giglamesh
- 14
- 0
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!
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?
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!

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?