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Suppose a random variable $Z_n=(X_n,Y_n)$ takes a value on $A= [0,1] \times B=[0,1]$. We can consider a conditional distribution of $Y_n$ given the realized value of $X_n$. Now suppose $Z_n$ converges to $Z$ in distribution. What can I say about the convergence of the conditional distribution $F_n(Y_n|X_n=x)$? Can I say, for example, for $x$ almost surely, $F_n(|X_n=x)$ converges to $F(|X=x)$ in distribution?

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Given that $Z_n$ converges to $Z$ in distribution, it has to be that the joint distribution of $Z_n$ converges to the joint distribution of $Z$ for all continuity set of $Z$. But since the joint distribution function is a nondecreasing function, it has to be that discontinuity takes place at most countable points. As such, for $x$ almost surely, $F_n(|X_n=x)$ converges to $F(|X=x) $ in distribution. Can it make sense? – webster Dec 25 '11 at 22:56
webster: your math comment above is strange. Assume you know the distribution of $(X,Y)$, how do you define $F(\ \mid X=x)$ for a given $x$? – Did Dec 26 '11 at 10:59
I meant conditional distributions. – webster Dec 28 '11 at 16:31

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