Reading a Joint Probability Table
A two-way probability table describes two categorical classifications of the same outcome. The entries inside the table are joint probabilities: each is the probability that one row category and one column category occur together. Row and column totals are the corresponding marginal probabilities. To check independence, compare a joint probability with the product of the two relevant marginal probabilities.
The table must be a valid probability table: every cell must be nonnegative, and all cells must sum to one. Its row and column totals must agree with the sums of their cells. A failed check of these basic requirements means the figures cannot describe a probability model, so an independence conclusion from them is not meaningful.
A test of one event pair is not automatically a test of the entire table. For example, \(R_1\) and \(C_1\) might be independent even if another row category is not independent of \(C_1\). Conversely, when checking whether two complete classifications are independent, every cell must satisfy the product rule.
Why the Cell-Product Test Works
Each cell represents the intersection of a row event and a column event. By the definition of independence, those two events are independent precisely when their intersection probability equals the product of their probabilities. That gives the test for an individual cell.
For the full classifications, independence means that each row category is independent of each column category. Therefore the product equality must hold in every cell. When it does, the result also applies to groups of categories. For row-index sets \(I\) and column-index sets \(J\), the probability of the grouped intersection is $$ P\left(\bigcup_{i\in I}R_i\;\cap\;\bigcup_{j\in J}C_j\right) =\sum_{i\in I}\sum_{j\in J}p_{ij} =\left(\sum_{i\in I}r_i\right)\left(\sum_{j\in J}c_j\right). $$ The first equality follows because the categories are mutually exclusive. The second follows by substituting \(p_{ij}=r_i c_j\) and distributing the sums. Thus independence of the complete classifications implies independence of any events formed by grouping their categories.
For two binary events, it is enough to check the cell where both occur. If \(P(A\cap B)=P(A)P(B)\), then, for example, \(P(A\cap B^c)=P(A)-P(A\cap B)=P(A)(1-P(B))=P(A)P(B^c)\). The other cells follow in the same way. This shortcut depends on having a valid two-event probability table; for larger tables, check all cells when testing the full classifications.
Testing a Particular Pair of Events
To assess a particular event pair, first identify the exact row and column categories—or unions of categories—that make up each event. Add the relevant cells to find the joint probability, and add the relevant row or column totals to find each marginal probability. Then compare the joint probability with the product of the marginals. A mismatch establishes dependence; equality establishes independence for that pair.
Worked Example: A Pair of Service and Claim Categories
Model. In a hypothetical probability model, one covered item is classified by service group, \(A\) or \(B\), and claim status, \(C\) or \(C^c\). The table gives the joint probabilities for a randomly selected item. There is no interest-rate or mortality assumption in this categorical model.
| Claim \(C\) | No claim \(C^c\) | Total | |
|---|---|---|---|
| Service group \(A\) | 0.18 | 0.12 | 0.30 |
| Service group \(B\) | 0.28 | 0.42 | 0.70 |
| Total | 0.46 | 0.54 | 1.00 |
Method. Check whether the two classifications are independent by comparing a joint cell with the product of its marginal probabilities.
The table is valid: its four cells sum to \(0.18+0.12+0.28+0.42=1.00\). The marginal probabilities are \(P(A)=0.30\) and \(P(C)=0.46\). Under independence, the joint probability for \(A\) and \(C\) would be $$ P(A)P(C)=0.30(0.46)=0.138. $$ The observed cell is \(P(A\cap C)=0.18\), which is not \(0.138\). Therefore \(A\) and \(C\) are dependent, and the full classifications are not independent.
The other cells confirm the conclusion. Their independence products would be \(P(A)P(C^c)=0.30(0.54)=0.162\), \(P(B)P(C)=0.70(0.46)=0.322\), and \(P(B)P(C^c)=0.70(0.54)=0.378\). These differ from the observed cells \(0.12\), \(0.28\), and \(0.42\), respectively.
As a check using conditional probability, \(P(C\mid A)=0.18/0.30=0.60\), whereas \(P(C)=0.46\). Because \(P(A)>0\), the conditional probability is defined; its difference from the marginal probability is consistent with dependence. The common mistake to avoid is deciding independence by looking only at whether the table “looks balanced.” Use the numerical product test instead.
A Full Table Can Be Dependent Even When One Pair Is Independent
The full-classification test and a test of one selected pair answer different questions. An individual cell may satisfy the product rule by coincidence while other cells do not. A valid table can therefore contain both independent and dependent event pairs.
Worked Example: One Independent Row Event in a Dependent Table
Model. For a hypothetical group of policies, each policy is assigned to exactly one risk category \(R_1,R_2,R_3\) and is classified as either \(H\) or \(H^c\). The entries are joint probabilities; no insurance pricing or mortality assumptions are needed.
| \(H\) | \(H^c\) | Total | |
|---|---|---|---|
| \(R_1\) | 0.08 | 0.12 | 0.20 |
| \(R_2\) | 0.13 | 0.17 | 0.30 |
| \(R_3\) | 0.19 | 0.31 | 0.50 |
| Total | 0.40 | 0.60 | 1.00 |
Method. Check the marginals and cell sums, then test the requested pair and the full classifications separately.
The \(H\) column sums to \(0.08+0.13+0.19=0.40\), and the \(H^c\) column sums to \(0.12+0.17+0.31=0.60\). Each row also sums to its stated total, and all cells sum to \(1.00\). Thus the table is internally consistent.
