Usually in the undergraduate level pmf associated with a discrete random variable and pdf is with continuous variable. But both can be used interchangeably ( I think ).
If I have a set of Probability distributions on a product space with marginal constraints, is there any way to (how to) express the same as a linear family of PD's ( i.e. all P s.t. E_P[ f_i] =a_i for some f_i, a_i )
I couldn't find this terminology in books. When I was reading a paper I-divergence geometry of distributions by I.CSISZAR, I came across this term. Initially I thought it could be mutually absolutely continuous; but its not. I couldn't make any guess.
When are two or more measures said to be measure theoretically equivalent?
I have spent sometime on searching it; I am not getting it. Please someone help.
Thank you very much Billy Bob! I completely got it now.
\int_{E_n}|f|^p\ge n\mu(E_n)
Therefore, \mu(E_n)\le 1/n\int_{E_n}|f|^p\le 1/n\int |f|^p
So \mu(E_n)=0 for some n or \mu(E_n)\to 0
The first one implies that f is bounded and the second one implies X contains sets of arbitrarily small...
Yes absolutely.
If E_n=\{x:|f(x)|^p>n\} then E_n=\{x:|f(x)|>n^{1/p}\} and E_1\subset E_2\subset...}
Moreover, since X does not contain sets of arbitrarily small measure, \exists \epsilon >0 s.t. \mu(E)\ge \epsilon for all E\subset X
From these facts I m unable to figure it out.
Thanks...
Yes using the fact that p<q, I proved that \int |f|^q<\infty on {|f|\le 1
But I don't know how to make use of the fact the X doesn't contain sets of arbitrarily small positive measure in proving (2).
I don't know how is the above trure.
We know only that there is a function f in L^p but not in L^q. From this we have to show that X contains sets of arbitrarily small measure.
Suppose 0 < p < q < \infty. Then L^p \nsubseteq L^q iff X contains sets of arbitrarily small positive measure.
I have proved one part, namely, if X contains sets of arbitrarily small positive measure then L^p \nsubseteq L^q
Can anyone give some hints to solve the other part?
Thanks