# Partial Derivative of Joint Distribution Function interpretation

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

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

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

\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?
• 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 $M$. – whuber Mar 9 '16 at 14:48

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.
• @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. – Artem Sobolev Mar 9 '16 at 17:33
• 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? – Kiran K. Mar 9 '16 at 19:51