Consider the k-variate random vector $Y\equiv (Y_1, Y_2, ..., Y_K)$ with cumulative distribution function (cdf) $F$.

How can I express in terms of $F$ the following probability: $$ \mathbb{P}(a_k\leq Y_k\leq b_k \text{ }\forall k) $$ ?

For example, if $K=2$, I know that

$$ \mathbb{P}(a_1\leq Y_1\leq b_1\text{, }a_2\leq Y_2\leq b_2)= F(a_2, b_2)+F(a_1, b_1)- F(a_1, b_2)-F(a_2, b_1) $$

(Sometimes this is also called rectangle formula)

Is there a way to generalise this to any $K>2$?

Thanks to the comment below, I now write the expression for $K=3$

1) I list all the $2^K=8$ vertices

$$ \begin{cases} a_1,a_2, a_3\\ b_1, a_2, a_3\\ a_1, b_2, a_3 \\ b_1, b_2, a_3\\ a_1,a_2, b_3\\ b_1, a_2, b_3\\ a_1, b_2, b_3 \\ b_1, b_2, b_3\\ \end{cases} $$

2) Algebraic sum of cdf at each vertex with +1 if the number of $a$'s is even and -1 otherwise

$$ \mathbb{P}(a_1\leq Y_1\leq b_1\text{, }a_2\leq Y_2\leq b_2\text{, }a_3\leq Y_3\leq b_3)= -F(a_1,a_2, a_3) +F(b_1, a_2, a_3) +F(a_1, b_2, a_3) -F(b_1, b_2, a_3) +F(a_1,a_2, b_3) -F(b_1, a_2, b_3) -F(a_1, b_2, b_3) +F(b_1, b_2, b_3) $$

  • 1
    $\begingroup$ Billingsley's Probability and Measure, Section 12 has a thorough discussion of this. $\endgroup$ Jul 30, 2018 at 8:50
  • $\begingroup$ Thank you. It is a bit overcomplicated for me. Any easier source? $\endgroup$
    – TEX
    Jul 30, 2018 at 8:56
  • 1
    $\begingroup$ Sorry to say that the expression for the probability in $\mathbb R^{k}$ is a bit complicated and I cannot think of a better source. $\endgroup$ Jul 30, 2018 at 8:59
  • $\begingroup$ @KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct? $\endgroup$
    – TEX
    Jul 30, 2018 at 9:20
  • 1
    $\begingroup$ I think you have got it right. $\endgroup$ Jul 30, 2018 at 9:23

1 Answer 1


In general $1_{A\cap B}=1_A\times 1_B$ and $1_{(a_i,b_i]}=1_{(-\infty,b_i]}-1_{(-\infty,a_i]}$.

This leads to:$$1_{(a_1,b_1]\times\cdots\times(a_n,b_n]}=\left(1_{(-\infty,b_1]}-1_{(-\infty,a_1]}\right)\times\cdots\times\left(1_{(-\infty,b_n]}-1_{(-\infty,a_n]}\right)$$

Working out the RHS we get $2^{n-1}$ terms with sign $+$ and $2^{n-1}$ with sign $-$.

The sign is $+$ if and only if the number of $a_i$'s that appear in the term is even.

Taking the integral on both sides gives the equality:$$P(a_1<Y_1\leq b_1,\dots, a_n<Y_n\leq b_n)=\sum_{(c_1,\dots,c_n)\in\prod_{i=1}^n\{a_i,b_i\}}(-1)^{e(c_1,\dots,c_n)}F(c_1,\dots,c_n)$$where $e(c_1,\dots,c_n)$ denotes the cardinality of the set $\{i\in\{1,\dots,n\}\mid c_i=a_i\}$


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