Proof of conditional copula relation to the marginal copulas

Question

Hello

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??

Answer 1

Have a look at: http://publications.rwth-aachen.de/record/59254/files/04_198.pdf

in Section 2.2, the derivation is done for you.

Answer 2

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.