Finite Partitions as Independence Certificates

Why Check Independence Through Partitions?

Independence is a statement about probabilities of combinations of events. When the events of interest come from finite categories, those combinations form cells: intersections of one category from each partition. Checking the probabilities of these cells gives a direct way to verify independence without testing every possible union of categories separately.

The crucial point is that a certificate must cover every tuple of cells in the partitions being studied. A matching probability for one cell, or even for one particular intersection, does not by itself establish independence.

Partitions and Their Cells

Definition: A finite measurable partition of a probability space $(\Omega,\mathcal F,P)$ is a finite collection $\mathcal P=\{A_1,\ldots,A_m\}$ of measurable events that are pairwise disjoint and whose union is $\Omega$. Its members $A_j$ are the atoms of the partition. Atoms may have probability zero.

Given $k$ finite partitions $\mathcal P_r=\{A_{r,1},\ldots,A_{r,m_r}\}$, for $r=1,\ldots,k$, a joint cell is an intersection formed by selecting one atom from each partition: $A_{1,j_1}\cap\cdots\cap A_{k,j_k}$. Some joint cells may be empty or have probability zero.

Definition: The partitions $\mathcal P_1,\ldots,\mathcal P_k$ are independent if the sigma-fields they generate are mutually independent. This means that for every choice of events $B_r\in\sigma(\mathcal P_r)$, $P(B_1\cap\cdots\cap B_k)=\prod_{r=1}^k P(B_r)$. Here $\sigma(\mathcal P_r)$ consists of all unions of atoms of $\mathcal P_r$.

For two partitions, independence asks for factorization for every event formed by unions of atoms in the first partition and every such event in the second. The finite-cell criterion below shows why it is enough to check the atoms themselves.

The Finite-Partition Criterion

Theorem: Finite measurable partitions $\mathcal P_1,\ldots,\mathcal P_k$ are mutually independent if and only if, for every choice $j_r\in\{1,\ldots,m_r\}$, $$ P(A_{1,j_1}\cap\cdots\cap A_{k,j_k}) =\prod_{r=1}^k P(A_{r,j_r}). $$

Proof. If the partitions are independent, each atom $A_{r,j_r}$ belongs to $\sigma(\mathcal P_r)$. Apply the definition of mutual independence to these atoms to obtain the displayed factorization.

Conversely, suppose the factorization holds for every joint cell. For each $r$, let $B_r\in\sigma(\mathcal P_r)$. Then $B_r$ is a union of some atoms; write $J_r$ for their indices, so $B_r=\bigcup_{j\in J_r}A_{r,j}$. Disjointness of the atoms gives

P(B_1\cap\cdots\cap B_k) =\sum_{j_1\in J_1}\cdots\sum_{j_k\in J_k} P(A_{1,j_1}\cap\cdots\cap A_{k,j_k}).

Using the assumed cell factorization and distributing the finite sums,

P(B_1\cap\cdots\cap B_k) =\sum_{j_1\in J_1}\cdots\sum_{j_k\in J_k} \prod_{r=1}^k P(A_{r,j_r}) =\prod_{r=1}^k\left(\sum_{j_r\in J_r}P(A_{r,j_r})\right) =\prod_{r=1}^k P(B_r).

Thus the generated sigma-fields are mutually independent. The argument uses only finite additivity over disjoint cells, so zero-probability atoms cause no exception. $\square$

Using the Criterion for Finite-Valued Variables

If $X_r$ takes values in a finite set, its value categories $\{X_r=x\}$ form a finite measurable partition after the possible values are listed. The theorem then says that $X_1,\ldots,X_k$ are mutually independent exactly when every joint probability factors into the corresponding marginal probabilities. Equivalently, the joint probability mass function at each tuple is the product of its marginal mass functions.

Worked Example: Checking a Nonuniform Joint Table

Suppose finite-valued variables $X$ and $Y$ have the following joint probability table. Determine whether they are independent.

	$Y=a$	$Y=b$	$Y=c$	Row total
$X=u$	0.08	0.12	0.20	0.40
$X=v$	0.12	0.18	0.30	0.60
Column total	0.20	0.30	0.50	1.00

The row and column totals give the marginal probabilities: $P(X=u)=0.40$, $P(X=v)=0.60$, and $P(Y=a)=0.20$, $P(Y=b)=0.30$, $P(Y=c)=0.50$. Check every cell against the product of its row and column marginals:

$P(X=u,Y=a)=0.08=0.40(0.20)$, $P(X=u,Y=b)=0.12=0.40(0.30)$, and $P(X=u,Y=c)=0.20=0.40(0.50)$.

For the second row, $P(X=v,Y=a)=0.12=0.60(0.20)$, $P(X=v,Y=b)=0.18=0.60(0.30)$, and $P(X=v,Y=c)=0.30=0.60(0.50)$. All six joint cells factor, so the two value partitions—and hence $X$ and $Y$—are independent.

The theorem also makes clear what a finite partition can certify. If the atoms are full value categories of finite-valued variables, the cell check establishes independence of those variables. If the atoms instead group several values together, the check establishes independence only for the information represented by those groups. It does not automatically establish independence for finer distinctions hidden within a group.

Collections of Events: Check the Binary Cells

For an event $E_r$, the associated binary partition is $\{E_r,E_r^c\}$, where $E_r^c$ is its complement. A collection of events $E_1,\ldots,E_k$ is mutually independent precisely when these binary partitions are independent. Thus the criterion requires factorization for every choice of $E_r$ or $E_r^c$, not just the cell $E_1\cap\cdots\cap E_k$.

