=0, which complicates the convexity assumption. Participants clarify that while a function can be non-differentiable at a point, it does not negate its convexity, and the misunderstanding stems from incorrectly assuming Y remains convex when X=0 is included. The conversation emphasizes the importance of correctly defining the function and its properties to apply Jensen's inequality accurately.
giglamesh
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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 X 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 Y=\frac{1}{X}

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

Now calculating E[Y]=E[\frac{1}{X}]=\sum_{i=1}^{3}{\frac{1}{i}Pr(i)}=0.6009615
then:
Y(\bar{X})=\frac{1}{\bar{X}}=\frac{1}{1.560308}=0.640899105

According to Jensen inequality what should happen:

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

Now also I have to mention the function Y=\frac{1}{X} 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?
 
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Without having looked at the details of your calculations, you certaintly have a problem with Y = 1/X for X=0
 
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
 
giglamesh said:
Yes you are right to assume it but in the calculation I assumed X>0 when calculating Y.

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.
 
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
 
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.
 
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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