From a joint density we can recover the density of each variable alone and the density of one given the other, just as rows and columns did for joint tables. Sums become integrals.
Key idea
$$f_X(x) = \int f(x, y)\,dy, \qquad f_{Y \mid X}(y \mid x) = \frac{f(x, y)}{f_X(x)},$$ integrating over the values of $y$ where $f(x, y) > 0$ for that $x$. $X$ and $Y$ are independent when $f(x, y) = f_X(x)\,f_Y(y)$ everywhere, which needs a support that is a rectangle.
Why it works
$$P(a \le X \le b) = \int_{a}^{b}\left(\int f(x, y)\,dy\right)dx$$ so the inner integral is the density of $X$: integrate out $y$. Dividing a slice $f(x, \cdot)$ by its total $f_X(x)$ makes it a density in $y$, the conditional density. If the support is not a rectangle, some pairs $(x, y)$ are impossible although $x$ and $y$ are each possible alone; there $f(x, y) = 0$ while $f_X(x)f_Y(y) > 0$, so independence fails.
Worked examples
Worked example: A constant on a triangle
$f(x, y) = 2$ for $0 \le x \le y \le 1$. Find $f_X$, $f_Y$ and $f_{Y \mid X}(y \mid x)$.
- For fixed $x$, $y$ runs from $x$ to 1: $f_X(x) = \int_{x}^{1} 2\,dy = 2(1 - x)$ on $[0, 1]$.
- For fixed $y$, $x$ runs from 0 to $y$: $f_Y(y) = \int_{0}^{y} 2\,dx = 2y$.
- $f_{Y \mid X}(y \mid x) = \frac{2}{2(1 - x)} = \frac{1}{1 - x}$ for $x \le y \le 1$: given $X = x$, $Y$ is uniform on $[x, 1]$.
Answer: $2(1 - x)$, $2y$, and uniform on $[x, 1]$.
Worked example: A sum on the square
$f(x, y) = x + y$ on the unit square. Find $f_X$ and decide whether $X$ and $Y$ are independent.
- $f_X(x) = \int_{0}^{1}(x + y)\,dy = x + \frac{1}{2}$; by symmetry $f_Y(y) = y + \frac{1}{2}$.
- $$f_X(x)f_Y(y) = xy + \frac{x + y}{2} + \frac{1}{4} \ne x + y$$ (try $x = y = 0$: $\frac{1}{4} \ne 0$).
Answer: $f_X(x) = x + \frac{1}{2}$; dependent.
Worked example: A product that is not independent
$f(x, y) = 8xy$ for $0 \le x \le y \le 1$. Are $X$ and $Y$ independent?
- The formula is a product, but the support is a triangle.
- $f_X(x) = \int_{x}^{1} 8xy\,dy = 4x(1 - x^{2})$ and $f_Y(y) = \int_{0}^{y} 8xy\,dx = 4y^{3}$; their product is not $8xy$.
- Directly: $f(0.9, 0.5) = 0$ although $f_X(0.9) > 0$ and $f_Y(0.5) > 0$.
Answer: No.
Common mistake
Calling $f(x, y) = 2$ on the triangle $0 < x < y < 1$ "a product of separate functions" because it is constant. The support is not a rectangle, so knowing $X$ limits $Y$ (if $X = 0.9$ then $Y \ge 0.9$). Check two things: the formula splits as (a function of $x$) × (a function of $y$), and the support is a rectangle.