Product Laws as Characterizations of Independence

From Event Independence to a Law of the Pair

Independence can be stated in terms of events: knowing whether one random element falls in a specified measurable set should not change the probability of the other doing so. There is an equivalent way to express this idea directly through probability laws. The joint law of the pair must assign to each measurable rectangle the product of the probabilities assigned by its two marginal laws.

This formulation is useful because joint laws describe the pair as a single random object, while marginal laws describe each component separately. The characterization below connects those descriptions without needing to construct a product measure.

Joint and Marginal Laws

Let $(\Omega,\mathcal F,P)$ be a probability space. Let $(S,\mathcal S)$ and $(T,\mathcal T)$ be measurable spaces, and let $X:\Omega\to S$ and $Y:\Omega\to T$ be measurable random elements. Write $\mathcal S\otimes\mathcal T$ for the product sigma-algebra on $S\times T$, generated by the measurable rectangles $A\times B$, where $A\in\mathcal S$ and $B\in\mathcal T$. The pair $(X,Y)$ is a measurable map into this product space.

Definition: The joint law of $(X,Y)$, denoted by $\mu_{X,Y}$, is the probability measure on $(S\times T,\mathcal S\otimes\mathcal T)$ given by

\mu_{X,Y}(C)=P\bigl((X,Y)\in C\bigr), \qquad C\in\mathcal S\otimes\mathcal T.

The marginal laws are the probability measures $\mu_X$ on $(S,\mathcal S)$ and $\mu_Y$ on $(T,\mathcal T)$, defined by

\mu_X(A)=P(X\in A), \qquad \mu_Y(B)=P(Y\in B).

For $A\in\mathcal S$ and $B\in\mathcal T$, the set $A\times B$ is a measurable rectangle. Its inverse image under the pair map is

\{(X,Y)\in A\times B\} = \{X\in A\}\cap\{Y\in B\}.

Consequently, the joint law on a rectangle records the probability that both component events occur. This elementary identity is the link between the event-based definition of independence and the law-based characterization.

Proposition: The marginals of the joint law are recovered by evaluating it on rectangles whose other side is the whole state space:

\mu_{X,Y}(A\times T)=\mu_X(A), \qquad \mu_{X,Y}(S\times B)=\mu_Y(B).

Proof: Since $Y$ takes values in $T$ with probability one,

\mu_{X,Y}(A\times T) =P(X\in A,\ Y\in T) =P(X\in A) =\mu_X(A).

Likewise, $X\in S$ with probability one, so

\mu_{X,Y}(S\times B) =P(X\in S,\ Y\in B) =P(Y\in B) =\mu_Y(B).

This also shows why a proposed joint law must be consistent with its proposed marginals. $\square$

The Product-Law Characterization

Independence of the random elements means independence of the sigma-algebras they generate. Equivalently, every event determined by $X$ is independent of every event determined by $Y$. The theorem makes this condition a statement about the joint law.

Theorem: The following conditions are equivalent:

The sigma-algebras $\sigma(X)$ and $\sigma(Y)$ are independent.
For every $A\in\mathcal S$ and $B\in\mathcal T$, $P(X\in A,\ Y\in B)=P(X\in A)P(Y\in B).$
For every $A\in\mathcal S$ and $B\in\mathcal T$, the joint law satisfies $\mu_{X,Y}(A\times B)=\mu_X(A)\mu_Y(B).$

Proof: By definition, $\sigma(X)$ and $\sigma(Y)$ are independent if $P(E\cap F)=P(E)P(F)$ for every $E\in\sigma(X)$ and $F\in\sigma(Y)$. Every event in $\sigma(X)$ has the form $\{X\in A\}$ for some $A\in\mathcal S$: the collection of all such inverse images is itself a sigma-algebra and is exactly the sigma-algebra generated by them. The same holds for $Y$. Thus independence of the two sigma-algebras is equivalent to the stated factorization for every measurable $A$ and $B$. This proves the equivalence of the first two conditions.

By the definition of the joint law and the identity for the inverse image of a rectangle,

\mu_{X,Y}(A\times B) =P(X\in A,\ Y\in B).

The definitions of the marginal laws give $\mu_X(A)=P(X\in A)$ and $\mu_Y(B)=P(Y\in B)$. Substitution shows that the second and third conditions are the same formula written in terms of laws. Hence all three conditions are equivalent. $\square$

The theorem’s product formula is a statement about every measurable rectangle. It says that on such a set, the joint law has the value obtained by multiplying the two marginal probabilities. It does not require constructing a measure on the product space from the marginals: the joint law is already defined as the law of $(X,Y)$, and its values on rectangles can be evaluated directly.

