When \(X\) and \(Y\) are continuous random variables, the following will happen:

1. It will be possible to write the joint pdf as a product of the individual pdf's over a rectangular region, possible after some change of coordinates. In particular, \(f(x,y) = f_X(x)f_Y(y)\) for all \((x,y) \in R\) for some possibly infinite rectangle \(R.\)
2. The conditional pdf's will be the same as the individual pdf's. In formulas, \(f_{Y|X}(y|x) = f_Y(y)\) and \(f_{X|Y}(x|y) = f_X(x)\)

Example

Let \(X\) and \(Y\) be independent random variables. The pdf of \(X\) is \(f_X(x) = \frac{1}{9}x^2\) for \(0 < x < 3\) and the pdf of \(Y\) is \(f_Y(y) = y+1\) for \(0 < y < \sqrt{3}-1.\) What is the probability that \((X,Y) \in (1,2) \times (0, \frac{1}{2})?\)

Since \(X\) and \(Y\) are independent, the joint pdf of \(X\) and \(Y\) is \[f(x,y) = f_X(x)f_Y(y) = \frac{1}{9}x^2(y+1).\] Therefore, \(P((X,Y) \in (1,2) \times (0, \frac{1}{2}))\) is \begin{align} \int_1^2 \int_0^{1/2} \frac{1}{9}x^2(y+1) dx dy & = \left(\int_1^2 \frac{1}{9}x^2 dx\right) \left(\int_0^{1/2} (y+1) dy\right)\\ & = \left(\left. \frac{x^3}{27} \right|_1^2\right)\left( \left. \frac{y^2}{2} + y \right|_0^{1/2}\right) \\ & = \left(\frac{8}{27} - \frac{1}{27}\right)\left(\frac{1}{8} + \frac{1}{2}\right) \\ & = \frac{35}{216} \end{align}