Formula for the multivariate cumulative distribution function (continuous case)

Question

Consider a random vector of dimension $5\times 1$, $X\equiv (X_1, X_2, X_3, X_4, X_5)$, with joint CDF denoted by $G$.

Consider the following probability $$ p\equiv Pr(X_1\geq a_1,X_2\geq a_2,X_3\geq a_3,X_4\geq a_4,X_5\geq a_5 ), $$ where $(a_1,_2,a_3,a_4,a_5)\in \mathbb{R}^5$.

I would like to write $p$ using the CDF of $X$, ($G$ introduced above). Could you advise on how to do that?

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My thoughts: At the moment, I have been able to derive the formula above for a vector with at most 3 components. For example, when $X$ is scalar $$ Pr(X_1\geq a_1)=P_{X_1}([a_1, \infty])=1-G(a_1) $$

When $X$ is bivariate $$ Pr(X_1\geq a_1, X_2\geq a_2)=P_{X_1,X_2}([a_1, \infty]\times [a_2, \infty])=1+G(a_1,a_2)-G(\infty, a_2)-G(a_1, \infty) $$ (see also this question).

When $X$ is trivariate $$ Pr(X_1\geq a_1, X_2\geq a_2, X_3\geq a_3)=P_{X_1,X_2,X_3}([a_1, \infty]\times [a_2, \infty]\times [a_3, \infty])=1-G(a_1,a_2,a_3)+G(\infty, a_2, a_3)+G(a_1, \infty, a_3)-G(\infty, \infty, a_3)+G(a_1, a_2, \infty)-G(\infty, a_2, \infty)-G(a_1, \infty, \infty) $$ How do I extend this to the $5$-variate case?

Answer 1

Let's denote $$\mathcal{P}(a_1,\cdots,a_n) = Pr(X_1\geqslant a_1, X_2\geqslant a_2, \cdots, X_n\geqslant a_n) =\\= \int\limits_{a_1}^{\infty}\int\limits_{a_2}^{\infty} \cdots \int\limits_{a_n}^{\infty}fdx_1dx_2\cdots dx_n$$ And known cumulativie one with $$\mathcal{G}(a_1,\cdots,a_n) = Pr(X_1\leqslant a_1, X_2\leqslant a_2, \cdots, X_n\leqslant a_n) =\\= \int\limits_{-\infty}^{a_1}\int\limits_{-\infty}^{a_2} \cdots \int\limits_{-\infty}^{a_n}fdx_1dx_2\cdots dx_n$$ Then $$1-\mathcal{P}(a_1,\cdots,a_n) = \sum\limits_{b_{i_k} \ne \infty}\mathcal{G}(b_{i_1},\cdots,b_{i_n}) - \sum\limits_{ b_{i_k} \ne \infty, b_{i_j} \ne \infty}\mathcal{G}(b_{i_1},\cdots,b_{i_n}) +\\+ \sum\limits_{ b_{i_k} \ne \infty, b_{i_j} \ne \infty \\b_{i_l} \ne \infty }\mathcal{G}(b_{i_1},\cdots,b_{i_n}) - \cdots +(-1)^{n-1}\mathcal{G}(a_1,\cdots,a_n) $$ Where in first sum all arguments are $\infty$ except only one, which equals some $a_i$, so we have $n$ such members. In second sum all arguments are $\infty$ except two ones, so we have all members with only 2 $a_i$ arguments, so $\frac{n(n-1)}{2}$ members and so on.

As far as I checked this by induction and by indicator as for inclusion-exclusion principle, then it seems correct, but I ask you check it carefully also, because I derived it for your question and never see before.

Shall be happy if it will be useful.