# Is $H=\min(t_1,…,t_n)$ a Copula?

n-Box is defined as $$B=[a_1,b_1]\times[a_2,b_2]\times[a_3,b_3]\times...\times[a_n,b_n]$$ Cartesian product of $$n$$ closed intervals, where $$a_i$$ and $$b_i$$ are all from $$R$$ and $$a_i\le b_i$$ for $$i=1,...,n$$. Vertices of the box are $$c=(c_1,...,c_n)$$ where $$c_k=a_k$$ or $$c_k=b_k$$.

We have a function defined as $$H=\min(t_1,...,t_n)$$.

We define $$H$$ volume of the box as $$V_H(B)=\sum \mathrm{sgn}(c)H(c)$$ where the sum is taken over all vertices of the box.

$$\mathrm{sgn}(c)=1$$ if $$c_k=a_k$$ for an even number of $$k$$'s

$$\mathrm{sgn}(c)=-1$$ if $$c_k=a_k$$ for an odd number of $$k$$'s

Prove that $$V_H(B)=\max(\min(b_1,...,b_n)-\max(a_1,...,a_n),0)$$

That would also be a proof that $$H$$ is Copula.

• Is this for some subject/course/study/self-study? An exercise from a book, etc? – Glen_b Jun 22 '14 at 2:13
• It is an exercise from Roger Nelsen "An introduction to Copulas" 2nd edition (Exercise 2.35). – Sergey Zykov Jun 22 '14 at 11:23
• Then you should probably add the self-study tag, and read its tag wiki info. – Glen_b Jun 22 '14 at 12:38

One way to prove a function $$H$$ on the $$n$$ cube $$[0,1]^n$$ is a copula is to exhibit a random variable whose distribution function restricted to the cube is $$H.$$
To that end, let $$X$$ be a univariate random variable with a uniform distribution on $$[0,1],$$ which means that for all $$t\in[0,1],$$ $$\Pr(X\le t)=t.$$ Define the $$n$$-vector-valued variable $$\mathbf X$$ as
$$\mathbf{X} = (X,X,\ldots, X).$$
Let $$(t_1,\ldots, t_n)\in [0,1]^n$$ and note (to justify the third equality below) that $$\min(t_1,\ldots, t_n)\in[0,1].$$ Successive application of the definitions of $$\mathbf X,$$ $$\min,$$ the uniform distribution, and $$H$$ justifies these four equalities:
\begin{aligned} \Pr(X_1\le t_1, \ldots, X_n\le t_n) &= \Pr(X \le t_1, \ldots, X\le t_n) \\ &= \Pr(X \le \min(t_1,\ldots,t_n)) \\ &= \min(t_1,\ldots, t_n) \\ &= H(t_1,\ldots, t_n), \end{aligned}