Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's consider a absolutely continuous random vector $V \equiv (X,Y)$ and its associated joint distribution function $F(x,y)=Pr(X<=x,Y<=y) = \int_{-\infty}^{x}\int_{-\infty}^{y} f(x,y)dxdy$.

If we take four points in the xy plane that are vertexes of a rectangle $R$, $A \equiv (x_1,y_1) \;, B \equiv (x_1,y_0) \;,C \equiv (x_0,y_0) \;,A \equiv (x_0,y_1)$, with $x_1 > x_0$ and $y_1>y_0$, it is well known that the probability that the values of the random vector $V$ are within the rectangle $R$ is given by the value of the distribution function $F(x,y)$ taken at those points according to the below formula:


Is there an explicit formula, generalizing the one above, that applies when we move from 2 to N dimensions ?

In other words, given the distribution function $F(x_1,...,x_N)=Pr(X1<=x1,..,X_N<=x_N)$, it is there a formula that allows to compute $Pr(a_1<X_1<=b_1,...,a_N<X_N<=b_N)$ by the values of the N-dimensional distribution function computed in the vertexes of the hyper-rectangle $[a1,b1]x..x[a_N,b_N]$ ?

What happens if values $a_i$ or $b_j$ are allowed to be $+\infty$ ?

In two dimensions we find the value of the 1-dimensional marginal distributions $F_i(x)$, what's found in the N-dimensional case ?

From computational point of view, is this formula applicable in practice for value of N equal to 10 ? I suppose it involves $2^{10}$ vertexes ...

share|cite|improve this question
Inclusion-exclusion – Henry Mar 28 '12 at 9:20

Here is the formula you are looking for: $$ \color{red}{\mathrm P(a_i\lt X_i\leqslant b_i\ \text{for all}\ i)=\sum_c(-1)^{n(c)}F(c_1,c_2,\ldots,c_n)} $$ The sum in the RHS runs over the $2^n$ vectors $c=(c_i)_{1\leqslant i\leqslant n}$ such that $c_i\in\{a_i,b_i\}$ for every $1\leqslant i\leqslant n$, and $n(c)$ denotes the number of indices $i$ such that $c_i=a_i$.

The simplest proof might be to note that the LHS if the expectation of the random variable $$ \prod_{i=1}^n\left(\mathbf 1_{X_i\leqslant b_i}-\mathbf 1_{X_i\leqslant a_i}\right)=\sum_c(-1)^{n(c)}\prod_{i=1}^n\mathbf 1_{X_i\leqslant c_i}, $$ and to take the expectation of the RHS, remembering that the expectation is a linear operator and that, for every $c$, $$ \mathrm E\left(\prod_{i=1}^n\mathbf 1_{X_i\leqslant c_i}\right)=\mathrm P(X_i\leqslant c_i\ \text{for all}\ i)=F(c_1,c_2,\ldots,c_n). $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.