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let's say I have a copula density function which I denote as $c(x,y)$. $X$ and $Y$ are uniformly distributed RVs.

I am curious if the following limit exists:

$\lim_{u\rightarrow 1^{-}} \int_0^u c(u,y) dy$. I have been told that it does not as $c(1,\cdot)$ is not defined while on the other hand $\int_0^u c(u,y) dy = P(Y \leq u|X=u)$ and that $\lim_{u \rightarrow 1} P(Y \leq u|X=u) = 1$.

Which one holds merit?

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Why would $P(Y\leqslant u\mid X=u)$ converge when $u\to1$ and why would it converge to $1$? –  Did Feb 16 '13 at 7:31
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it was argued to me that this conditional distribution can be regarded as another distrubitionand for a distribution $Z$ we have $P(Z \leq 1) =1$ –  Tomas Jorovic Feb 16 '13 at 8:01

1 Answer 1

There is no reason for $P(Y\leqslant u\mid X=u)$ to converge when $u\to1$ nor that it converges to $1$.

There is no reason for a random variable $Z$ to exist such that $P(Z\leqslant u)=P(Y\leqslant u\mid X=u)$ for every $u$ in $(0,1)$.

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May I know why? Do you have some examples? –  Tomas Jorovic Feb 18 '13 at 16:30
    
Yes, plenty. Say, don't you think the burden of the proof is rather on you? As I mention in the answer, you might begin by explaining why you think such a random variable $Z$ exists. And to do that, you might want to explain why the function $u\mapsto\mathbb P(Y\leqslant u\mid X=u)$ is always nondecreasing, which I doubt. –  Did Feb 18 '13 at 17:19
    
@user1237300 Did you downvote by any chance? –  Did Feb 19 '13 at 21:34

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