Marginal Distributions

Definition

Given a joint distribution of $X$ and $Y$, the marginal distribution of $X$ gives the distribution of $X$ alone, ignoring $Y$.

The term “marginal” comes from writing joint distributions in tables with totals in the margins.

Discrete Case

$p_X(x) = \sum_y p(x, y)$

$p_Y(y) = \sum_x p(x, y)$

Example: Joint Distribution Table

$X \backslash Y$	$Y = 0$	$Y = 1$	$Y = 2$	$p_X(x)$
$X = 0$	0.10	0.05	0.05	0.20
$X = 1$	0.15	0.20	0.10	0.45
$X = 2$	0.05	0.15	0.15	0.35
$p_Y(y)$	0.30	0.40	0.30	1.00

To find $p_X(0)$: sum row $X = 0$: $0.10 + 0.05 + 0.05 = 0.20$.

To find $p_Y(1)$: sum column $Y = 1$: $0.05 + 0.20 + 0.15 = 0.40$.

Continuous Case

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

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

Example

$f(x, y) = 6xy \text{ on } 0 \leq x \leq 1, 0 \leq y \leq \sqrt{x}$

Find $f_X(x)$:

$f_X(x) = \int_0^{\sqrt{x}} 6xy \, dy = 6x \cdot \frac{x}{2} = 3x^2, \quad 0 \leq x \leq 1$

Find $f_Y(y)$:

First, determine the region. For fixed $y$, $x$ ranges from $y^2$ to 1.

$f_Y(y) = \int_{y^2}^1 6xy \, dx = 6y \cdot \frac{1 - y^4}{2} = 3y(1 - y^4), \quad 0 \leq y \leq 1$

Why Marginals Matter

Marginals let us analyse one variable while ignoring others:

1. Computing $P(X \in A)$: Use the marginal of $X$
2. Finding $E[X]$: Compute from the marginal
3. Testing independence: Compare joint to product of marginals

Information Loss

Marginals lose information about dependence:

• Many different joints can give the same marginals
• Knowing marginals does not determine the joint

Example: Same Marginals, Different Joints

Two joint distributions with identical marginals:

Joint 1 (independent):

$X \backslash Y$	$Y = 0$	$Y = 1$
$X = 0$	0.25	0.25
$X = 1$	0.25	0.25

Joint 2 (perfect positive correlation):

$X \backslash Y$	$Y = 0$	$Y = 1$
$X = 0$	0.5	0
$X = 1$	0	0.5

Both have $X \sim \text{Bern}(0.5)$ and $Y \sim \text{Bern}(0.5)$ as marginals, but completely different dependence structures.

Marginal Expectation

To compute expectations from marginals:

$E[X] = \sum_x x \cdot p_X(x) \quad \text{or} \quad \int x f_X(x) dx$

Using the joint directly (unnecessary but valid):

$E[X] = \sum_x \sum_y x \cdot p(x, y)$

Summary

Concept	Discrete	Continuous
Marginal of $X$	$\sum_y p(x,y)$	$\int f(x,y) dy$
Marginal of $Y$	$\sum_x p(x,y)$	$\int f(x,y) dx$
Loss	Joint $\to$ marginal loses dependence info