Worked Example: Certifying Three Independent Events

Let $\Omega=\{0,1\}^3$ consist of the eight binary triples, each with probability $1/8$. For $r=1,2,3$, let $E_r$ be the event that coordinate $r$ equals 1. The binary partitions are determined by whether each coordinate is 0 or 1.

Each tuple of coordinate values specifies exactly one outcome. Therefore every joint cell, such as $E_1\cap E_2^c\cap E_3$, has probability $1/8$. Each event $E_r$, and each complement $E_r^c$, contains four outcomes, so each has probability $4/8=1/2$. For any choices $C_r\in\{E_r,E_r^c\}$,

P(C_1\cap C_2\cap C_3)=\frac18 =\frac12\cdot\frac12\cdot\frac12 =P(C_1)P(C_2)P(C_3).

This verifies all eight joint cells, one for each choice of event or complement. The finite-partition criterion therefore proves mutual independence of $E_1,E_2,E_3$.

For more than two events, checking the probability of their full intersection alone is not enough: it tests only one of the $2^k$ binary cells. Likewise, checking all pairs alone is not the same as checking every joint cell for the entire collection. The cell criterion specifies the complete set of checks required for mutual independence.

A Cell Mismatch Refutes Independence

To disprove independence, only one failing cell is needed. Its joint probability must differ from the product of its marginal probabilities. When outcomes are equally likely, the probability of a cell is its number of underlying outcomes divided by the total number of outcomes. The underlying outcomes must be counted, even when several outcomes receive the same values of the variables.

Worked Example: Repeated Coordinate Values Produce Dependence

Let $\Omega=\{\omega_1,\ldots,\omega_6\}$, with each outcome having probability $1/6$. Assign the values of $(U,V)$ as follows:

Outcome	$U$	$V$
$\omega_1$	1	0
$\omega_2$	1	1
$\omega_3$	2	0
$\omega_4$	2	0
$\omega_5$	3	1
$\omega_6$	3	1

Each of $U=1$, $U=2$, and $U=3$ occurs twice, so each has probability $2/6=1/3$. The event $V=0$ occurs for $\omega_1,\omega_3,\omega_4$, and $V=1$ occurs for $\omega_2,\omega_5,\omega_6$; hence each has probability $3/6=1/2$. The cell $U=2,V=0$ contains $\omega_3$ and $\omega_4$, so

P(U=2,V=0)=\frac26=\frac13, \qquad P(U=2)P(V=0)=\frac13\cdot\frac12=\frac16.

These probabilities are unequal. This single cell mismatch proves that the value partitions of $U$ and $V$ are not independent, and therefore $U$ and $V$ are dependent.

Coarsening and the Limits of a Certificate

A useful consequence of the finite-cell theorem is that independence is preserved when each partition is coarsened by joining some of its atoms. In that case, each new cell is a union of old atoms, and its probability is a sum of their probabilities.

Result: If finite partitions $\mathcal P_1,\ldots,\mathcal P_k$ are independent, then any partitions obtained by grouping atoms separately within each $\mathcal P_r$ are also independent.

Proof. Let $D_r$ be a cell in the coarsened partition derived from $\mathcal P_r$, and let $J_r$ index the original atoms that make it up. Then $D_r=\bigcup_{j\in J_r}A_{r,j}$. By disjointness and independence of the original partitions,

\begin{aligned} P(D_1\cap\cdots\cap D_k) &=\sum_{j_1\in J_1}\cdots\sum_{j_k\in J_k} \prod_{r=1}^k P(A_{r,j_r})\\ &=\prod_{r=1}^k\left(\sum_{j_r\in J_r}P(A_{r,j_r})\right) =\prod_{r=1}^k P(D_r). \end{aligned}

Every coarsened joint cell therefore factors, so the criterion proves independence of the coarsened partitions. The converse need not hold: merging categories can hide dependence that would be visible in the finer cells. $\square$

Common Mistakes

Counting value pairs instead of underlying outcomes. Distinct outcomes can map to the same pair of variable values. Count all outcomes in each cell, then divide by the total probability as appropriate.
Checking only one cell. A cell that factors does not certify independence; the criterion requires every joint cell. A single cell that fails to factor does, however, refute independence.
Checking only the event intersection. For a collection of events, mutual independence requires the binary cells formed from every combination of events and complements.
Confusing a grouped certificate with a full one. Independence for coarse categories establishes independence for those categories, not necessarily for finer values within them.

Key takeaway: For finite partitions, independence is exactly the factorization of every joint-cell probability into its marginal cell probabilities. Use a complete grid of cells to certify independence; use any one failing cell to disprove it.

Check Your Understanding

For each question, identify which cell probabilities need to be checked and explain what the check can establish.

Two partitions have three atoms and two atoms, respectively. How many joint cells must be checked to certify their independence?
Three events satisfy $P(E_1\cap E_2\cap E_3)=P(E_1)P(E_2)P(E_3)$. Why does this equality alone not certify mutual independence?
A finite-valued variable has value categories that are grouped into two larger categories. If those larger categories are independent of another partition, what has been established—and what has not necessarily been established?
In an equally likely finite sample, two outcomes have the same pair of variable values. How should the probability of that joint cell be calculated?
What does one joint cell with $P(A\cap B)\ne P(A)P(B)$ imply about the independence of the two partitions?

Outcome	\(U\)	\(V\)
\(\omega_1\)	1	0
\(\omega_2\)	1	1
\(\omega_3\)	2	0
\(\omega_4\)	2	0
\(\omega_5\)	3	1
\(\omega_6\)	3	1

	\(Y=a\)	\(Y=b\)	\(Y=c\)	Row total
\(X=u\)	0.08	0.12	0.20	0.40
\(X=v\)	0.12	0.18	0.30	0.60
Column total	0.20	0.30	0.50	1.00