Why Convert a Joint Table?
A selection factorization describes the outcome distribution first and then the response probability conditional on the outcome. A pattern-mixture factorization describes the response pattern first and then the outcome distribution within that pattern. These are two ways of writing one joint probability law, not competing claims about the data. A small binary example makes the conversion concrete: every factor can be calculated directly from the joint table, and every cell can be reconstructed to check the arithmetic.
Let \(Y\) be a binary outcome, coded \(0\) or \(1\), and let \(R\) be a binary response indicator, with \(R=1\) meaning the outcome is observed and \(R=0\) meaning it is missing. The four joint probabilities \(P(Y=y,R=r)\) must be nonnegative and sum to one. The table represents the full joint law, including the outcome values for records with \(R=0\); it is not merely a table of observed outcomes.
For a given joint table, obtain the selection factors by summing over \(R\) to find \(P(Y=y)\), then dividing each joint cell by that outcome margin to find \(P(R=r\mid Y=y)\). Obtain the pattern-mixture factors by summing over \(Y\) to find \(P(R=r)\), then dividing each cell by that response-pattern margin to find \(P(Y=y\mid R=r)\). The denominators differ because the conditioning variables differ.
Why the Two Products Agree
The conversion follows directly from the definition of conditional probability. If the relevant marginal probabilities are positive, then
and, in the other order,
Thus both factorizations reconstruct the same joint cell. This is an algebraic identity, not an additional assumption about how the response process operates. The calculation does require care with zero margins: a conditional probability given an event of probability zero is undefined. In that case, the corresponding product is zero regardless of how a conditional distribution is assigned on that null event, so it cannot be recovered from the joint table alone.
The calculations below use finite probabilities, so summing rows or columns gives the relevant marginal probabilities. As a basic validity check, every joint cell must be between zero and one, and all four cells must sum to one. Conditional probabilities within each positive-margin group must also sum to one.
Worked Example: Converting a Joint Table With Unequal Response Rates
Suppose a hypothetical study has the following joint probability table. The entries are probabilities, not sample counts.
| \(R=1\) | \(R=0\) | Marginal \(P(Y=y)\) | |
|---|---|---|---|
| \(Y=0\) | 0.30 | 0.10 | 0.40 |
| \(Y=1\) | 0.20 | 0.40 | 0.60 |
| Marginal \(P(R=r)\) | 0.50 | 0.50 | 1.00 |
First check the total: \(0.30+0.10+0.20+0.40=1.00\). The outcome margins are \(P(Y=0)=0.30+0.10=0.40\) and \(P(Y=1)=0.20+0.40=0.60\). The response margins are \(P(R=1)=0.30+0.20=0.50\) and \(P(R=0)=0.10+0.40=0.50\).
Selection factors. Divide each cell by its outcome margin. For \(Y=0\), \(P(R=1\mid Y=0)=0.30/0.40=0.75\), and \(P(R=0\mid Y=0)=0.10/0.40=0.25\). For \(Y=1\), \(P(R=1\mid Y=1)=0.20/0.60=1/3\approx0.3333\), and \(P(R=0\mid Y=1)=0.40/0.60=2/3\approx0.6667\). Each pair sums to one. The selection factorization is the outcome margin multiplied by these response probabilities conditional on outcome.
Pattern-mixture factors. Divide each cell by its response margin. Within \(R=1\), \(P(Y=0\mid R=1)=0.30/0.50=0.60\) and \(P(Y=1\mid R=1)=0.20/0.50=0.40\). Within \(R=0\), \(P(Y=0\mid R=0)=0.10/0.50=0.20\) and \(P(Y=1\mid R=0)=0.40/0.50=0.80\). The pattern-mixture factorization is therefore \(P(R=r)\) multiplied by the corresponding outcome distribution within response pattern \(r\).
Reconstruction check. From the selection factors, the four products are \(0.40(0.75)=0.30\), \(0.40(0.25)=0.10\), \(0.60(1/3)=0.20\), and \(0.60(2/3)=0.40\), in the order \((Y=0,R=1)\), \((Y=0,R=0)\), \((Y=1,R=1)\), \((Y=1,R=0)\). From the pattern-mixture factors, they are \(0.50(0.60)=0.30\), \(0.50(0.20)=0.10\), \(0.50(0.40)=0.20\), and \(0.50(0.80)=0.40\). Both reproduce every original cell.
In this hypothetical law, the probability of response differs by outcome: it is \(0.75\) for \(Y=0\) and about \(0.3333\) for \(Y=1\). The pattern-mixture representation expresses the same association by showing that the outcome distributions differ across \(R=1\) and \(R=0\). Neither representation changes the underlying joint probabilities.
Two Further Checks on the Conversion
Worked Example: Recognizing Independence in the Factors
Consider a second hypothetical joint table with all four cells equal to \(0.25\). Its row margins and column margins are all \(0.50\), so the total is \(1.00\).
For the selection factorization, \(P(Y=0)=P(Y=1)=0.50\). Within either outcome group, \(P(R=1\mid Y=y)=0.25/0.50=0.50\) and \(P(R=0\mid Y=y)=0.25/0.50=0.50\). Thus the response distribution is the same for both outcome values.
For the pattern-mixture factorization, \(P(R=0)=P(R=1)=0.50\). Within either response group, \(P(Y=0\mid R=r)=0.25/0.50=0.50\) and \(P(Y=1\mid R=r)=0.25/0.50=0.50\). The outcome distribution is also the same across response patterns.
Each of the four cells is reconstructed as \(0.50(0.50)=0.25\) from either factorization. The equality of the conditional distributions across groups is consistent with independence of \(Y\) and \(R\) in this table. It is not a requirement for conversion: the first example converted successfully despite unequal conditional distributions.
