Conditional Expectations of Indicators

From an Event to a Random Variable

An event has a probability, but it can also be represented by a random variable: its indicator records whether it occurred. This representation makes conditional event probabilities part of the general theory of conditional expectation. It also gives a precise way to discuss how the probability of an event changes with an observed variable.

Definition: For an event $A$, its indicator is the random variable $\mathbf{1}_A$, equal to $1$ on $A$ and $0$ on $A^c$. If $\mathcal{G}$ is a sigma-field, the conditional probability of $A$ given $\mathcal{G}$ is the random variable $$ P(A\mid\mathcal{G})=E[\mathbf{1}_A\mid\mathcal{G}]. $$ In particular, $P(A\mid X)$ denotes $E[\mathbf{1}_A\mid\sigma(X)]$, where $\sigma(X)$ is the sigma-field generated by $X$. This is a random variable, defined up to equality almost surely.

The notation $P(A\mid X=x)$ is often used for a function of $x$ that represents $P(A\mid X)$. For discrete $X$, its value at each point with positive probability is an ordinary conditional probability. For continuous $X$, it is typically calculated from a conditional density. Its value at a particular $x$ can depend on the chosen version at points that have probability zero under the distribution of $X$.

Why the Indicator Gives Conditional Probability

The key fact is a direct consequence of the averaging characterization of conditional expectation. Conditional expectation preserves the event's probability when averaged over any information in the conditioning sigma-field.

Theorem: Let $A$ be an event and $\mathcal{G}$ a sigma-field. Then $E[\mathbf{1}_A\mid\mathcal{G}]$ is a $\mathcal{G}$-measurable random variable $Q$ such that, for every $B\in\mathcal{G}$, $$ E[\mathbf{1}_B Q]=P(A\cap B). $$ Equivalently, $P(A\mid\mathcal{G})=E[\mathbf{1}_A\mid\mathcal{G}]$ is the conditional probability of $A$ given $\mathcal{G}$.

Proof. Since $0\leq \mathbf{1}_A\leq 1$, its conditional expectation exists and is $\mathcal{G}$-measurable. By the defining averaging property of conditional expectation, for every $B\in\mathcal{G}$,

E[\mathbf{1}_B E[\mathbf{1}_A\mid\mathcal{G}]] =E[\mathbf{1}_B\mathbf{1}_A] =P(A\cap B).

Thus it has precisely the required averaging property. Uniqueness of conditional expectation up to equality almost surely establishes the claim. $\square$

Taking $B$ to be the whole sample space also gives $E[P(A\mid\mathcal{G})]=P(A)$. More generally, if $B$ describes a condition that can be determined from the available information, then

P(A\mid B)=\frac{E[\mathbf{1}_B P(A\mid\mathcal{G})]}{P(B)},\qquad P(B)>0.

This formula averages the conditional probability over the event $B$ before dividing by $P(B)$. The event $B$ must belong to $\mathcal{G}$ for this use of the defining property.

Conditioning on a Discrete Variable

Suppose $X$ is discrete. At a value $x$ with $P(X=x)>0$, the conditional probability of $A$ given $X=x$ is

P(A\mid X=x)=\frac{P(A\cap\{X=x\})}{P(X=x)}.

These values specify a version of the random variable $P(A\mid X)$: on the event $\{X=x\}$, it takes the value $P(A\mid X=x)$. Values at $x$ with $P(X=x)=0$ do not affect the conditional expectation.

Result: For discrete $X$, if $P(X=x)>0$, then $$ E[\mathbf{1}_A\mid X=x]=P(A\mid X=x). $$ For any set $C$ of values of $X$ with $P(X\in C)>0$, $$ P(A\mid X\in C) =\frac{\sum_{x\in C}P(A\mid X=x)P(X=x)} {\sum_{x\in C}P(X=x)}. $$

To verify the averaging formula, partition $A\cap\{X\in C\}$ by the possible values of $X$. Each term satisfies $P(A\cap\{X=x\})=P(A\mid X=x)P(X=x)$, so summing and dividing by $P(X\in C)$ gives the result.