Worked Examples

Worked Example: Two Independent Bernoulli Random Elements

Let $\Omega=\{0,1\}\times\{0,1\}$, give each of its four points probability $1/4$, and let $X$ and $Y$ be the first and second coordinate. Take $S=T=\{0,1\}$ with all subsets measurable. For any $A\subseteq S$ and $B\subseteq T$, there are $|A||B|$ points of $A\times B$, so

\mu_{X,Y}(A\times B)=\frac{|A||B|}{4}.

Each coordinate takes each value with probability $1/2$. Therefore

\mu_X(A)\mu_Y(B) =\frac{|A|}{2}\frac{|B|}{2} =\frac{|A||B|}{4} =\mu_{X,Y}(A\times B).

The equality holds for every subset $A$ and $B$, including the empty set and the whole state space. The theorem therefore shows that $X$ and $Y$ are independent.

Worked Example: Equal Marginals Do Not Ensure Independence

Let $\Omega=\{0,1\}$, with probability $1/2$ on each outcome, and set $X(\omega)=\omega$ and $Y(\omega)=\omega$. Both marginal laws are Bernoulli with probability $1/2$ at $1$. But for $A=B=\{1\}$,

\mu_{X,Y}(A\times B) =P(X=1,Y=1) =\frac12, \qquad \mu_X(A)\mu_Y(B) =\frac12\cdot\frac12 =\frac14.

This one rectangle violates the required product formula, so $X$ and $Y$ are not independent. The example illustrates why knowing both marginals is not enough to determine whether the pair is independent: the joint law must also have the required values on rectangles.

Worked Example: Independent Uniform Coordinates

Let $\Omega=[0,1]^2$ with the usual uniform area probability, and let $U(x,y)=x$ and $V(x,y)=y$ be the coordinate random elements. Take measurable sets $A,B\subseteq\mathbb R$. Their inverse images in the square form $(A\cap[0,1])\times(B\cap[0,1])$. Using the indicator functions of these sets and Fubini’s theorem,

\begin{aligned} P(U\in A,V\in B) &=\int_{[0,1]^2} \mathbf{1}_{A\cap[0,1]}(x)\mathbf{1}_{B\cap[0,1]}(y)\,dx\,dy\\ &=\left(\int_0^1\mathbf{1}_{A\cap[0,1]}(x)\,dx\right) \left(\int_0^1\mathbf{1}_{B\cap[0,1]}(y)\,dy\right)\\ &=P(U\in A)P(V\in B). \end{aligned}

The integrands are bounded and measurable, so Fubini applies. Since the calculation holds for every measurable $A$ and $B$, the rectangle formula holds for every measurable rectangle, and the coordinates are independent.

Worked Example: A Constant Random Element

Suppose $X$ is constant, with $X(\omega)=s_0$ for every $\omega\in\Omega$, and let $Y$ be any measurable random element. For $A\in\mathcal S$, the probability $P(X\in A)$ is $1$ if $s_0\in A$, and $0$ otherwise. For any $B\in\mathcal T$, if $s_0\in A$, then

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

If $s_0\notin A$, then the event $\{X\in A,Y\in B\}$ is empty, and

P(X\in A,Y\in B)=0=P(X\in A)P(Y\in B).

The factorization holds in both cases and for every measurable $B$. Hence a constant random element is independent of every random element.

What the Rectangle Formula Does—and Does Not—Say

The condition must hold for all measurable choices of $A$ and $B$. Checking a few convenient rectangles can reveal dependence, as in the equal-coordinate example, but successful checks on a small selection of rectangles alone do not establish independence.

It is also important to distinguish the rectangle formula from a claim about arbitrary measurable subsets of $S\times T$. A general measurable set need not itself be a rectangle. The result proved here characterizes independence by the joint law’s values on rectangles; it does not claim that the joint law and some other measure agree on every measurable set. Such a broader measure-identification statement requires an additional uniqueness argument.

The distinction is useful in practice. To disprove independence, one rectangle with a failed product identity suffices. To prove independence directly from the definition, the identity must hold for every pair of measurable sets in the component spaces. When the random elements take values in familiar spaces, the choice of measurable sets may be simple; for general measurable spaces, retaining the full quantifier is essential.

Key takeaway: Independence of $X$ and $Y$ is equivalent to the identity $\mu_{X,Y}(A\times B)=\mu_X(A)\mu_Y(B)$ for every $A\in\mathcal S$ and $B\in\mathcal T$. The theorem concerns measurable rectangles and uses only the joint and marginal laws already defined.

Check Your Understanding

Use the definitions and product-law characterization to answer these questions.

How is the joint law of $(X,Y)$ defined on a measurable set $C\subseteq S\times T$?
Why does $\mu_{X,Y}(A\times T)=\mu_X(A)$ hold?
Which event identity connects the joint law on a rectangle to independence of $\sigma(X)$ and $\sigma(Y)$?
In the equal-coordinate example, which rectangle demonstrates that the variables are dependent?
Why does verifying the product identity for only a few rectangles generally not prove independence?