Expected values in Probability space

  • Thread starter Thread starter Lily@pie
  • Start date Start date
  • Tags Tags
    Probability Space
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Lily@pie
Messages
108
Reaction score
0

Homework Statement



Let a probability space be [itex](Ω, \epsilon, P)[/itex]. A set of random variables X1,...,Xn

Give an example where [itex]I_{p}(lim inf_{n -> ∞}X_{n}[/itex]) < [itex]lim inf_{n -> ∞}I_{p}(X_{n})[/itex]


The attempt at a solution

I know that [itex]I_{p}(lim inf_{n -> ∞}X_{n}[/itex])=[itex]E[lim inf_{n -> ∞}X_{n}[/itex]]
and [itex]lim inf_{n -> ∞}I_{p}(X_{n})[/itex] = [itex]lim inf_{n -> ∞}E[X_{n}][/itex]

I think I need to find a sequence of Xn such that lim inf Xn will have a smaller value than all the individual expected value, E[Xn].

Am I on the correct path? I'm kind of stuck here and not sure how to proceed.

Would be really really thankful for the help.
 
Physics news on Phys.org
How is ##\displaystyle \liminf_{n \to \infty} X_n## defined? The limit of the "smallest value" of those random variables?
In that case, I don't see how both can be equal apart from trivial Xi.
 
micromass said:
try to make a nonconvergent sequence ##X_n## such that ##E[X_n]## are all constant.

I saw this example...

Ω=N
P(ω)=2, ω in Ω
Xn(ω)=2nδω,n

I can see that Xn is not convergent.

But I'm not quite sure in computing the integrals... I'm very bad with the lim inf concept >_< So, I have written my attempt on evaluating the integrals and reasoning that I've used.

My attempt,
lim inf Xn
=lim inf [2nδω,n]
Since this takes in values 0 or 2n
=0

E[Xn]
=IP[2nδω,n]
= ∫2nδω,n2
When ω=n, this reduces to ∫δω,n
ω,n

I'm not very sure about the way I evaluate the integral. Would be very helpful for some guidance.