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 $$\begin{align} \int_1^2 \int_0^{1/2} \frac{1}{9}x^2(y+1) dx dy & = \left(\int_1^2 \frac{1}{9}x^2 dx\right) \left(\int_0^{1/2} (y+1) dy\right)\\ & = \left(\left. \frac{x^3}{27} \right|_1^2\right)\left( \left. \frac{y^2}{2} + y \right|_0^{1/2}\right) \\ & = \left(\frac{8}{27} - \frac{1}{27}\right)\left(\frac{1}{8} + \frac{1}{2}\right) \\ & = \frac{35}{216} \end{align}$$