Multi-dimensional CDF on a discrete support

Suppose I have two discrete-support random variables, $$X$$ and $$Y$$.

They have joint CDF $$F(X,Y)$$.

If I want to find

$$\Pr(a \leq X \leq b , c \leq Y \leq d)$$.

It is obviously not:

$$F(b ,d)-F(a-1 ,c-1)$$.

What is it then, in simplest terms with respect to the CDF?

I believe, for two dimensions, it is:

$$F(b ,d)-F(a-1 ,d)-F(b,c-1)+F(a-1,c-1)$$.

What is it for an arbitrary $$d$$ dimensions?

• By definition of $F,$ $$\Pr(a \lt X \le b,\ c \lt Y \le d) = F(b,d) - F(a,d) - F(b,c) + F(a,c).$$ Whether that answers your question depends on what you mean by "discrete-support:" would that mean with support on an integral lattice (which is a very narrow meaning) or would it literally mean discrete (nowhere continuous)? – whuber Jan 9 at 19:44
• On lattice, natural numbers, for example. – wolfsatthedoor Jan 9 at 19:50
• Good: I hope the answer is now clear. – whuber Jan 9 at 19:53
• What about for dimensions > 2? – wolfsatthedoor Jan 9 at 19:54
• Consider drawing a picture of the four planar regions represented by the four terms. Its generalization to three dimensions might then become clearer, from which you ought to be able to see the pattern. – whuber Jan 9 at 19:55

The distribution function (CDF) of a random variable $$X=(X_1,X_2,\ldots, X_n)$$ with values in $$\mathbb{R}^n$$ is defined to be

$$F_X(x_1,x_2,\ldots, x_n) = \Pr(X_1\le x_1,\, X_2\le x_2,\ \ldots,\ X_n\le x_n).$$

To find the chance that $$X$$ lies within some nonempty (half-open) parallelipiped $$\mathcal{I} = (i_1,j_1] \times (i_2,j_2] \times \cdots \times (i_n,j_n]$$

(or equivalently, to find the chance that simultaneously $$i_k \lt X_k \le j_k$$ for $$k=1,2,\ldots,n$$), simplify the problem by choosing coordinate axes centered on $$(i_1,i_2,\ldots,i_n)$$ with unit distances $$j_k-i_k.$$ This makes all the $$i_k$$ equal to $$0$$ and all the $$j_k$$ equal to $$1:$$ $$\mathcal I$$ becomes the unit cube in $$\mathbb{R}^n.$$ (This is really just a notational convenience rather than a change of any mathematical substance.)

By partitioning the last coordinate into non-positive values and values less than or equal to $$1,$$ observe that the unit cube -- merely as a set -- can be expressed as the set difference

$$\mathcal{I} = \mathcal{I}^{(1)}\setminus\mathcal{I}^{(0)}$$

where (for any real number $$s$$) $$\mathcal{I}^{(s)} = (0,1]\times(0,1]\times \cdots \times (0,1] \times (-\infty, s].$$

The axioms of probability (specifically, that probabilities of non-intersection regions will add) imply

\eqalign{ \Pr(X\in\mathcal{I}) &= \Pr(X\in\mathcal{I}^{(1)}) - \Pr(X\in\mathcal{I}^{(0)}) \\ &= F(1,1,\ldots,1,1) - F(1,1,\ldots,1,0). }

Repeating this reduction at coordinates $$n-1,$$ $$n-2, ...$$ down to $$1$$ will double the number of terms at each iteration. The resulting combination therefore has $$2^n$$ terms. All terms are of the form $$\pm F(s_1,s_2,\ldots,s_n)$$ with $$s_j\in\{0,1\},$$ and therefore are distinct: there is no algebraic simplification. The sign is determined by the number of zeros in the arguments of $$F$$: negative for an odd number, positive for an even number.

This result can be written as an iterated sum over $$n$$ variables,

$$\Pr(X\in (0,1]^n) = \sum_{s_1\in\{0,1\}}\sum_{s_2\in\{0,1\}}\cdots \sum_{s_n\in\{0,1\}}(-1)^{n-\sum_{j=1}^n s_j}\ F(s_1,s_2, \ldots, s_n).$$

In a more succinct and memorable vector notation, where $$\mathbf{s} = (s_1,s_2,\ldots,s_n)$$ and $$|\mathbf{s}| = s_1+s_2+\cdots+s_n,$$ we may write

$$\Pr(X\in (0,1]^n) = \sum_{\mathbf{s}\in\{0,1\}^n} (-1)^{n-|\mathbf{s}|} \ F(\mathbf{s}).$$

When $$n=3,$$ for instance, this is

$$\Pr(X\in \mathcal (0,1]^3) = F(1,1,1)-F(1,1,0)-F(1,0,1)-F(0,1,1)+F(1,0,0)+F(0,1,0)+F(0,0,1)-F(0,0,0).$$

In terms of the original coordinates this would be written

$$\Pr(X\in \mathcal I) = F(j_1,j_2,j_3)-F(j_1,j_2,i_3)-F(j_1,i_2,j_3)-F(i_1,j_2,j_3)+F(j_1,i_2,i_3)+F(i_1,j_2,i_3)+F(i_1,i_2,j_3)-F(i_1,i_2,i_3).$$

Finally, when $$X$$ is supported on the integral lattice $$\mathbb{Z}^n$$ (of vectors all of whose coordinates are integers), because

$$\Pr(X_j \lt x) = \Pr(X_j \le x-1),$$

you may replace each $$i_k$$ by $$i_k-1$$ to find the chance that $$X$$ lies in the closed parallelipiped $$\bar{\mathcal{I}} = [i_1,j_1]\times \cdots \times [i_n,j_n].$$