$P(u_1\leq\alpha,u_2\leq\beta)$ and $P(u_1>\alpha,u_2>\beta)$ and copulas

Question

I know this is meant to be elementary, but I don't know why I cannot convince myself -- given a copula $C(u_1,u_2)=P(u_1\leq \alpha,u_2\leq\beta)$, why is $P(u_1>\alpha,u_2>\beta)=1-\alpha-\beta-C(\alpha,\beta)$? Surely it's just integrating the density over $[0,1]^2\setminus\{[0,\alpha]\times[0,\beta]\}$? If it is, where did the $\alpha$ and $\beta$ terms come from? I am probably overcomplicating something that should be simple, please help!

Answer 1

First of all, note that since both $U_1, U_2$ has a marginal distribution of $\text{Uniform}(0,1)$, the support of their joint distribution must be a subset of $(0, 1) \times (0, 1)$

Next, as depicted from the picture, for any $\alpha, \beta$, $U_1 = \alpha$ and $U_2 = \beta$ will divide the $U_1$-$U_2$ plane into $4$ quadrants. And by the complementary probability we have

$$ \begin{align} &\Pr\{U_1 > \alpha, U_2 > \beta\} \\ = &~ 1 - (\Pr\{U_1 < \alpha, U_2 < \beta\} + \Pr\{U_1 > \alpha, U_2 < \beta\} + \Pr\{U_1 < \alpha, U_2 > \beta\}) \end{align} $$

Note that the sum of the $3$ terms in the bracket is corresponding to the area of the lower left $3$ quadrants. It can be also expressed as

$$ \begin{align} &\Pr\{U_1 < \alpha, U_2 < \beta\} + \Pr\{U_1 > \alpha, U_2 < \beta\} + \Pr\{U_1 < \alpha, U_2 > \beta\} \\ = &~ \Pr\{U_1 < \alpha\} + \Pr\{U_2 < \beta\} - \Pr\{U_1 < \alpha, U_2 < \beta\}\end{align} $$

You may view it by inclusion-exclusion principle $\Pr\{U_1 < \alpha \cup U_2 < \beta\}$; graphically means the area of that $L$-shaped area can be decomposed as the sum of two rectangles, minus the overlapping part (the lower left quadrant)

Therefore, combining together $$ \begin{align} &\Pr\{U_1 > \alpha, U_2 > \beta\} \\ = &~ 1 - (\Pr\{U_1 < \alpha\} + \Pr\{U_2 < \beta\} - \Pr\{U_1 < \alpha, U_2 < \beta\}) \end{align} $$

In particular, when $\alpha, \beta \in (0, 1)$, we have

$$ \Pr\{U_1 < \alpha\} = \alpha, \Pr\{U_2 < \beta\} = \beta $$

by the marginal uniform distribution. So the result reduce to

$$ \Pr\{U_1 > \alpha, U_2 > \beta\} = 1 - \alpha - \beta + \Pr\{U_1 < \alpha, U_2 < \beta\}$$