Worked Example: A Discrete Risk Table

Let $X$ be a three-level score with $P(X=-1)=0.2$, $P(X=0)=0.5$, and $P(X=1)=0.3$. An event $A$, such as a positive test result, has conditional probabilities $P(A\mid X=-1)=0.1$, $P(A\mid X=0)=0.4$, and $P(A\mid X=1)=0.8$. Find $P(A)$ and $P(A\mid X\geq 0)$.

The conditional expectation of the indicator takes the corresponding event probability at each score:

E[\mathbf{1}_A\mid X]= \begin{cases} 0.1,&X=-1,\\ 0.4,&X=0,\\ 0.8,&X=1. \end{cases}

Average over all three score values to obtain the event's unconditional probability:

$$ P(A)=0.1(0.2)+0.4(0.5)+0.8(0.3) =0.02+0.20+0.24=0.46. $$

For the condition $X\geq 0$, the probability of the condition is $0.5+0.3=0.8$. The probability of both $A$ and $X\geq0$ is $0.4(0.5)+0.8(0.3)=0.44$. Therefore,

P(A\mid X\geq0)=\frac{0.44}{0.8}=0.55.

Equivalently, this is the weighted average $[0.4(0.5)+0.8(0.3)]/(0.5+0.3)$. The conditional probabilities are weighted by how often each score occurs within the selected group.

Conditioning on a Continuous Variable

For a continuous variable, $P(X=x)$ is typically zero, so the discrete ratio cannot be used. Instead, a joint density gives a conditional probability through a ratio of densities. Let $(X,Y)$ have joint density $f_{X,Y}$ and marginal density $f_X$. For a Borel set $D$, at values $x$ where $0<f_X(x)<\infty$, define

q_D(x)=\frac{\int_D f_{X,Y}(x,y)\,dy}{f_X(x)}.

Result: Under the joint-density conditions above, $q_D(X)$ is a version of $P(Y\in D\mid X)$. More generally, for a measurable event $A$ determined by $(X,Y)$, replace the numerator by $$ \int_{\mathbb{R}}\mathbf{1}_A(x,y)f_{X,Y}(x,y)\,dy. $$ The ratio is defined wherever $f_X(x)>0$; its values where $f_X(x)=0$ may be assigned arbitrarily.

Proof. Write $A_D=\{Y\in D\}$, and let $C$ be any Borel set of possible values of $X$. Using the definition of $q_D$, the marginal density, and integration of the joint density,

\begin{aligned} E[\mathbf{1}_{\{X\in C\}}q_D(X)] &=\int_C q_D(x)f_X(x)\,dx\\ &=\int_C\int_D f_{X,Y}(x,y)\,dy\,dx\\ &=P(X\in C,\ Y\in D). \end{aligned}

The integrands are nonnegative, so the exchange of integration is justified by Tonelli's theorem. The last expression is $E[\mathbf{1}_{\{X\in C\}}\mathbf{1}_{A_D}]$. Since events of the form $\{X\in C\}$ generate $\sigma(X)$, this averaging identity identifies $q_D(X)$ as a version of $E[\mathbf{1}_{A_D}\mid\sigma(X)]$. The same argument with $\mathbf{1}_A(x,y)$ in the integrand proves the general event formula. $\square$

Worked Example: A Conditional Probability from a Triangular Density

Suppose $(X,Y)$ has joint density $f_{X,Y}(x,y)=2$ for $0<y<x<1$, and zero elsewhere. Find $P(Y<1/2\mid X=x)$, then check that averaging this conditional probability gives $P(Y<1/2)$.

First find the marginal density of $X$:

f_X(x)=\int_0^x 2\,dy=2x,\qquad 0<x<1.

