Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
Still no progress on the content of my answer? Or are we still in the ultra constructive approach from two years ago? – Did Feb 25 at 15:25

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 $(0,1)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.