Conditional Independence in Discrete Tables

Why Read Tables Within Strata?

A contingency table records how often combinations of categorical values occur. When the table is stratified by a third variable, the question is whether two variables are independent within each stratum. For example, a researcher might ask whether treatment assignment \(X\) and an outcome \(Y\) are independent among patients at each clinic \(Z=z\). The answer must be assessed separately in the clinic-specific tables; comparing only totals can conceal how the joint distribution behaves within a stratum.

The earlier material on conditional independence gives the general factorization definition. Here we make that definition operational for finite tables: identify the relevant conditional cell probabilities, compare them with products of their conditional margins, and use equivalent probability or cross-product checks.

Write \(p_{ij\mid z}=P(X=x_i,Y=y_j\mid Z=z)\), \(p_{i+\mid z}=P(X=x_i\mid Z=z)\), and \(p_{+j\mid z}=P(Y=y_j\mid Z=z)\). The plus sign indicates that probabilities have been summed over that index. Conditional independence says that every cell in a stratum equals the product of its row and column margins:

For a count table in stratum \(z\), let \(n_{ijz}\) be the cell count, \(n_{i+z}=\sum_j n_{ijz}\) the row total, \(n_{+jz}=\sum_i n_{ijz}\) the column total, and \(n_{++z}=\sum_i\sum_j n_{ijz}\) the stratum total. The corresponding empirical conditional probabilities are the cell and margin counts divided by \(n_{++z}\). Thus the empirical factorization condition can also be written as

The quantity on the right is the cell count implied by independence for the observed margins. A count table can be checked exactly this way, though observed counts from a random sample will not generally match those quantities exactly. Exact equality describes the table’s empirical distribution; it is not, by itself, an inferential test of a population model.

Equivalent Checks Within a Stratum

It is often easier to compare conditional probabilities than to calculate every product of margins. If independence holds, the distribution of \(X\) given \(Y=y\) and \(Z=z\) must equal the distribution of \(X\) given \(Z=z\), whenever the conditioning event has positive probability. The same statement holds with \(X\) and \(Y\) exchanged.

Proof. Fix a stratum \(z\) with positive probability. If the factorization holds and \(P(Y=y\mid Z=z)>0\), then

Conversely, suppose the conditional-probability equality holds for every \(y\) with positive conditional probability. For such \(y\), the multiplication rule gives

If instead \(P(Y=y\mid Z=z)=0\), then \(P(X=x,Y=y\mid Z=z)=0\) for every \(x\), so the same factorization holds because its right side is also zero. This proves the factorization for all \(x,y\) in the stratum, and hence conditional independence. Interchanging \(X\) and \(Y\) proves the other version. \(\square\)

For a two-by-two table, a further shortcut is available. Label its conditional cell probabilities \(p_{11},p_{12},p_{21},p_{22}\), omitting \(z\) temporarily. Independence requires

For a table with more rows or columns, the corresponding check is that every two-by-two minor is zero.

Proof. If the table factors, then \(p_{ij}p_{k\ell}=(p_{i+}p_{+j})(p_{k+}p_{+\ell})\), which equals \(p_{i\ell}p_{kj}\). For the converse, choose a cell \((r,s)\) with \(p_{rs}>0\); one exists because all entries are nonnegative and sum to one. The zero-minor condition gives

Set \(u_i=p_{is}\), \(v_j=p_{rj}\), \(U=\sum_i u_i\), and \(V=\sum_j v_j\). Summing \(p_{ij}=u_iv_j/p_{rs}\) over all \(i,j\) gives \(1=UV/p_{rs}\), so \(p_{rs}=UV\). Therefore \(p_{ij}=(u_i/U)(v_j/V)\). Summing this expression across a row gives \(p_{i+}=u_i/U\), and summing down a column gives \(p_{+j}=v_j/V\). Hence \(p_{ij}=p_{i+}p_{+j}\), as required. This argument permits zero cells. \(\square\)

Worked Table Checks

A Reliable Diagnostic and Common Pitfalls

For a stratified table, a consistent check is to inspect every positive-total stratum, compute its conditional margins, and compare the cells with the products of those margins. Conditional probability comparisons can make the discrepancy especially clear; cross-products are a compact alternative for two-by-two tables. A zero cell does not invalidate these checks: use the factorization or minor equalities rather than dividing by that cell.

A common mistake is to check whether \(P(X=x\mid Y=y,Z=z)\) is the same for different \(z\). That is not the within-stratum criterion: compare \(P(X=x\mid Y=y,Z=z)\) with \(P(X=x\mid Z=z)\) while holding the same \(z\) fixed. Conditional independence does not require the distribution of \(X\) to be identical across strata. Another mistake is to regard approximate agreement in a sample as proof of population independence. The exact calculations here diagnose a probability table—or describe an empirical table. Statistical evidence about a population requires an inferential procedure beyond exact table factorization.