Independence of Continuous Random Variables

Defines independence of two continuous random variables via factorization of the joint PDF, and develops the factorization criterion (including the rectangular-support requirement) for checking independence and computing joint probabilities.

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Defining Independence for Continuous Random Variables

Recall that two discrete random variables XX and YY are independent if and only if their joint PMF factors as p(x,y)=pX(x)pY(y)p(x,y)=p_X(x)\,p_Y(y) for all x,y.x,y. The same idea applies in the continuous case, with PDFs replacing PMFs.

Two continuous random variables XX and YY are independent if their joint PDF satisfies

f(x,y)=fX(x)fY(y)f(x,y)=f_X(x)\,f_Y(y)

for all x,yR.x,y\in\mathbb{R}.

Intuitively, knowing the value of one variable provides no information about the other — the joint behavior is completely determined by the two marginal distributions. If this equation fails for some (x,y),(x,y), then XX and YY are dependent.

For instance, suppose XX and YY have joint PDF

f(x,y)={x+y0x1, 0y10otherwisef(x,y)=\begin{cases} x+y & 0\le x\le 1,\ 0\le y\le 1 \\ 0 & \text{otherwise}\end{cases}

with marginals fX(x)=x+12f_X(x)=x+\tfrac12 and fY(y)=y+12f_Y(y)=y+\tfrac12 on [0,1].[0,1]. Then

fX(x)fY(y)=(x+12)(y+12)=xy+x2+y2+14,f_X(x)\,f_Y(y)=\left(x+\tfrac12\right)\left(y+\tfrac12\right)=xy+\tfrac{x}{2}+\tfrac{y}{2}+\tfrac14,

which is not equal to x+y.x+y. Therefore XX and YY are dependent.

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