Independence of random variables — formal definition: Question bank

The core slogan we keep testing: the joint object must factor into the product of marginals at every point, AND the support must be a rectangle (a product set). Break either half and independence fails.

True or false — justify

Independence means

X

and

Y

can never both be large at the same time.

False. Independence says knowing one gives no information about the other — they can still coincidentally be large together; that joint probability just equals the product of the two separate probabilities.

X ⊥ Y

then

P (X \in A, Y \in B) = P (X \in A) P (Y \in B)

must hold only for intervals.

False. The definition demands it for all Borel sets

A, B

; it happens that checking intervals

(- \infty, x]

is enough because they generate every Borel set, but the property itself is required everywhere.

Cov (X, Y) = 0

then

X

and

Y

are independent.

False. Zero covariance only kills linear association;

X \sim U (- 1, 1)

with

Y = X^{2}

has

Cov = 0

yet

Y

is fully determined by

X

. See Covariance and correlation.

X ⊥ Y

then

Cov (X, Y) = 0

True. Independence gives

E [X Y] = E [X] E [Y]

, so

Cov = E [X Y] - E [X] E [Y] = 0

. The arrow only runs this direction.

A joint density that factors as

g (x) h (y)

always implies independence.

False. It implies independence only if the support is a product set (rectangle); on the triangle

0 \leq x \leq y \leq 1

the constant density

2

"factors" as

2 \cdot 1

yet

X, Y

are dependent.

For jointly Gaussian

X, Y

, zero correlation does imply independence.

True. Gaussians are the special family where uncorrelated

\Leftrightarrow

independent; setting the off-diagonal covariance to zero makes the joint density factor. See Jointly Gaussian random variables.

X ⊥ Y

then

X^{2}

and

Y^{3}

are also independent.

True. Any (measurable) functions of independent variables are independent, because a question about

X^{2}

is just a question about

X

, and those were assumed independent of all questions about

Y

If every pair among

X, Y, Z

is independent, then all three are mutually independent.

False. Pairwise independence is strictly weaker; three variables can be pairwise independent yet fail the three-way product rule

P (X \in A, Y \in B, Z \in C) = P (X \in A) P (Y \in B) P (Z \in C)

F_{X, Y} (x, y) = F_{X} (x) F_{Y} (y)

at one convenient point, that establishes independence.

False. The factorisation must hold for all

x, y

; one matching point (or one matching cell in a PMF table) can happen by coincidence.

Two independent variables must have the same distribution.

False. Independence is about the relationship between them, not their individual laws; an

Exp (1)

and a Bernoulli can be independent.

Spot the error

f_{X, Y} (x, y) = 2

0 \leq x \leq y \leq 1

factors as

2 \cdot 1

, so

X ⊥ Y

The support isn't a rectangle: the constraint

x \leq y

ties the variables. Knowing

X = 0.9

forces

Y \geq 0.9

, and indeed

f_{X} f_{Y} = 4 y (1 - x) \neq = 2

"They're uncorrelated, and correlation measures dependence, so they're independent."

Correlation measures only linear dependence. Non-linear links (like

Y = sin X

) can be invisible to covariance while making the variables far from independent.

p_{X, Y} (0, 0) = p_{X} (0) p_{Y} (0)

checks out, so the PMF table shows independence."

You must verify the product rule in every cell of the table; a single matching entry can occur even when other cells fail.

E [X Y] = E [X] E [Y]

, therefore

X

and

Y

are independent."

This equality is equivalent to zero covariance, which is only necessary for independence, not sufficient — same trap as the

Y = X^{2}

counterexample.

"Since

f_{X, Y} (x, y) = x + y

[0, 1]^{2}

has a rectangular support,

X ⊥ Y

The support is fine, but

x + y

does not factor into a product

g (x) h (y)

— a sum is not a product, so the density fails the factorisation half.

"The conditional density equals the marginal at

x = 0.5

, so

X ⊥ Y

Independence requires

f_{Y ∣ X} (y ∣ x) = f_{Y} (y)

for all

x

, not just one slice. See Conditional distributions and conditional independence.

X

and

Y

are independent, so

X

and

X + Y

are too."

False reasoning:

X + Y

contains

X

, so knowing

X

shifts the distribution of

X + Y

. Only the original independent pieces are independent, not derived combinations that reuse a variable.

Why questions

Why is checking intervals

(- \infty, x]

enough to guarantee independence for all sets?

Because those half-infinite intervals generate the whole Borel

σ

-algebra, so the product rule on them propagates to every measurable

A, B

Why must the support be a rectangle for the factorisation shortcut to work?

A non-rectangular support means the allowed values of

Y

depend on

X

(or vice versa), which is itself information transfer — independence forbids that regardless of the formula's shape.

Why do we say "the joint factors into marginals" rather than "the joint equals the marginals"?

The joint lives on the plane while each marginal lives on a line; independence is precisely the statement that the 2-D object rebuilds as the product of the two 1-D objects.

Why does independence imply

E [X Y] = E [X] E [Y]

Because

f_{X, Y} = f_{X} f_{Y}

lets the double integral

\iint x y f_{X, Y}

split into

(\int x f_{X}) (\int y f_{Y})

, i.e. the product of the separate expectations.

Why isn't zero covariance enough to recover independence in general?

Covariance is a single number capturing average linear co-movement; independence is an infinite family of constraints (one per pair of sets), so one number cannot encode all of them.

Why does convolution appear when adding independent variables?

Because independence makes the joint density factor, and summing forces an integral over all splits

x + (s - x) = s

, which is exactly the convolution

f_{X} * f_{Y}

. See Sums of independent random variables — convolution.

Why is independence of random variables built from independence of events?

A random variable is a machine turning outcomes into numbers, so "

X

tells nothing about

Y

" must mean every event

{X \in A}

is independent of every event

{Y \in B}

. See Independence of events.

Edge cases

Is a constant random variable

X = c

independent of every other variable

Y

Yes.

{X \in A}

is either the whole sample space or empty, and both are independent of any event, so the product rule holds trivially.

Can

X

be independent of itself?

Only if

X

is (almost surely) constant. Otherwise

P (X \leq x, X \leq x) = P (X \leq x) \neq = P (X \leq x)^{2}

, since a non-trivial probability

p

has

p \neq = p^{2}

If the joint density is zero on a set of measure zero (a curve), does that break independence?

No. Densities are defined "almost everywhere," so behaviour on a measure-zero set is irrelevant — the factorisation only needs to hold a.e.

Two discrete variables each take a single value with probability 1. Independent?

Yes, trivially — each is constant, and constants are independent of everything, since every event about them is the whole space or empty.

X

and

Y

have a rectangular support but the density vanishes in one corner, are they still possibly independent?

Only if the vanishing corner is consistent with a product

g (x) h (y)

; a "hole" that can't be written as

{g = 0} \cup {h = 0}

(a full row/column) means the density no longer factors, so they're dependent.

Is a mixture of two independent joint laws still independent?

Not generally. Averaging two product densities gives

\frac{1}{2} f_{1} g_{1} + \frac{1}{2} f_{2} g_{2}

, which usually does not factor — mixing induces dependence even from independent ingredients.

Recall One-line survival kit

Independence = factor everywhere + rectangular support. Zero covariance is a symptom, not a diagnosis. Functions of independent things stay independent; recombinations that reuse a variable do not.

Connections

Independence of Random Variables — Formal Definition
Joint distribution and marginals
Conditional distributions and conditional independence
Covariance and correlation
Jointly Gaussian random variables
Sums of independent random variables — convolution
Independence of events