Functions of Independent Blocks

Why Group Variables Into Blocks?

Many derived quantities use more than one observation. A statistic might be a function of the first two variables in a sample, while another is calculated from the next three. If the underlying variables are mutually independent, the two groups have no random inputs in common. The key question is whether applying different functions to those groups preserves independence. It does, provided the groups themselves are independent and the functions are measurable.

The result is useful because it lets us identify independence without calculating a full joint distribution for the derived quantities. We can establish independence from how the quantities are constructed, then calculate their marginal distributions separately.

Definition: Random elements $X_1,\ldots,X_n$ are mutually independent if their joint distribution is the product of their marginal distributions. Equivalently, for all Borel sets $C_1,\ldots,C_n$, $$ P(X_1\in C_1,\ldots,X_n\in C_n) = \prod_{i=1}^n P(X_i\in C_i). $$ A block is a vector formed from a specified group of these random elements. A function of a block is assumed measurable, so its value is a random element.

Here, each $X_i$ may be real-valued or a finite-dimensional random vector. A block can contain one variable or many. We consider nonempty, disjoint index sets $I_1,\ldots,I_m$, where $I_j$ specifies the indices used by the $j$th function. The sets need not include every index in the original collection.

The Block-Function Theorem

First, grouping mutually independent random elements into disjoint blocks preserves independence between the blocks. In fact, the joint distribution of all the original elements is the product of their marginal distributions. Reordering coordinates and grouping them therefore gives the product of the marginal distributions of the block vectors. Equality of these distributions can be checked on measurable rectangles, which determine probability measures.

Theorem: Let $X_1,\ldots,X_n$ be mutually independent random elements, and let $I_1,\ldots,I_m$ be pairwise disjoint, nonempty subsets of $\{1,\ldots,n\}$. Write $X_{I_j}$ for the vector of variables indexed by $I_j$. Let $g_j$ be a measurable function on the space of $X_{I_j}$, with values in a Euclidean space, and define $Y_j=g_j(X_{I_j})$. Then $Y_1,\ldots,Y_m$ are mutually independent.

The block functions do not need to be linear, one-to-one, or of the same form. The theorem permits each $g_j$ to produce a scalar or a vector. Dependence between different quantities calculated from the same block is not ruled out.

Form each block.
Collect the variables indexed by $I_j$ into the random vector $X_{I_j}$.

Apply its function.
Use the measurable map $g_j$ to obtain $Y_j=g_j(X_{I_j})$.

Use independence of the blocks.
Disjoint blocks of a mutually independent family are mutually independent; measurable functions of those blocks remain independent.

Proof. The vectors $X_{I_1},\ldots,X_{I_m}$ are mutually independent by the product form of the joint distribution of $X_1,\ldots,X_n$: grouping its coordinates into disjoint collections gives the product of the block distributions. Now choose arbitrary Borel sets $A_1,\ldots,A_m$ in the respective output spaces. Measurability ensures that $g_j^{-1}(A_j)$ is a Borel set in the space of $X_{I_j}$. Consequently,

\begin{aligned} P(Y_1\in A_1,\ldots,Y_m\in A_m) &=P\bigl(X_{I_1}\in g_1^{-1}(A_1),\ldots, X_{I_m}\in g_m^{-1}(A_m)\bigr)\\ &=\prod_{j=1}^m P\bigl(X_{I_j}\in g_j^{-1}(A_j)\bigr)\\ &=\prod_{j=1}^m P(Y_j\in A_j). \end{aligned}

This is the defining factorization for mutual independence of $Y_1,\ldots,Y_m$. Thus the derived quantities are mutually independent. $\square$

Worked Examples

Worked Example: Functions of Two Bernoulli Blocks

Let $X_1,X_2,X_3,X_4$ be mutually independent, with each $X_i$ taking the values 0 and 1 with probability $1/2$. Define $U=1\{X_1=X_2\}$ and $V=X_3X_4$. Find the distributions of $U$ and $V$, and verify their independence.

The blocks $(X_1,X_2)$ and $(X_3,X_4)$ are disjoint. The first function returns 1 when the two values match. Of the four equally likely pairs $(0,0),(0,1),(1,0),(1,1)$, two match. Therefore $P(U=1)=1/2$ and $P(U=0)=1/2$.

The product $V=X_3X_4$ is 1 only when both inputs equal 1. By independence, $P(V=1)=(1/2)(1/2)=1/4$, so $P(V=0)=3/4$.

The block-function theorem gives independence. For a direct numerical check, $P(U=1,V=1)=P(U=1)P(V=1)=(1/2)(1/4)=1/8$. The other joint probabilities also factor: $P(U=1,V=0)=(1/2)(3/4)=3/8$, $P(U=0,V=1)=(1/2)(1/4)=1/8$, and $P(U=0,V=0)=(1/2)(3/4)=3/8$. They sum to 1, and each is the product of the corresponding marginal probabilities.

