The Joint CDF of Two Continuous Random Variables

The joint cumulative distribution function of two continuous random variables, defined as FX,Y(x,y)=P(Xx,Yy)F_{X,Y}(x,y) = P(X \le x, Y \le y). Compute the joint CDF by integrating the joint PDF, recover the joint PDF via the mixed partial derivative 2F/xy\partial^2 F/\partial x\,\partial y, extract marginal CDFs by sending one argument to infinity, and check independence using the factorization FX,Y(x,y)=FX(x)FY(y)F_{X,Y}(x,y) = F_X(x)\,F_Y(y).

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Introduction

The joint cumulative distribution function (CDF) of two continuous random variables XX and YY is defined exactly as in the discrete case:

FX,Y(x,y)=P(Xx,Yy).F_{X,Y}(x,y) = P(X \le x,\, Y \le y).

If XX and YY have joint PDF fX,Y,f_{X,Y}, then the joint CDF is recovered by integrating the PDF over the region {(s,t):sx,ty}:\{(s,t) : s \le x,\, t \le y\}{:}

FX,Y(x,y)=xyfX,Y(s,t)dtds.F_{X,Y}(x,y) = \int_{-\infty}^{x} \int_{-\infty}^{y} f_{X,Y}(s,t)\,dt\,ds.

Geometrically, FX,Y(x,y)F_{X,Y}(x,y) is the volume under the joint density surface lying over the region to the lower-left of the point (x,y).(x,y).

For example, suppose fX,Y(s,t)=1f_{X,Y}(s,t) = 1 on the unit square [0,1]2[0,1]^2 (and zero elsewhere). Then for (x,y)[0,1]2,(x,y) \in [0,1]^2,

FX,Y(x,y)=0x0y1dtds=0xyds=xy.F_{X,Y}(x,y) = \int_0^x \int_0^y 1\,dt\,ds = \int_0^x y\,ds = xy.
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