Re: Convergence of random variables

quantalfred wrote:

quantalfred wrote:



When talking about almost sure convergence, do we


have to clarify the underlying sample space
beforehand?

Say for example, let X_n be iid such that P(X_n)=1=P(X_n)=0=1/2,
then what's lim X_n?



Declaring the randomv variables to be independent
implies they are defined on the same probability space.



Then what could you say about this example: ([0,1], B[0,1], Lebesgue measure), set X_n(x) = 1 if 0<=x<1/2, X_n(x) = 0 if 1/2<=x<=1, we have P({x: limsup X_n(x) = 1}) = 1/2.  But we do have P(X_n=1)=P(X_n=0)=1/2 for each n.

What's wrong?


With this explicit definition, the X_i are no longer independent, in contrast to your original statement.



In this example, the sequence diverges almost surely.




Why?  By Borel-Cantelli?  But isn't it counter-intuitive?  All random variables are identical, but the limit doesn't exist.  And why it's not true that it converges to X_1?



Cite souces of assistance in submitted homework.




Unfortunately, it's not a homework. I couldn't find any books which deals with the stuff very clearly.



Ah, I thought you were taking a class. Why do you find it counterintuitive? You are flipping a fair coin over and over; the sequence will surely oscillate.

Are you confusing almost sure convergence with convergence in *distribution*? Convergence in distribution is also called weak convergence; do not confuse it with convergence in measure (also called convergence in probability). It is really a statement about the sequence of distributions rather than the random variables. Convergence in distribution is the type of convergence that appears the Central Limit Theorem. It is trivial that your X_n converges to X1 in distribution.

As to showing that X_n does not converge to X_1 almost surely, use the fact that almost sure convergence implies convergence in probability.

References:
Billingsley, Probability and Measure.
Drake, Fundamentals of Applied Probablity Theory.
Ross, Stochastic Processes.

By the way, you do not need an infinite-product space to define a sequence of i.i.d. r.v.'s. Let U be the identity map on [0,1] under Lebesgue measure. Let X_n = [2^n U] mod 2 (i.e., the nth digit in the binary expansion of a single uniform random variable), where [.] indicates the integer floor.

--
Stephen J. Herschkorn                        sjherschko@xxxxxxxxxxxx
.

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