Joint Distributions for Continuous Random Variables

Joint PDF

For continuous random variables $X$ and $Y$, the joint PDF $f(x, y)$ satisfies:

$P((X, Y) \in A) = \iint_A f(x, y) \, dx \, dy$

Properties

1. $f(x, y) \geq 0$ for all $(x, y)$
2. $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dx \, dy = 1$

Interpretation

The joint PDF is a surface over the $xy$-plane. Probabilities correspond to volumes under the surface.

$P(a \leq X \leq b, c \leq Y \leq d) = \int_a^b \int_c^d f(x, y) \, dy \, dx$

Joint CDF

The joint CDF is:

$F(x, y) = P(X \leq x, Y \leq y) = \int_{-\infty}^x \int_{-\infty}^y f(u, v) \, dv \, du$

Recovering the PDF:

$f(x, y) = \frac{\partial^2 F(x, y)}{\partial x \partial y}$

Example: Uniform on Unit Square

$(X, Y)$ uniform on $[0, 1] \times [0, 1]$:

$f(x, y) = \begin{cases} 1 & 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$

$P(X + Y \leq 1) = \int_0^1 \int_0^{1-x} 1 \, dy \, dx = \int_0^1 (1 - x) \, dx = \left[x - \frac{x^2}{2}\right]_0^1 = \frac{1}{2}$

Marginal PDFs

$f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy$

$f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx$

Intuition: Integrate out one variable to get the distribution of the other.

Example

$f(x, y) = \begin{cases} 2e^{-x} e^{-2y} & x > 0, y > 0 \\ 0 & \text{otherwise} \end{cases}$

Find $f_X(x)$:

$f_X(x) = \int_0^{\infty} 2e^{-x} e^{-2y} \, dy = 2e^{-x} \int_0^{\infty} e^{-2y} \, dy = 2e^{-x} \cdot \frac{1}{2} = e^{-x}, \quad x > 0$

So $X \sim \text{Exp}(1)$.

Similarly: $f_Y(y) = 2e^{-2y}$, $y > 0$. So $Y \sim \text{Exp}(2)$.

Independence for Continuous Variables

$X$ and $Y$ are independent if:

$f(x, y) = f_X(x) \cdot f_Y(y) \quad \text{for all } x, y$

Equivalently:

$F(x, y) = F_X(x) \cdot F_Y(y)$

Example

For the PDF above:

$f_X(x) f_Y(y) = e^{-x} \cdot 2e^{-2y} = 2e^{-x} e^{-2y} = f(x, y)$

Thus $X$ and $Y$ are independent.

Counterexample

Let $f(x, y) = 8xy$ on $0 \leq x \leq y \leq 1$.

$f_X(x) = \int_x^1 8xy \, dy = 8x \cdot \frac{1 - x^2}{2} = 4x(1 - x^2)$

$f_Y(y) = \int_0^y 8xy \, dx = 8y \cdot \frac{y^2}{2} = 4y^3$

$f_X(x) f_Y(y) = 16xy^3(1 - x^2) \neq 8xy$

Not independent.

Example: Computing Probabilities

For $f(x, y) = 2e^{-x} e^{-2y}$ ($x, y > 0$), find $P(X < Y)$.

$P(X < Y) = \int_0^{\infty} \int_0^y 2e^{-x} e^{-2y} \, dx \, dy$

Inner integral: $\int_0^y 2e^{-x} \, dx = 2(1 - e^{-y})$

$P(X < Y) = \int_0^{\infty} 2e^{-2y} \cdot 2(1 - e^{-y}) \, dy = 4\int_0^{\infty} (e^{-2y} - e^{-3y}) \, dy$

$= 4\left(\frac{1}{2} - \frac{1}{3}\right) = 4 \cdot \frac{1}{6} = \frac{2}{3}$

Summary

Concept	Formula
Joint PDF	$P((X,Y) \in A) = \iint_A f(x,y) dx dy$
Marginal	$f_X(x) = \int f(x,y) dy$
Independence	$f(x,y) = f_X(x) f_Y(y)$
Joint CDF	$F(x,y) = P(X \leq x, Y \leq y)$