Conditional Distributions in Discrete Models

From a Joint Mass Function to a Conditional Law

A joint mass function describes two discrete random variables together. A conditional mass function answers a more focused question: once one variable is known to equal a particular value, how is probability distributed across the possible values of the other? The essential operation is to restrict attention to the relevant row or column of the joint law and divide by its total mass.

Let \(X\) and \(Y\) be discrete random variables on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), taking values in countable sets \(\mathcal{X}\) and \(\mathcal{Y}\). Their joint mass function is \(p_{X,Y}(x,y)=\mathbb{P}(X=x,Y=y)\), and their marginal mass functions are obtained by summing over the other variable. In particular, \(p_Y(y)=\sum_{x\in\mathcal{X}}p_{X,Y}(x,y)\).

The denominator is the probability of the conditioning event, not the probability of the particular joint outcome. Because the joint mass in the numerator is part of the total mass in the conditioning row or column, this division makes the conditional masses add to one.

Proof. Each joint mass \(p_{X,Y}(x,y)\) is nonnegative and \(p_Y(y)>0\), so the ratio is nonnegative. Summing over all \(x\) gives \(\sum_x p_{X\mid Y}(x\mid y)=\sum_x p_{X,Y}(x,y)/p_Y(y)=p_Y(y)/p_Y(y)=1\). For any \(C\subseteq\mathcal{X}\), the event \(\{X\in C,Y=y\}\) is the disjoint union of the events \(\{X=x,Y=y\}\) for \(x\in C\). Thus its probability is \(\sum_{x\in C}p_{X,Y}(x,y)\). Dividing this probability by \(\mathbb{P}(Y=y)=p_Y(y)\), as required by conditional probability on a positive-probability event, gives \(\mathbb{P}(X\in C\mid Y=y)=\sum_{x\in C}p_{X\mid Y}(x\mid y)\). \(\square\)

A useful consequence is that a conditional probability for a set is found by summing conditional masses over that set. For example, \(p_{X\mid Y}(x\mid y)\) gives the law of \(X\) after \(Y=y\) is known; it is not the joint probability of \((X,Y)=(x,y)\). Conversely, multiplying by the marginal recovers the joint mass: \(p_{X,Y}(x,y)=p_{X\mid Y}(x\mid y)p_Y(y)\) whenever \(p_Y(y)>0\).

Proof. The event \(\{g(X)=u\}\) is the disjoint union of \(\{X=x\}\) over the values \(x\) satisfying \(g(x)=u\). Intersecting each event with \(\{Y=y\}\) preserves disjointness. Therefore \(\mathbb{P}(g(X)=u,Y=y)=\sum_{x:g(x)=u}p_{X,Y}(x,y)\). Since \(p_Y(y)>0\), divide both sides by \(p_Y(y)\). Each term on the right becomes \(p_{X\mid Y}(x\mid y)\), and the left is the stated conditional probability. \(\square\)

Reading Conditional Laws from a Joint Table

In a joint probability table, each entry is a joint mass. To find \(X\) given \(Y=y\), select the column for \(y\) and divide every entry in that column by its column total. To find \(Y\) given \(X=x\), use the row for \(x\) and divide by its row total. First check the joint masses sum to one, and then check each conditional distribution sums to one. A conditioning value with zero marginal mass cannot be used in the ratio.

Conditional Distributions from Counts and Sampling

A contingency table often records counts rather than probabilities. If each observation is treated as equally likely to be selected from the table, divide every count by the grand total to get the empirical joint mass function. Conditional probabilities then come from dividing a cell count by the total count in the conditioning row or column. The direction matters: conditioning on a group and conditioning on an outcome usually use different totals.

Sampling without replacement produces conditional laws by the same rule. The joint masses come from multiplying the probabilities of successive draws, updating the available composition after each draw. The conditional distribution can then be read from the joint table by dividing by the appropriate marginal.

Denominators, Zero-Mass Values, and a Common Error

A conditional mass function is a probability distribution only after normalization by the marginal mass of the conditioning value. A common mistake is to divide a joint cell by the grand total even though the question conditions on a row or column. In the contingency table above, the grand total is 80, but the event \(R=1\) has total 26. Thus \(18/80\) is the joint probability of \(G=A,R=1\), not \(\mathbb{P}(G=A\mid R=1)\). The correct calculation is \(18/26=9/13\), because the conditional probabilities over the possible groups must sum to one: \(18/26+8/26=1\). The incorrect values \(18/80\) and \(8/80\) sum to \(26/80\), not one.

If \(p_Y(y)=0\), then every joint mass \(p_{X,Y}(x,y)\) is zero, and the ratio defining \(p_{X\mid Y}(x\mid y)\) has zero denominator. There is no conditional mass function specified by this ratio at that value. Do not divide by zero or infer a conditional law from a zero-probability column. By contrast, a zero joint cell within a positive-mass column simply gives a zero conditional probability, as in the urn example.

The central checks are straightforward: verify the joint masses or counts are consistent, identify the conditioning value and its marginal total, divide every entry in that slice by the same total, and check the resulting conditional masses sum to one. Reversing the roles of \(X\) and \(Y\) changes the conditioning slice and its denominator.