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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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You have asked five questions till now, all in a span of two days and all of them about convergence of random variables. More to the point, none of these questions show any effort or thought from you. (I must admit this one looks marginally better than the previous ones.) Are these questions homework? If so, please add the [homework] tag. And do attempt the problems yourself for some time and show your efforts here. Regards, –  Srivatsan Dec 25 '11 at 22:50
    
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
    
I misspoke the previous time: only three of them are about convergence of random variables, and one more is on convergence of real numbers. My apologies. [However, I do hope that the spirit of the comment is clear.] –  Srivatsan Dec 26 '11 at 0:03
    
Yes I appreciate. –  webster Dec 26 '11 at 1:18
    
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
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