# What is $P([a_1,b_1]\times [a_2,b_2]\times…\times[a_k,b_k])$

How can I obtain an expression of $P([a_1,b_1]\times [a_2,b_2]\times...\times[a_k,b_k])$ knowing that $F(x_1,...,x_k)=P((-\infty,x_1]\times (-\infty,x_2]\times...\times(-\infty,x_k])$?

I solve the problem for $k=2$. $P([a_1,b_1]\times [a_2,b_2])=F(b_1,b_2)+F(a_1-,a_2-)-F(a_1-,b_2-)-F(b_1-,a_2-)$.

But I dont know how to do it for any $k$.

Thanks a lot :)

• @yanko No, $[a,b]=(-\infty,b]\setminus(-\infty,a)$. – Did Oct 11 '17 at 17:17

## 1 Answer

Let $A=[a_1,b_1]\times [a_2,b_2]\times...\times[a_k,b_k]$ and $N=\{1,2,\ldots,n\}$. For every $I\subseteq N$, let $c_I$ denote the $n$tuple made of the entries $a_i^-$ for every $i$ in $I$ and $b_i$ for every $i$ in $N\setminus I$. For example, $$c_\varnothing=(b_1,b_2,\ldots,b_n)\quad c_{\{1\}}=(a_1^-,b_2,\ldots,b_n)$$ and $$c_{\{1,3\}}=(a_1^-,b_2,a_3^-,b_4,\ldots,b_n)\qquad c_N=(a_1^-,a_2^-,\ldots,a_n^-)$$ Then, $$P(A)=\sum_{I \subseteq N}(-1)^{\#I}F(c_I)$$ where the sum on the RHS has $2^n$ terms. This follows from an identity between random variables. Namely, for every $I\subseteq N$, let $$A_I=\{(x_1,x_2,\ldots,x_n)\mid \forall i\in I,x_i<a_i,\forall j\in N\setminus I,x_j\leqslant b_j]$$ then, for every $I\subseteq N$, $$F(c_I)=P(A_I)$$ and $$\mathbf 1_A(x_1,x_2,\ldots,x_n)=\prod_{i=1}^n(\mathbf 1_{x_i\leqslant b_i}-\mathbf 1_{x_i<a_i})$$ hence, expanding the product in the RHS, $$\mathbf 1_A=\sum_{I \subseteq N}(-1)^{\#I}\mathbf 1_{A_I}$$

• And how can I proof this? :( – Alopiso Oct 11 '17 at 17:19
• But this doesn't work if $k=2$ because I get that $P([a_1,b_1]\times [a_2,b_2])=F(b_1,b_2)+F(a_1-,a_2-)-F(a_1-,b_2-)-F(b_1-,a_2-)$, but you don't consider if some $b_i$ is $b_i-$. – Alopiso Oct 13 '17 at 13:25
• Sorry but this does "work" (to use your parlance) and no $b^-$ is involved. – Did Oct 13 '17 at 13:28
• Then, Am I making a mistake if $k=2$? – Alopiso Oct 13 '17 at 15:05
• Sorry but I cannot say, actually I am not sure what your objection is. – Did Oct 14 '17 at 11:02