Worked Example: A Zero Cell With Positive Margins
In a third hypothetical table, suppose the joint probabilities are \(P(Y=0,R=1)=0.40\), \(P(Y=1,R=1)=0.10\), \(P(Y=0,R=0)=0\), and \(P(Y=1,R=0)=0.50\). They are nonnegative and sum to one. The outcome margins are \(0.40\) and \(0.60\), and the response margins are \(0.50\) and \(0.50\), so every conditional distribution needed here has a positive conditioning margin.
For the selection factors, \(P(R=1\mid Y=0)=0.40/0.40=1\) and \(P(R=0\mid Y=0)=0/0.40=0\). For \(Y=1\), \(P(R=1\mid Y=1)=0.10/0.60=1/6\), while \(P(R=0\mid Y=1)=0.50/0.60=5/6\). Reconstruction gives \(0.40(0)=0\) for \((Y=0,R=0)\), as required.
For the pattern-mixture factors, the \(R=1\) conditional probabilities are \(P(Y=0\mid R=1)=0.40/0.50=0.80\) and \(P(Y=1\mid R=1)=0.10/0.50=0.20\). For \(R=0\), they are \(P(Y=0\mid R=0)=0/0.50=0\) and \(P(Y=1\mid R=0)=0.50/0.50=1\). Multiplication by the response margins again gives \(0.40\), \(0.10\), \(0\), and \(0.50\) in the appropriate cells. A zero cell is therefore not itself a problem; a zero conditioning margin is the special case requiring care.
Checks, Comparisons, and Limits
A reliable calculation has three layers of checks. First, verify the joint table is a probability distribution: its cells are nonnegative and sum to one. Second, verify that each conditional distribution sums to one within every group with a positive margin. Third, multiply each marginal by its conditional probabilities and compare the resulting cells with the starting table. Checking only that the reconstructed cells sum to one is weaker: an incorrect allocation across cells can still have the correct total.
| Representation | First factor | Conditional factor | Useful way to read it |
|---|---|---|---|
| Selection | \(P(Y=y)\) | \(P(R=r\mid Y=y)\) | Response probabilities across outcome values |
| Pattern-mixture | \(P(R=r)\) | \(P(Y=y\mid R=r)\) | Outcome distributions across response patterns |
The selection form is convenient when the response process is described conditional on the outcome; the pattern-mixture form is convenient when the outcome distribution is described separately by response pattern. The table conversion itself does not select one as more correct, nor does it establish that either set of conditional probabilities follows a particular parametric model.
A common mistake is to use the wrong marginal as a denominator. For example, the selection probability \(P(R=1\mid Y=0)\) is the cell \(P(Y=0,R=1)\) divided by \(P(Y=0)\), not by \(P(R=1)\). In the first example these give \(0.30/0.40=0.75\) and \(0.30/0.50=0.60\), respectively; the latter is \(P(Y=0\mid R=1)\), a different conditional probability. The correction is to name the event after the conditioning bar first, then use its marginal in the denominator.
A second pitfall is treating the conditional factors as if they were themselves the joint cells. Conditional probabilities are distributions within a group; to return to the population joint law, multiply by that group’s marginal. Finally, when a conditioning margin is zero, do not divide by zero or report an empirical conditional probability from that margin. The joint table determines the product for that pattern as zero, but it does not determine a unique conditional distribution within an event that never occurs.
These calculations establish equivalence of representations for a specified binary joint law. They do not, by themselves, determine missing outcomes from observed data or resolve which assumptions about unobserved outcomes are appropriate. The conversion is a bookkeeping and interpretation tool: its central test is whether the factors reproduce the stated joint probabilities.
Check Your Understanding
Use \(R=1\) for observed and \(R=0\) for missing. Each question gives a hypothetical joint table; calculate or check the requested factors, then compare with the solutions below.
- A table has cells \(P(0,1)=0.20\), \(P(0,0)=0.30\), \(P(1,1)=0.10\), and \(P(1,0)=0.40\), where each ordered pair is \((Y,R)\). Find \(P(R=1\mid Y=0)\) and \(P(Y=0\mid R=1)\).
- For the same table, calculate \(P(Y=1)\), \(P(R=0)\), and \(P(Y=1\mid R=0)\).
- Suppose all four cells of a binary joint table are \(0.25\). What are \(P(R=1\mid Y=1)\) and \(P(Y=1\mid R=1)\)? Reconstruct \(P(Y=1,R=1)\).
- Suppose \(P(Y=0,R=1)=0.50\), \(P(Y=1,R=1)=0.50\), and both cells with \(R=0\) are zero. Is \(P(Y\mid R=0)\) determined by the joint table? What is the joint probability for either \(R=0\) cell?
Solutions. For question 1, \(P(Y=0)=0.20+0.30=0.50\), so \(P(R=1\mid Y=0)=0.20/0.50=0.40\). Also \(P(R=1)=0.20+0.10=0.30\), so \(P(Y=0\mid R=1)=0.20/0.30=2/3\approx0.6667\). These use different denominators because they condition on different events. For question 2, \(P(Y=1)=0.10+0.40=0.50\), \(P(R=0)=0.30+0.40=0.70\), and \(P(Y=1\mid R=0)=0.40/0.70\approx0.5714\). For question 3, both margins are \(0.50\), giving \(P(R=1\mid Y=1)=0.25/0.50=0.50\) and \(P(Y=1\mid R=1)=0.25/0.50=0.50\). Either factorization reconstructs the cell as \(0.50(0.50)=0.25\). For question 4, \(P(R=0)=0\), so \(P(Y\mid R=0)\) is undefined from the joint table. Each \(R=0\) joint cell is nevertheless zero, since its probability is specified as zero.