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Joint CDF of Discrete Random Variables

Definition

The joint CDF of a random pair (X,Y)(X,Y) is

FX,Y(x,y)=P(X≤x,Y≤y).\begin{aligned} F_{X,Y}(x,y)=\mathbb{P}(X\leq x,Y\leq y). \end{aligned}

The definition applies to any random pair; this page develops its discrete form, and Joint CDF of Continuous Random Variables develops the form with a joint density.


Interpretation

The joint CDF measures the probability that two variables simultaneously lie below specified thresholds: FX,Y(x,y)F_{X,Y}(x,y) is the probability that XX is at most xx and YY is at most yy. For a discrete pair it is the sum of the joint masses in the lower-left corner below (x,y)(x,y).

A 6 by 6 grid of two-dice outcomes with the lower-left block of x at most 2 and y at most 3 highlighted; its six outcomes give the joint CDF value 1/6.

For two fair dice the highlighted corner {X≤2, Y≤3}\{X\le2,\,Y\le3\} contains 2⋅3=62\cdot3=6 of the 3636 equally likely outcomes, so FX,Y(2,3)=6/36=16F_{X,Y}(2,3)=6/36=\tfrac{1}{6}. Because the dice are independent, this also equals FX(2)FY(3)=26⋅36F_X(2)F_Y(3)=\tfrac{2}{6}\cdot\tfrac{3}{6}.

The joint CDF is defined at every point (x,y)(x,y) of the plane, not only at possible values. Between possible values it is constant, so for the dice FX,Y(2.5,3.7)=FX,Y(2,3)F_{X,Y}(2.5,3.7)=F_{X,Y}(2,3). It is nondecreasing in each argument, and letting y→∞y\to\infty removes the condition on YY and leaves the CDF of XX. In general the joint CDF is not the product of the two marginal CDFs; it is for independent variables. For X=Y∼Bern⁡(0.4)X=Y\sim\operatorname{Bern}(0.4), FX,Y(0,0)=0.6F_{X,Y}(0,0)=0.6, not 0.620.6^2.

Probabilities of rectangles follow from four corner values by inclusion–exclusion: for x1<x2x_1<x_2 and y1<y2y_1<y_2,

P(x1<X≤x2, y1<Y≤y2)=FX,Y(x2,y2)−FX,Y(x1,y2)−FX,Y(x2,y1)+FX,Y(x1,y1).\mathbb{P}(x_1<X\le x_2,\,y_1<Y\le y_2) =F_{X,Y}(x_2,y_2)-F_{X,Y}(x_1,y_2) -F_{X,Y}(x_2,y_1)+F_{X,Y}(x_1,y_1).
P(x1<X≤x2, y1<Y≤y2)=FX,Y(x2,y2)−FX,Y(x1,y2)−FX,Y(x2,y1)+FX,Y(x1,y1).\begin{aligned} &\mathbb{P}(x_1<X\le x_2,\,y_1<Y\le y_2)\\ &=F_{X,Y}(x_2,y_2)-F_{X,Y}(x_1,y_2)\\ &\quad-F_{X,Y}(x_2,y_1)+F_{X,Y}(x_1,y_1). \end{aligned}

For the dice, FX,Y(2,3)−FX,Y(1,3)−FX,Y(2,2)+FX,Y(1,2)=(6−3−4+2)/36=136=P(X=2,Y=3)F_{X,Y}(2,3)-F_{X,Y}(1,3)-F_{X,Y}(2,2)+F_{X,Y}(1,2)=(6-3-4+2)/36=\tfrac{1}{36}=\mathbb{P}(X=2,Y=3). The joint CDF of continuous variables has the same definition, with the sum replaced by an integral of a joint density.


Key Results

  • FX,Y(x,y)=∑u≤x∑v≤ypX,Y(u,v)\displaystyle F_{X,Y}(x,y)=\sum_{u\le x}\sum_{v\le y}p_{X,Y}(u,v).
  • FX(x)=lim⁡y→∞FX,Y(x,y)F_X(x)=\lim_{y\to\infty}F_{X,Y}(x,y) and similarly for YY.

Examples

  • For independent fair dice, FX,Y(2,3)=(2/6)(3/6)=1/6F_{X,Y}(2,3)=(2/6)(3/6)=1/6.
  • If X=YX=Y is Bernoulli with parameter 0.40.4, then FX,Y(0,0)=0.6F_{X,Y}(0,0)=0.6, rather than 0.620.6^2.

Prerequisites

Sources

Keywords: joint CDF, joint distribution function Symbols: infinity, sum

Last updated: 20 October 2026 03:02