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Using the Copula operator $C$, which for any (possibly dependent) RVs $U$ and $V$ represents the joint cumulative DF of their inverse probability transform. That is, $U^* = F^{-1}_U (U) \sim \mbox{Uniform}(0,1)$ and similarly for $V$. Therefore, $C(u, v) = P(U^* < u, V^* < v)$.

Copulas generally allow one to visualize the dependence relation between two random variables on a square.

I would like to formulate the conditional probability $P(U | V \ge v)$ as a function of the copula of $U$ and $V$.

I am starting with the notion that:

$P(U \le u|V \ge v) = 1-C(u,v)/v$.

But don't know where to go.

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  • $\begingroup$ Can you give some context, e.g., actually define the symbols you're using? $\endgroup$
    – dsaxton
    Jul 28, 2015 at 18:14

1 Answer 1

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Your problem boils down to pure "conditional probabilities": $$P(U \leq u | V \geq v) = \frac{P(U \leq u, V \geq v)}{P(V \geq v)}$$

Let's assume two uniform random variables $U,V \sim {\rm U}(0,1)$ with copula $C$. Think of the problem geometrically supposing you have a large sample of your copula $C$. The probability of $P(U \leq u | V \geq v)$ is the amount of points in the top-left rectangle of the unit square spanned by $(0,v)$ and $(u,1)$ (assuming u is on your x-axis and v on your y-axis) divided by all points above the line at $v$. Instead of "number of points", we can also think of the probability mass based on the copula density $c$. To calculate the mass of the top-left rectangle, we take the entire mass 1 and subtract the mass of the area of points larger than $u$ (which is $1-u$) and the mass of the area of the lower left rectangle (spanned by $(0,0)$ and $(u,v)$) that is the copula value (i.e., $C(u,v)$). The divisor is the mass of the area larger than $v$ (which is $1-v$)):

$$P(U \leq u | V \geq v) = \frac{1-\big((1-u) + C(u,v)\big)}{1-v} = \frac{u-C(u,v)}{1-v}$$

Initially, you asked for the conditional density. That is, driving the above ideas to infinite small rectangles, for every $u$ the integral of the copula density $c$ from $v$ to 1, re-scaled by the entire probability mass still to be distributed in the conditioning area:

\begin{eqnarray} f(u | V \geq v) &=& \frac{1}{{1-v}} \int\limits_{v}^{1} c(u,x) ~ {\rm d}x \\ &=& \frac{1}{{1-v}} \left(\frac{\partial}{\partial u} C(u,1) - \frac{\partial}{\partial u} C(u,v) \right) \\ &=& \frac{1 - \frac{\partial}{\partial u} C(u,v) }{{1-v}} \end{eqnarray}

Note that the conditional density is merely the derivative with respect to $u$ of the above conditional CDF. See below for a R-code snippet that implements (and empirically double-checks) the above.

library(copula)
library(spcopula)

conFunLarger <- function(u, v=0.25, copula=gumbelCopula(4)) {
  (1-dduCopula(cbind(u,rep(v,length(u))), copula))/(1-v)
}

plot(1:100/101,conFunLarger(1:100/101),type="l")
sum(conFunLarger(1:1000/1001)*1/1001)
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