# Applying Prokhorov's theorem to collection of random variables

Suppose we have a collection of random variables $X_{i,n}$ for $i \in [0,1]$ and $n=1,2,3....$ Suppose this collection of random variables is tight. Then, can we construct a subsequence $n'$ such that along this subsequence for every $i$, $X_{i,n}$ converges to some $X_{i}$ in distribution?

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Can we regard this double sequence as a sequence in the space of the product space of random variables and apply Tychonoff's theorem? – user12586 Dec 25 '11 at 21:16
Do you mean sth like this: $X_{i,n}: \Omega \to \mathbb{X}$ and treat the family of $X_{i,n}$ as a sequence of $Y_n: \Omega \times [0,1] \to \mathbb{X} \times [0,1]$? An interesting approach, but even if it may lead us to convergence of the joint distribution, I'm afraid that the condition "from every $i$" from the question will not be satisfied. – savick01 Dec 26 '11 at 0:16
But will it be that weak convergence of joint distribution will lead to weak convergence of every $i$ with at most a countable number of exceptions? – webster Dec 26 '11 at 1:24
No. In my answer I wrote "There are $2^{\aleph_0}$ numbers between 0 and 1.". I could have written "There are $2^{\aleph_0} \cdot 2^{\aleph_0}$ numbers between 0 and 1." and then the number of exceptions would be at least $2^{\aleph_0}$. – savick01 Dec 26 '11 at 11:08
But will it be still measure zero? – webster Dec 28 '11 at 16:30

It looks like we can create some bad distributions/random variables. There are $2^{\aleph_0}$ subsequences of $\mathbb{N}$. There are $2^{\aleph_0}$ numbers between 0 and 1. So we can associate every number $i$ with one subsequence $(n')$. Then we define $X_{i,n}$ to be whatever (but uniformely bounded) for $n \notin (n')$. For $n \in (n')$ $X_{i,n}$ can be $\delta_0$ and $\delta_1$ by turns (to prevent convergence).
EDIT: For $i$ belonging to a countable set $I$ (instead of $[0,1]$) we can use the diagonal method - we don't even need tightness of the whole family - tightness of each family $(X_{i,n})_{n\in \mathbb{N}}$ ($i$ fixed) would be enough.