Worked Example: A Sign Function and a Squared Normal

Let $X_1,X_2,X_3$ be mutually independent with $X_i\sim N(0,1)$. Define $U=1\{X_1X_2>0\}$ and $V=X_3^2$. Find their marginal distributions and a joint probability.

The standard normal distribution is symmetric about zero and assigns probability zero to the value 0. Thus $X_1$ and $X_2$ each have probability $1/2$ of being positive and probability $1/2$ of being negative. Their product is positive when both have the same sign, so $P(U=1)=(1/2)(1/2)+(1/2)(1/2)=1/2$. Hence $U$ is Bernoulli with success probability $1/2$.

For $t\geq 0$, the event $V\leq t$ is $-\sqrt{t}\leq X_3\leq\sqrt{t}$. If $\Phi$ denotes the standard normal CDF, then $P(V\leq t)=\Phi(\sqrt{t})-\Phi(-\sqrt{t})=2\Phi(\sqrt{t})-1$, using symmetry. For $t<0$, $P(V\leq t)=0$.

The variable $U$ is a function of the block $(X_1,X_2)$, while $V$ is a function of the disjoint block $X_3$. Therefore they are independent. For example, at $t=1$, $$ P(U=1,V\leq 1) = P(U=1)P(V\leq 1) = \frac{1}{2}\bigl(2\Phi(1)-1\bigr) = \Phi(1)-\frac{1}{2}. $$

Worked Example: Minimum From One Block, Comparison From Another

Let $X_1,X_2,X_3,X_4$ be independent Uniform$(0,1)$ variables. Set $A=\min(X_1,X_2)$ and $B=1\{X_3<X_4\}$. Find the distribution of $A$, the distribution of $B$, and $P(A\leq 0.6,B=1)$.

For $0\leq t\leq 1$, the minimum exceeds $t$ exactly when both inputs exceed $t$. Thus $P(A>t)=P(X_1>t,X_2>t)=(1-t)^2$, and $P(A\leq t)=1-(1-t)^2$. Outside this range the CDF is 0 for $t<0$ and 1 for $t\geq 1$.

The two independent continuous uniform variables $X_3,X_4$ are equally likely to occur in either order. Since a tie has probability zero, $P(B=1)=P(X_3<X_4)=1/2$. The blocks $(X_1,X_2)$ and $(X_3,X_4)$ are independent, so $A$ and $B$ are independent. In particular, $P(A\leq 0.6)=1-(1-0.6)^2=1-0.16=0.84$, and $P(A\leq 0.6,B=1)=0.84(0.5)=0.42$.

What the Theorem Does—and Does Not—Say

The important condition is independence of the blocks, not simply a difference between their index sets. Mutual independence of all the original variables is a sufficient condition for independence of disjoint blocks. If the original collection is only pairwise independent, disjoint blocks need not be independent.

For example, let $X$ and $Y$ be independent Rademacher variables, each taking $-1$ and $1$ with probability $1/2$, and set $Z=XY$. Each pair among $X,Y,Z$ is independent: for fixed signs $x,z$, the event $X=x,Z=z$ requires $Y=xz$, which has probability $1/4=P(X=x)P(Z=z)$; the same argument applies to the other pairs. But the blocks $(X,Y)$ and $Z$ are not independent. The function $U=XY$ of the first block equals $Z$ exactly, so $P(U=Z)=1$, whereas independent Rademacher variables would agree with probability $1/2$. Pairwise independence of the inputs did not make the blocks independent.

There is another boundary to keep in mind: functions of the same block can be dependent, even if each function is well-defined and measurable. For instance, if $X$ is a nonconstant random variable, then $X$ and $X^2$ both use the same input. The block-function theorem makes no claim that they are independent. It guarantees independence across outputs only when each output is formed from a separate block and those blocks are mutually independent.

Do check the input blocks. Disjoint indices alone do not establish independence.
Do check measurability. It ensures that inverse images of Borel output sets are valid events.
Do not require linear functions. Indicators, minima, products, and vector-valued maps are all permitted.
Do not infer independence within a block. Separate outputs from the same inputs may share randomness.

Key takeaway: If a collection of random elements is mutually independent, then functions of disjoint groups from that collection are mutually independent. To use the result, identify each function's inputs, verify that the groups are disjoint, and confirm independence of the groups.

Check Your Understanding

For each question, identify the relevant blocks before deciding whether the conclusion follows.

If $X_1,X_2,X_3$ are mutually independent, are $X_1+X_2$ and $X_3^2$ independent? State the theorem condition that applies.
Let $X_1,\ldots,X_4$ be independent Uniform$(0,1)$ variables. Which blocks produce $1\{X_1<X_2\}$ and $\max(X_3,X_4)$? What justifies independence of the two quantities?
Why does measurability of each block function matter in the proof?
Can two different functions of the same random variable be declared independent by the block-function theorem? Explain.
In the Rademacher construction $Z=XY$, why does pairwise independence fail to establish independence between the blocks $(X,Y)$ and $Z$?

Functions of Independent Blocks

What You'll Learn