Independence Between Coordinate Blocks
A random vector can contain several coordinates, and a statistical question may concern whether one group of coordinates is independent of another group. This is a stronger and more organized question than checking whether individual coordinates are independent in pairs. The key is to treat each block as one random object and ask whether events about the first block factor from events about the second.
For example, let \(X=(X_1,\ldots,X_r)\) and \(Y=(Y_1,\ldots,Y_s)\). An event such as \(X\in A\) can impose a joint condition on every coordinate of \(X\); it need not split into separate conditions on \(X_1,\ldots,X_r\). Independence of the blocks requires factorization for all such events, not just for events involving one coordinate at a time.
This definition concerns the full vectors. If a vector \(Z=(Z_1,\ldots,Z_k)\) is divided into disjoint coordinate blocks indexed by \(I\) and \(J\), write \(Z_I=(Z_i:i\in I)\) and \(Z_J=(Z_j:j\in J)\), keeping a fixed order within each block. The blocks are independent when \(Z_I\) and \(Z_J\) satisfy the definition above. A partition into more than two blocks is mutually independent when the probability of every choice of one Borel event per block equals the product of the individual probabilities.
The events \(X\in A\) and \(Y\in B\) are events in separate coordinate spaces. The definition does not require either event to be a simple interval or a single-coordinate condition. A set \(A\), for instance, can describe a curved region or a constraint involving several coordinates at once.
Equivalent Factorization Criteria
A useful way to express block independence is through the joint distribution of the concatenated vector \((X,Y)\). Let \(\mu\) be its probability law, and let \(\mu_X\) and \(\mu_Y\) be the marginal laws of \(X\) and \(Y\). The product measure \(\mu_X\otimes\mu_Y\) assigns \(A\times B\) the probability \(\mu_X(A)\mu_Y(B)\). Thus independence says that the joint law agrees with this product law.
- \(X\) and \(Y\) are independent.
- The joint law of \((X,Y)\) equals \(\mu_X\otimes\mu_Y\).
- For every Borel \(A\subseteq\mathbb{R}^r\) and \(B\subseteq\mathbb{R}^s\), \(\mu(A\times B)=\mu_X(A)\mu_Y(B)\).
Proof. The first and third statements are the definition written using the joint law. If the joint law is the product measure, then for every such \(A,B\), $$ P(X\in A,\;Y\in B) =\mu(A\times B) =(\mu_X\otimes\mu_Y)(A\times B) =\mu_X(A)\mu_Y(B), $$ so the first statement holds. Conversely, suppose the displayed factorization holds for all Borel \(A,B\). The sets \(A\times B\) form a \(\pi\)-system: the intersection of \(A_1\times B_1\) and \(A_2\times B_2\) is \((A_1\cap A_2)\times(B_1\cap B_2)\). They generate the Borel sets of \(\mathbb{R}^{r+s}\). The two probability measures \(\mu\) and \(\mu_X\otimes\mu_Y\) agree on this \(\pi\)-system. By the \(\pi\)-\(\lambda\) theorem, they agree on the generated Borel sets. Hence the joint law is the product law. \(\square\)
There is also a joint-CDF version of the criterion. For \(x=(x_1,\ldots,x_r)\) and \(y=(y_1,\ldots,y_s)\), use coordinatewise inequalities to define \(X\leq x\) and \(Y\leq y\).
Proof. Independence implies the equality by taking \(A=\prod_{i=1}^r(-\infty,x_i]\) and \(B=\prod_{j=1}^s(-\infty,y_j]\) in the definition. For the converse, the lower orthants \(A\times B\) of this form are a \(\pi\)-system generating the Borel sets of \(\mathbb{R}^{r+s}\). The assumed equality says that the joint law and \(\mu_X\otimes\mu_Y\) agree on every such orthant. The same \(\pi\)-\(\lambda\) uniqueness argument shows that the two laws agree on all Borel sets, which gives independence. \(\square\)
Worked Examples
Worked Example: A Discrete Vector and a Binary Block
Let \(X\) take the three vector values \(a_1=(0,0)\), \(a_2=(0,1)\), and \(a_3=(1,1)\), with probabilities \(1/2\), \(1/3\), and \(1/6\), respectively. Let \(Y\) take values 0 and 1. Suppose the joint probabilities are $$ \begin{array}{c|cc} & Y=0 & Y=1\\ \hline X=a_1 & 1/8 & 3/8\\ X=a_2 & 1/12 & 1/4\\ X=a_3 & 1/24 & 1/8 \end{array} $$ Test whether the vector \(X\) and the block \(Y\) are independent.
First find the marginal probabilities for \(Y\): $$ P(Y=0)=\frac18+\frac1{12}+\frac1{24} =\frac3{24}+\frac2{24}+\frac1{24} =\frac14, $$ and therefore \(P(Y=1)=3/4\). The listed row sums are \(1/2,1/3,1/6\), as specified for \(X\).