For $0<x\leq 1/2$, every possible value $0<y<x$ is less than $1/2$, so the numerator in the conditional-density ratio is $2x$. For $1/2<x<1$, only $0<y<1/2$ qualifies, giving numerator $2(1/2)=1$. Thus a version of the conditional probability is

P(Y<1/2\mid X=x)= \begin{cases} 1,&0<x\leq 1/2,\\[2pt] \dfrac{1}{2x},&1/2<x<1. \end{cases}

Now average against $f_X(x)=2x$:

\begin{aligned} P(Y<1/2) &=\int_0^{1/2}1(2x)\,dx +\int_{1/2}^1\frac{1}{2x}(2x)\,dx\\ &=\left[x^2\right]_0^{1/2} +\left[x\right]_{1/2}^1 =\frac14+\frac12=\frac34. \end{aligned}

The result also follows directly by integrating the joint density over $0<y<\min(x,1/2)$: $\int_0^{1/2}2x\,dx+\int_{1/2}^1 1\,dx=3/4$. Both calculations agree.

Using Conditional Probabilities to Answer a Group Question

A conditional probability given $X$ varies with $X$; a conditional probability given a group such as $X>c$ is a single average over that group. The indicator identity makes the connection explicit: if $P(X\in C)>0$, then

P(A\mid X\in C) =\frac{E[\mathbf{1}_{\{X\in C\}}P(A\mid X)]}{P(X\in C)}.

This follows because $\{X\in C\}\in\sigma(X)$, so the averaging characterization gives $E[\mathbf{1}_{\{X\in C\}}P(A\mid X)]=P(A\cap\{X\in C\})$. Dividing by the group probability produces the ordinary conditional probability.

Worked Example: Averaging a Probability Over a Continuous Group

Let $X\sim\operatorname{Uniform}(0,2)$. Given $X=x$, suppose an event $A$ occurs when an independent $U\sim\operatorname{Uniform}(0,1)$ satisfies $U\leq x/2$. Find $P(A\mid X>1)$.

For fixed $x\in(0,2)$, the interval of $U$-values satisfying $U\leq x/2$ has length $x/2$. Therefore $P(A\mid X=x)=x/2$. Since $X$ has density $1/2$ on $(0,2)$,

P(A\mid X>1) =\frac{\int_1^2 (x/2)(1/2)\,dx}{P(X>1)} =\frac{\frac14\left[\frac{x^2}{2}\right]_1^2}{1/2} =\frac{3/8}{1/2} =\frac34.

The numerator is $P(A\cap\{X>1\})$, and $P(X>1)=(2-1)/2=1/2$. As a check, conditional on $X>1$, $X$ is uniform on $(1,2)$, so the average of $X/2$ is $((1+2)/2)/2=3/4$, the same answer.

Common Pitfall and Why the Identity Matters

A common mistake is to treat $P(A\mid X=x)$ for a continuous $X$ as the ratio $P(A\cap\{X=x\})/P(X=x)$. Both numerator and denominator are generally zero, so this ratio is undefined. The conditional probability is instead a version obtained from a conditional density or another conditional law; its values are meaningful almost everywhere with respect to the distribution of $X$, not as a ratio of probabilities of point events.

The indicator formulation is useful because it unifies event probabilities with conditional expectations. It lets a statistician represent a changing event probability as a random variable, recover an overall probability by averaging, and compute a group-specific probability by averaging only over the relevant values of the conditioning variable. These operations rely on the same identity, whether the conditioning variable is discrete or continuous.

Key takeaway: The conditional probability of an event given available information is the conditional expectation of its indicator. For discrete conditioning, use conditional probability ratios at positive-mass points; for continuous conditioning, use conditional-density ratios and interpret the result as a version defined almost surely.

Check Your Understanding

Use the indicator identity and the conditioning formulas to answer these questions.

Why is $E[\mathbf{1}_A\mid\mathcal{G}]$ between zero and one almost surely, and what averaging property identifies it?
If $P(X=0)=0.4$, $P(X=1)=0.6$, $P(A\mid X=0)=0.2$, and $P(A\mid X=1)=0.7$, calculate $P(A)$.
For a continuous $X$, why is $P(A\mid X=x)$ generally not computed by dividing $P(A\cap\{X=x\})$ by $P(X=x)$?
State the joint-density ratio for $P(Y\in D\mid X=x)$, and identify the marginal-density condition needed to use it.
Explain how $P(A\mid X\in C)$ can be obtained by averaging $P(A\mid X)$ over the group $X\in C$.