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I can't seem to find this kind of result anywhere so let me ask here.

Let $\{X_n\}_{n \geq 0}$ random variables be independent of $Y$ for all $n$. Under which types of convergence $X_n \to X$ is $X$ independent of $Y$? And why?

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Can you clarify "independent of Y"? Is it that, for each $n$, $X_n$ is independent of $Y$? Or that the sequence $\{X_n\}$ is independent of $Y$? (They are not the same.) –  Nate Eldredge Mar 30 '13 at 15:56
    
Ehm, for what I need it for they are simultaneously independent, but I'm curious about the other as well. –  Henrik Mar 30 '13 at 16:29
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2 Answers

up vote 4 down vote accepted

Suppose that $\{X_n\}$ is independent of $Y$ in the sense that for any $n$, the random vector $(X_1, \dots, X_n)$ is independent of $Y$. Then it follows that the $\sigma$-field $\sigma(X_1, X_2, \dots)$ generated by all $X_n$ is also independent of $Y$. (Our assumption gives us that for each $n$, the $\sigma$-field $\sigma(X_1, \dots, X_n)$ is independent of $Y$. A monotone class or $\pi$-$\lambda$ argument lets us conclude that $\sigma\left(\bigcup_n \sigma(X_1, \dots, X_n)\right) = \sigma(X_1, X_2, \dots)$ is also independent of $Y$.)

If $X_n \to X$ surely, then $X$ is itself $\sigma(X_1, X_2, \dots)$-measurable (a limit of measurable functions is measurable). Hence $X$ is independent of $Y$. The same holds if $X_n \to X$ almost surely.

If $X_n \to X$ in probability, we may find a subsequence $X_{n_k} \to X$ almost surely, so by the previous result $X$ is again independent of $Y$. This also covers the case when $X_n \to X$ in $L^p$ for some $p \ge 1$.

Weak convergence (aka convergence in distribution) is not sufficient, since it only determines the limiting random variable up to equality in distribution. For example, let $X_1, X_2, \dots, Y$ be iid with a non-constant distribution. It follows from the definition that $X_n \to Y$ weakly, but $Y$ is not independent of itself.

Edit: In his comment below, Did points out the following alternate proof, which also covers the case where we merely assume that for each $n$, $X_n$ is independent of $Y$. Suppose $u,v$ are bounded continuous functions; by independence, $E[u(X_n) v(Y)] = E[u(X_n)] E[v(y)]$. Using the dominated convergence theorem, we pass to the limit to obtain $E[u(X) v(Y)] = E[u(X)] E[v(Y)]$. By a monotone class argument, this also holds for $u,v$ any bounded measurable functions, and hence $X$ and $Y$ are independent. As before, this argument works directly when $X_n \to X$ almost surely, and by passing to a subsequence it also works when $X_n \to X$ in probability or in $L^p$.

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Thank you for your very elaborate answer it helped a lot. –  Henrik Mar 30 '13 at 19:01
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No doubt you "don't have a counterexample at present", since, if $X_n$ is independent of $Y$ for each $n$ and $X_n\to X$ almost surely, then $X$ is independent of $Y$. To wit, for every bounded continuous functions $u$ and $v$, $E[u(X_n)v(Y)]\to E[u(X)v(Y)]$ by dominated convergence, $E[u(X_n)]\to E[u(X)]$ by dominated convergence, and $E[u(X_n)v(Y)]=E[u(X_n)]E[v(Y)]$ by independence. Thus, $E[u(X)v(Y)]=E[u(X)]E[v(Y)]$, hence the independence. –  Did Apr 2 '13 at 7:50
    
@Did: Of course, how simple! You should post this as an answer. –  Nate Eldredge Apr 2 '13 at 13:21
    
Thanks. Rather, I suggest you add this to your answer. –  Did Apr 2 '13 at 15:10
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Just to complete Nate's answer, note that the convergence of $\{X_n\}$ itself can depend on $Y$, even if every $X_n$ is individually independent of $Y$.

For a trivial example, let $X_0$ and $Y$ be independent and uniformly distributed over $\{-1, 1\}$, and let $X_n = YX_{n-1}$ for $n > 0$. It is easy to see that every $X_n$ is individually independent of $Y$. Clearly, if $Y = 1$, then $X_n = X_{n-1} = X_0$ for all $n > 0$, and the sequence converges in any sense you care to name; whereas, if $Y = -1$, then $X_n = -X_{n-1}$ and it does not.

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That is a nice addition. Thanks! –  Henrik Apr 2 '13 at 18:15
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