Conditional Expectation Given an Event

Conditioning on an Event

When an event \(A\) has occurred, probabilities and averages should be computed relative to the outcomes inside \(A\). This changes the relevant probability scale: the original probability assigned to a subset of \(A\) must be divided by \(\mathbb{P}(A)\). The same normalization applies to an expectation restricted to \(A\). It is essential that \(\mathbb{P}(A)>0\); conditioning on a probability-zero event cannot be defined by this ratio.

Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, so \(\mathbb{P}(\Omega)=1\). No completeness assumption is needed. Events belong to \(\mathcal{F}\), and random variables are real-valued and \(\mathcal{F}\)-measurable unless stated otherwise. For conditional means, assume the random variable \(X\) is integrable, meaning \(\mathbb{E}[|X|]<\infty\). Equalities between random variables identified as conditional expectations hold \(\mathbb{P}\)-almost surely; a conditional mean given a fixed event is a number.

The Normalized Restricted Probability Measure

The formula is not just a convenient rule. It arises by making a new probability measure that assigns probability only to outcomes in \(A\). For each \(B\in\mathcal{F}\), define \(\mathbb{P}_A(B)=\mathbb{P}(B\cap A)/\mathbb{P}(A)\). Then \(\mathbb{P}_A(A)=1\), and \(\mathbb{P}_A\) is the probability law of outcomes viewed under the condition \(A\).

Proof. For every \(B\in\mathcal{F}\), \(\mathbb{P}_A(B)\ge0\), and \(\mathbb{P}_A(\Omega)=\mathbb{P}(A)/\mathbb{P}(A)=1\). If \(B_1,B_2,\ldots\) are pairwise disjoint events, then \(B_n\cap A\) are pairwise disjoint and their union is \((\bigcup_n B_n)\cap A\). Countable additivity of \(\mathbb{P}\) therefore gives \(\mathbb{P}_A(\bigcup_n B_n)=\sum_n\mathbb{P}_A(B_n)\). Thus \(\mathbb{P}_A\) is a probability measure.

For a nonnegative simple function \(\varphi=\sum_{j=1}^k a_j\chi_{B_j}\), where the \(B_j\) are disjoint events and \(a_j\ge0\), the definition of the integral gives \(\mathbb{E}_{\mathbb{P}_A}[\varphi]=\sum_j a_j\mathbb{P}_A(B_j)=\mathbb{E}_{\mathbb{P}}[\varphi\chi_A]/\mathbb{P}(A)\). The same identity holds for every nonnegative measurable function by approximating it from below by nonnegative simple functions and applying the Monotone Convergence Theorem on both probability spaces. Apply this identity to \(|X|\): it gives \(\mathbb{E}_{\mathbb{P}_A}[|X|]=\mathbb{E}_{\mathbb{P}}[|X|\chi_A]/\mathbb{P}(A)<\infty\). Hence \(X\) is integrable under \(\mathbb{P}_A\). Applying the identity separately to \(X^+\) and \(X^-\), both of which have finite expectations, and subtracting proves the claimed formula for \(X\). \(\square\)

Taking \(X=\chi_B\) in the formula shows that \(\mathbb{E}_{\mathbb{P}}[\chi_B\mid A]=\mathbb{P}(B\mid A)\). Thus conditional probability is the special case of conditional mean for an indicator. For a finite-valued random variable, the expectation formula also gives the familiar weighted average over the outcomes that satisfy \(A\).

Connection with Conditional Expectation

There are two related objects to distinguish. The number \(\mathbb{E}[X\mid A]\) is one average, defined only when \(\mathbb{P}(A)>0\). In contrast, \(\mathbb{E}[X\mid\sigma(A)]\) is a random variable measurable with respect to the sigma-algebra \(\sigma(A)=\{\varnothing,A,A^c,\Omega\}\). When both \(A\) and \(A^c\) have positive probability, it takes one constant value on each of these two events.

Proof. The right-hand side, denoted by \(Y\), is \(\sigma(A)\)-measurable because it is constant on \(A\) and on \(A^c\). To verify the defining integral condition, it is enough to check the four possible events \(C\in\sigma(A)\). For \(C=A\), \(\mathbb{E}[Y\chi_A]=\mathbb{E}[X\mid A]\mathbb{P}(A)=\mathbb{E}[X\chi_A]\). For \(C=A^c\), the corresponding identity is \(\mathbb{E}[Y\chi_{A^c}]=\mathbb{E}[X\chi_{A^c}]\). For \(C=\varnothing\), both integrals are zero; for \(C=\Omega\), add the two identities for \(A\) and \(A^c\). Thus \(\mathbb{E}[Y\chi_C]=\mathbb{E}[X\chi_C]\) for every \(C\in\sigma(A)\), which is the defining property of conditional expectation. \(\square\)

If \(\mathbb{P}(A)=1\), the conditional expectation given \(\sigma(A)\) equals \(\mathbb{E}[X]\) almost surely; its value on \(A^c\), a null event, does not affect that equality. If \(\mathbb{P}(A)=0\), the scalar \(\mathbb{E}[X\mid A]\) is undefined by the event formula. A conditional expectation given \(\sigma(A)\) can still be specified almost surely, but its value on the null atom \(A\) is not determined by the defining integral condition. These boundary cases are why the positive-probability condition matters.

Bayes’ Formula and Conditional Means

Bayes’ formula follows by expressing the event being conditioned on as a disjoint union. Let \(H\) be an event with \(0<\mathbb{P}(H)<1\), and let \(T\) be an event with \(\mathbb{P}(T)>0\). Then \(\mathbb{P}(T)=\mathbb{P}(T\mid H)\mathbb{P}(H)+\mathbb{P}(T\mid H^c)\mathbb{P}(H^c)\). Dividing \(\mathbb{P}(H\cap T)=\mathbb{P}(T\mid H)\mathbb{P}(H)\) by this positive denominator yields \[ \mathbb{P}(H\mid T)= \frac{\mathbb{P}(T\mid H)\mathbb{P}(H)} {\mathbb{P}(T\mid H)\mathbb{P}(H)+\mathbb{P}(T\mid H^c)\mathbb{P}(H^c)}. \] The denominator must be positive; this is exactly the requirement \(\mathbb{P}(T)>0\).

A Normalization Pitfall

A frequent error is to report \(\mathbb{E}[X\chi_A]\) as the conditional mean. This quantity includes the probability of \(A\) in its scale; it is not an average under the condition \(A\). In the die example, \(\mathbb{E}[X\chi_A]=2\), but the conditional mean is \(2/\mathbb{P}(A)=2/(1/2)=4\). The incorrect value 2 is only half the correct conditional mean because the event has probability \(1/2\).

Another error is to condition on an event of probability zero by using the same ratio. If \(\mathbb{P}(A)=0\), then \(\mathbb{E}[X\chi_A]=0\) for integrable \(X\), so the proposed ratio is \(0/0\), not a defined conditional mean. Conditioning on such events requires a different framework and is not given by the event formula here.

When the random variable is an indicator, always identify the event represented by that indicator before calculating. For example, \(\mathbb{E}[\chi_H\mid T]=\mathbb{P}(H\mid T)\): the condition is \(T\), while the indicator records whether \(H\) occurs. Reversing these events changes the question; in general \(\mathbb{P}(H\mid T)\) and \(\mathbb{P}(T\mid H)\) are not equal.