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I am trying to derive the second equation displayed in section 7.1 (or p. 41) of this article or equation (6.3) of this book. I've seen this in many documents discussing conditional sampling with copulae.

Suppose $C(u_1,...u_d)$ is a d dimensional copula and $C_j(u_1,...u_j)=C(u_1,...u_j,1,...,1)$, $j \in {2,...,d-1}$ is the $j$ dimensional marginal copula.

I am trying to prove the following relationship as in page 420 of this book (search inside the book for "conditional inverse method" in google books to see this page),

$\mathbb{P}\left(U_{j}\leq u_{j},U_1=u_{1},\ldots,U_{j-1}=u_{j-1}\right)=\frac{\partial^{j-1}C_{j}\left(u_{1},\ldots,u_{j}\right)}{\partial u_{1},\ldots,\partial u_{j-1}}$

Maybe something along the lines of,

$\mathbb{P}\left(U_{j}\leq u_{j},U_1=u_{1},\ldots,U_{j-1}=u_{j-1}\right)$

$=\lim_{\triangle u_{1},\ldots,\triangle u_{j-1}\rightarrow0}\mathbb{P}\left(U_{j}\leq u_{j},u_{1}\leq U_{1}\leq u_{1}+\triangle u_{1},\ldots,u_{j-1}\leq U_{j-1}\leq u_{j-1}+\triangle u_{j-1}\right) $

then the paper i'm reading tells me if this is right it should equal

$=\lim_{\triangle u_{1},\ldots,\triangle u_{j-1}\rightarrow0}\frac{C_{j}\left(u_{1}+\triangle u_{1},\ldots,u_{j-1}+\triangle u_{j-1},u_{j}\right)-C\left(u_{1},\ldots,u_{j}\right)}{\triangle u_{1}\cdots\triangle u_{j-1}}$ ... (*)

Where the fact the copula $C_j$ is the joint distribution of $j$ uniform random variables, $U_1,...,U_j$ is used. But I dont quite understand where the denominator in (*) $\triangle u_{1}\cdots\triangle u_{j-1}$ comes from??

In fact the derivation in the paper goes against my understanding of probability.

For example how does the author divide by $\mathbb{P}\left(U_{j}\leq u_{j},U=u_{1},\ldots,U_{j-1}=u_{j-1}\right)$ in the derivation. I would have imagined since the $u_i$'s are continuous this quantity would be 0... but how come its not??

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Rephrasing your question, you are asking why is the conditional d.f. of a random vector given by partial derivatives. The answer is available, for example, here.

Just as a side note: what you have on mind is a conditioning. Therefore you should write $$ \mathbb{P}\left[U_{j}\le u_{j}|U_{1}=u_{1},\ldots, U_{j-1}=u_{j-1}\right] $$ instead of $$ \mathbb{P}\left[U_{j}\le u_{j},U_{1}=u_{1},\ldots, U_{j-1}=u_{j-1}\right] $$ the latter being trivially $0$, given the uniformity of all marginals.

Edit: you asked where and why did the denominator appear in (*). It is there just because the conditional distribution is a conditioning (by definition). You obtained the strange expression (without denominator) because you started from a wrong definition of the conditional distribution. The correct definition is the first expression in this answer and is computed as the limit $$ \lim_{d_{1}\to 0,\;\ldots\;,\;d_{j-1}\to0} \mathbb{P}\left[U_{j}\le u_{j}\;|\;u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right] $$ which boils down to $$ \lim_{d_{1}\to 0,\;\ldots\;,\;d_{j-1}\to0} \frac{\mathbb{P}\left[U_{j}\le u_{j},u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right]}{\mathbb{P}\left[u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right]} $$ where you can already see the denominator.

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Can you elaborate a little please? I dont see how the link above provides the answer. Have a look at this ... ... article. See section 7.1 on p41, It is the second equation displayed in that I am trying to derive. It seems $\mathbb{P}\left[U_{j}\le u_{j},U_{1}=u_{1},\ldots, U_{j-1}=u_{j-1}\right] \neq 0$ since the author is dividing by this quantity. – mark May 14 '11 at 2:52
The writing of Section 7.1 page 41 is seriously faulty, for the reason explained by Peter. This is the only example of its kind in the linked reference. It is odd that, up to page 41, the authors wrote carefully and correctly every conditioning they had to introduce. – Did May 14 '11 at 6:29
By the way, your definition of marginal copulas in your question is wrong as well since you use the random variables $U_j$ instead of the real numbers $u_j$. – Did May 14 '11 at 6:32
@Didier Piau, thanks for pointing out the typo. Why do you think the author of the MSc is wrong? I'm pretty sure he/she copied these lines directly from a book such as the ones I have mentioned (just now) in my question. – mark May 14 '11 at 22:45
@Peter Sarkoci. I'm not sure if this is how the derivation proceeds. For $\mathbb{P}\left[u_{1}\le U_{1}\le u_{1}+d_{1}\,\ldots,u_{j-1}\le U_{j-1}\le u_{j-1}+d_{j-1}\right]$ to equal to $d_{1}\cdots d_{j-1}$. The Ui's would have to be jointly uniform or else independent. But they are neither. Their joint distribution is described by the copula function C. Also regarding the side note. I did not make a mistake here; See the third reference listed in my post. Also thanks for your input – mark May 14 '11 at 22:54

Have a look at:

in Section 2.2, the derivation is done for you.

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