For the particular pair \(R_1,H\), the product is \(P(R_1)P(H)=0.20(0.40)=0.08\), exactly the observed joint probability. Therefore \(R_1\) and \(H\) are independent. The complementary pair is also independent: \(P(R_1\cap H^c)=0.12\), and \(P(R_1)P(H^c)=0.20(0.60)=0.12\).
The full classifications are nevertheless dependent. For \(R_2,H\), the product is \(0.30(0.40)=0.12\), but the cell is \(0.13\). For \(R_3,H\), the product is \(0.50(0.40)=0.20\), but the cell is \(0.19\). The \(H^c\) cells also fail the full test: \(0.17\ne0.30(0.60)=0.18\) and \(0.31\ne0.50(0.60)=0.30\). The mistake to avoid is concluding that the entire table is independent because one pair, \(R_1,H\), passes the test.
Pooling Categories Before Testing
An event may combine several categories. To test it, sum the relevant cells to obtain the joint probability and sum the relevant marginal totals to obtain the event probability. Do not test just one of the component cells when the event includes multiple categories. Deviations from independence in different cells can also cancel when categories are pooled.
Worked Example: Independence After Pooling Two Categories
Model. A hypothetical portfolio outcome is classified into one of three mutually exclusive groups \(A,B,C\), and into either \(X\) or \(X^c\). The probabilities below form the entire model.
| \(X\) | \(X^c\) | Total | |
|---|---|---|---|
| \(A\) | 0.20 | 0.20 | 0.40 |
| \(B\) | 0.21 | 0.14 | 0.35 |
| \(C\) | 0.09 | 0.16 | 0.25 |
| Total | 0.50 | 0.50 | 1.00 |
Method. First test whether the full classifications are independent; then separately test the requested pooled event \(B\cup C\) against \(X\).
The table is valid because the row totals are \(0.40,0.35,0.25\), the column totals are \(0.50,0.50\), and the grand total is \(1.00\). The full classifications are dependent: for example, \(P(B\cap X)=0.21\), while \(P(B)P(X)=0.35(0.50)=0.175\).
Now pool \(B\) and \(C\). The event probability is \(P(B\cup C)=0.35+0.25=0.60\). The joint probability with \(X\) is \(P((B\cup C)\cap X)=0.21+0.09=0.30\). The product of the pooled marginal probabilities is \(0.60(0.50)=0.30\). Hence \(B\cup C\) and \(X\) are independent, even though the full classifications are dependent.
For contrast, \(A\cup B\) has probability \(0.40+0.35=0.75\), and its joint probability with \(X\) is \(0.20+0.21=0.41\). The independence product is \(0.75(0.50)=0.375\), not \(0.41\), so \(A\cup B\) and \(X\) are dependent. The common mistake is to infer that a pooled event is dependent just because one of its component categories is dependent; calculate the pooled joint probability and pooled product directly.
Practical Checks and Common Traps
Check that cells are nonnegative, the total is one, and each stated marginal equals the sum of its cells.
Specify whether the question concerns a single row-column pair, grouped categories, or the entire classifications.
Add cells or marginal totals as needed when an event combines categories.
Independence requires exact equality in the probability model: \(P(A\cap B)=P(A)P(B)\). For the full table, make the comparison in every cell.
Do not confuse independence with mutual exclusivity: two events that cannot occur together have joint probability zero, which generally does not equal the product of their positive marginal probabilities. Also, conditional probability is not a substitute when the conditioning event has probability zero; \(P(A\mid B)\) is undefined if \(P(B)=0\), though the factorization criterion \(P(A\cap B)=P(A)P(B)\) remains meaningful.
Check Your Understanding
For each valid table or stated model, identify the relevant joint and marginal probabilities before deciding. Answers are included for checking.
- A two-category classification has row totals \(0.40\) and \(0.60\), and a binary column event \(D\) has probability \(0.25\). The joint probability of the first row and \(D\) is \(0.10\). Is the first row event independent of \(D\), and does that establish independence of the full binary classifications? (Answer: Yes; \(0.40(0.25)=0.10\), and for valid binary classifications this one cell establishes independence of the full pair.)
- A valid table has row totals \(P(P)=0.30\), \(P(Q)=0.70\), column total \(P(S)=0.40\), and cells \(P(P\cap S)=0.09\), \(P(Q\cap S)=0.31\). Are the row and column classifications independent? (Answer: No; \(0.30(0.40)=0.12\ne0.09\), and \(0.70(0.40)=0.28\ne0.31\).)
- A three-row table has row totals \(0.20,0.30,0.50\), and a column event \(H\) has total \(0.40\). Its joint cells with \(H\) are \(0.08,0.12,0.20\). Is the full row classification independent of \(H\)? (Answer: Yes; the products are \(0.20(0.40)=0.08\), \(0.30(0.40)=0.12\), and \(0.50(0.40)=0.20\), and the cells sum to \(0.40\).)
- A binary row event \(A\) has probability \(0.45\), and a column event \(C\) has probability \(0.30\). The joint probability is \(P(A\cap C)=0.12\). Are \(A\) and \(C\) independent? (Answer: No; \(P(A)P(C)=0.45(0.30)=0.135\ne0.12\).)
- A valid table has row totals \(0.25,0.35,0.40\), column probability \(P(K)=0.30\), and \(K\)-column cells \(0.04,0.14,0.12\). Are the full classifications independent, and is the pooled event “row 1 or row 2” independent of \(K\)? (Answer: The full classifications are dependent because \(0.04\ne0.25(0.30)=0.075\); the pooled event is independent because its joint probability is \(0.04+0.14=0.18\), equal to \((0.25+0.35)(0.30)=0.18\).)