Each entry is the product of its row and column marginals. For example, $$ P(X=a_2,Y=1)=\frac14 =\frac13\cdot\frac34. $$ The other entries also factor: \(1/8=(1/2)(1/4)\), \(3/8=(1/2)(3/4)\), \(1/12=(1/3)(1/4)\), \(1/24=(1/6)(1/4)\), and \(1/8=(1/6)(3/4)\).
For arbitrary subsets \(S\subseteq\{a_1,a_2,a_3\}\) and \(T\subseteq\{0,1\}\), summing the factored entries gives $$ P(X\in S,Y\in T) =\sum_{a_i\in S}\sum_{y\in T}P(X=a_i)P(Y=y) =P(X\in S)P(Y\in T). $$ Every event for these finite-valued variables is such a subset, so \(X\) and \(Y\) are independent.
Worked Example: Checking Independence With a Joint Density
Suppose \((X_1,X_2,Y)\) has density $$ f(x_1,x_2,y)= \begin{cases} 2x_1, & 0<x_1<1,\;0<x_2<1,\;0<y<1,\\ 0, & \text{otherwise}. \end{cases} $$ Test independence between \(X=(X_1,X_2)\) and \(Y\).
The density integrates to one: $$ \int_0^1\int_0^1\int_0^1 2x_1\,dy\,dx_2\,dx_1 =\int_0^1 2x_1\,dx_1=1. $$ On the support, it factors as \(g(x_1,x_2)h(y)\), where \(g(x_1,x_2)=2x_1\) on the unit square and \(h(y)=1\) on \((0,1)\). Each factor is a probability density: \(\int_0^1\int_0^1 2x_1\,dx_2\,dx_1=1\), and \(\int_0^1 1\,dy=1\).
For any Borel \(A\subseteq\mathbb{R}^2\) and \(B\subseteq\mathbb{R}\), integration of the factored density gives $$ P(X\in A,Y\in B) =\left(\int_A g(x_1,x_2)\,dx_1\,dx_2\right) \left(\int_B h(y)\,dy\right) =P(X\in A)P(Y\in B). $$ The integrands are nonnegative, so the separation of the integrals is justified by Tonelli's theorem. Thus the blocks are independent.
As a numerical check, take \(A=[1/2,1]\times[0,1]\) and \(B=[0,1/4]\). Then $$ P(X\in A)=\int_{1/2}^1 2x_1\,dx_1=\left[x_1^2\right]_{1/2}^1=\frac34, \qquad P(Y\in B)=\frac14. $$ The joint probability is \((3/4)(1/4)=3/16\), which also follows by integrating \(2x_1\) over \(A\times B\).
Worked Example: Pairwise Independence Does Not Ensure Block Independence
Let \(U\) and \(V\) be independent fair binary random variables, and define \(W=U\mathbin{\mathrm{xor}}V\), where \(\mathrm{xor}\) is 1 when its inputs differ and 0 otherwise. The four equally likely outcomes for \((U,V,W)\) are $$ (0,0,0),\quad(0,1,1),\quad(1,0,1),\quad(1,1,0). $$ Consider the coordinate blocks \(X=(U,V)\) and \(Y=W\).
The event \(A=\{(0,0)\}\) for \(X\) has probability \(1/4\), and \(P(Y=0)=1/2\). But \(X=(0,0)\) forces \(Y=0\), so $$ P(X\in A,Y=0)=\frac14 \ne \frac14\cdot\frac12=\frac18. $$ Therefore the vector block \(X\) and the scalar block \(Y\) are not independent.
This does not contradict pairwise independence of the three individual binary coordinates. For instance, \(P(U=0,V=0)=1/4=P(U=0)P(V=0)\); the same check holds for the other coordinate pairs. Block independence tests events about the entire vector, including events that combine coordinates, and can fail even when every pair of individual coordinates is independent.
Common Pitfall and Why Block Independence Matters
A common mistake is to check only pairs of coordinates and conclude that two blocks are independent. Pairwise checks test events such as \(X_i\in C\) and \(Y_j\in D\) one coordinate at a time. The definition instead requires the factorization \(P(X\in A,Y\in B)=P(X\in A)P(Y\in B)\) for every pair of Borel sets \(A,B\), including sets whose conditions involve multiple coordinates jointly. The parity example shows precisely how those stronger events can expose dependence missed by pairwise checks.
Block independence is useful when a model groups observations or variables into meaningful units: for example, a vector of measurements from one process and a second vector from another. It lets a question be phrased at the level of the blocks actually used in an analysis. The event definition is the fundamental criterion; joint-CDF factorization is a convenient test, and density factorization is a convenient test when densities exist. None of these tests should be replaced by matching marginals alone.
Check Your Understanding
Use the event definition and factorization criteria to answer these questions.
- State what must be true for random vectors \(X\in\mathbb{R}^r\) and \(Y\in\mathbb{R}^s\) to be independent.
- Why does factorization of the joint CDF on all coordinatewise lower orthants imply independence?
- If a joint density factors as \(f_X(x)f_Y(y)\) almost everywhere, what integral property gives event factorization?
- In the parity example, identify an event for each block that demonstrates failure of independence.
- Explain why pairwise independence of all individual coordinates does not, in general, establish independence between coordinate blocks.