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Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [1] \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2] \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [4] \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b) db$ is the marginal of this density, lets call it $g(u)$.

  • I don't think that $g(u)$ equals 1?
  • Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1.
  • With respect to copula's, I'm not sure what the interpretation of $\int_{0}^1 c(u,b) db$ even means? Because the copula captures dependency between random variables, I don't know if looking for meaning there is fruitful?
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    $\begingroup$ Your question (1) assumes $\int f(x,b) db = 1$. Usually that is not the case, as you can see by looking at almost any copula, such as $W$ or $M$. $\endgroup$ – whuber Mar 9 '16 at 14:48
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No, as you discovered in (2) that's incorrect: $\frac{\partial F(x, y)}{\partial x} \not= \mathbb{P}(Y \le y | X = x)$ because for $y = +\infty$ we have $\frac{\partial F(x, y)}{\partial x} = f(x)$ while $\mathbb{P}(Y \le +\infty \;|\; \text{whatever}) = 1$

It could mean $\mathbb{P}(Y \le y, X = x)$ but the problem is that for continuous $X$ event $\{X = x\}$ has zero probability.

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  • $\begingroup$ Your logic makes sense. I've added equation #'s to the above post, and Equation 4 above is given in Nelsen's "Introduction to Copulas" book. I don't think it is an incorrect formula, so that leads me to wonder: Is the interpretation given in Equation 4 only valid for copulas, and not joint distribution functions in general? $\endgroup$ – Kiran K. Mar 9 '16 at 14:18
  • $\begingroup$ @KiranK. [4] itself is correct (provided no negative values are possible), but your probabilistic interpretation isn't. Or is it from Nelsen's, too? If so, please tell me the page number. $\endgroup$ – Artem Sobolev Mar 9 '16 at 14:56
  • $\begingroup$ I got both the probabilistic interpretation and the formula from Nelsen, it is on page 41, Equation 2.9.1. The page numbers and equation numbers correspond to the 2nd edition of the book (2006 version). $\endgroup$ – Kiran K. Mar 9 '16 at 14:59
  • $\begingroup$ @KiranK. sorry, my bad. Indeed, for copula the formulas are correct, and $g(u) = 1$ for any $u \in [0, 1]$ because copula has uniform marginals, that is, $g(u)$ is a density of a uniformly distributed r.v. $\endgroup$ – Artem Sobolev Mar 9 '16 at 17:33
  • $\begingroup$ That's the link I was missing, $g(u) = 1$ for any $u \in [0,1]$ because of the uniform marginal property... So in summary, is it correct to say, the probabilistic interpretation only holds for copulas, but not for joint distributions in general? $\endgroup$ – Kiran K. Mar 9 '16 at 19:51

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