Conditional Expectation Given a Discrete Variable

From Conditioning on an Event to Conditioning on a Variable

A discrete variable can take several possible values, and each value identifies an event on which we can calculate a conditional mean. Conditioning on the variable means assembling those event-specific means into one random variable: its value depends on which value of the conditioning variable is observed. The key is to distinguish the number obtained at a particular value from the random variable that records the appropriate number across all outcomes.

Let \(X\) and \(Y\) be discrete random variables, and let \(g\) be a real-valued function. We will define the conditional expectation of \(g(X)\) given \(Y\). At any value \(y\) with \(P(Y=y)>0\), this is the ordinary conditional mean of \(g(X)\) on the event \(\{Y=y\}\). Collecting those means over the possible values of \(Y\) gives a function of \(Y\).

The Conditional-Mass Formula

The conditional distribution at \(y\) is obtained by restricting the joint distribution to outcomes with \(Y=y\), then dividing by the probability of that event. For any positive-probability value \(y\), its masses sum to one:

Consequently, the conditional mean is an ordinary expectation under this conditional probability distribution. Since \(E[|g(X)|]<\infty\), the sum defining the conditional mean is absolutely convergent at every \(y\) with positive probability. Indeed,

The denominator matters: this is a conditional average, so the joint masses for the selected value of \(Y\) must be rescaled to sum to one. Equivalently, the formula can be written directly in terms of joint probabilities:

Proof. The support of a discrete random variable is countable. For each positive-probability \(y\), the conditional-mass formula defines \(h(y)\), and the preceding absolute-summability bound makes it finite. Assign any finite value to \(h(y)\) at points with \(P(Y=y)=0\). The composition \(h(Y)\) is a function of a discrete random variable, so it is measurable.

To verify integrability, sum over the positive-probability values of \(Y\). The triangle inequality and the conditional-mass formula give

This establishes that the pointwise conditional means form an integrable random variable depending only on \(Y\). For uniqueness at positive-probability values, suppose \(h(Y)=k(Y)\) with probability one. If there were a positive-probability value \(y\) with \(h(y)\ne k(y)\), then the two random variables would differ on the entire event \(\{Y=y\}\), an event of positive probability. That contradicts equality with probability one. Thus they agree at every positive-probability value. At zero-probability values, their assigned values have no effect on either random variable. \(\square\)

Worked Examples

Zero-Probability Values and Common Pitfalls

The pointwise formula requires \(P(Y=y)>0\). If \(P(Y=y)=0\), the quotient by \(P(Y=y)\) is undefined, and the distribution of \(X\) conditional on that event is not determined by ordinary event conditioning. No value at such a point can change \(h(Y)\), because the event on which that value would be used has probability zero. For a countable support, the union of all zero-probability points also has probability zero.

• Using a joint mass without normalizing. The sum \(\sum_x g(x)P(X=x,Y=y)\) is not the conditional mean. Divide it by \(P(Y=y)\) to obtain the conditional distribution and its mean.
• Confusing a number with a random variable. \(E[g(X)\mid Y=y]\) is a number for a specified \(y\). \(E[g(X)\mid Y]=h(Y)\) is the random variable that selects the relevant number.
• Assigning a meaningful conditional law at a null point. If \(P(Y=y)=0\), do not apply the positive-probability formula. Any finite assigned value there gives the same random variable with probability one.

This construction is useful because it turns conditioning on a variable into a collection of familiar conditional means on events. Once the conditional distribution is specified, a nonlinear function causes no special difficulty: apply the function to each possible value of \(X\), weight by its conditional probability, and sum. The same procedure works for finite and countably infinite supports, provided the expectation is integrable as stated.