Event Indicators as Independence Tests

Why Encode Events With Indicators?

An event records whether a particular condition holds; its indicator turns that yes-or-no outcome into a random variable taking the numerical values 1 and 0. This simple encoding makes event probabilities visible in a joint distribution table. It also gives an exact bridge between two forms of independence: independence of events and independence of random variables.

For two events \(A\) and \(B\), the essential comparison is whether the probability that both occur equals the product of their individual probabilities. The indicators make this comparison one cell of a two-by-two probability table. The other three cells provide useful checks on complement and intersection calculations.

Indicators and Their Joint Distribution

Definition: The indicator random variable of an event \(A\), denoted \(\mathbf{1}_A\), is defined by \(\mathbf{1}_A(\omega)=1\) when \(\omega\in A\), and \(\mathbf{1}_A(\omega)=0\) when \(\omega\notin A\). It has a Bernoulli distribution with success probability \(P(A)\): \(P(\mathbf{1}_A=1)=P(A)\) and \(P(\mathbf{1}_A=0)=1-P(A)\).

For two events, write \(I=\mathbf{1}_A\) and \(J=\mathbf{1}_B\). Their four possible value pairs correspond exactly to whether each event occurs. In particular, \((I,J)=(1,1)\) precisely on \(A\cap B\). The other cells correspond to \(A\cap B^c\), \(A^c\cap B\), and \(A^c\cap B^c\).

\begin{aligned} P(I=1,J=1)&=P(A\cap B),\\ P(I=1,J=0)&=P(A)-P(A\cap B),\\ P(I=0,J=1)&=P(B)-P(A\cap B),\\ P(I=0,J=0)&=1-P(A)-P(B)+P(A\cap B). \end{aligned}

The second and third identities subtract the shared outcome \(A\cap B\) from the probability of each event. The final identity follows because the four cells are disjoint and cover the sample space; equivalently, it is the inclusion-exclusion formula for \(P(A^c\cap B^c)\).

The Equivalence Theorem

Theorem: Two events \(A\) and \(B\) are independent if and only if their indicator random variables \(\mathbf{1}_A\) and \(\mathbf{1}_B\) are independent. Here event independence means \(P(A\cap B)=P(A)P(B)\), and random-variable independence means that for every pair of Borel sets \(C,D\subseteq\mathbb{R}\), \(P(\mathbf{1}_A\in C,\mathbf{1}_B\in D)=P(\mathbf{1}_A\in C)P(\mathbf{1}_B\in D)\).

Proof. Set \(p=P(A)\), \(q=P(B)\), and \(r=P(A\cap B)\). The indicators only take values in \(\{0,1\}\). Thus, for any Borel set \(C\), the event \(\{\mathbf{1}_A\in C\}\) is one of \(\varnothing\), \(A\), \(A^c\), or the whole sample space. The corresponding statement holds for \(B\). It therefore suffices to check independence for the four possible pairs of indicator values: if their joint cell probabilities factor, the factorization also holds for any unions of cells, and hence for all \(C,D\).

If \(A\) and \(B\) are independent, then \(r=pq\). Substituting this into the four cell probabilities gives

\begin{aligned} P(I=1,J=1)&=pq,\\ P(I=1,J=0)&=p-pq=p(1-q),\\ P(I=0,J=1)&=q-pq=(1-p)q,\\ P(I=0,J=0)&=1-p-q+pq=(1-p)(1-q). \end{aligned}

These are exactly the products of the corresponding marginal probabilities: \(P(I=1)=p\), \(P(I=0)=1-p\), \(P(J=1)=q\), and \(P(J=0)=1-q\). Hence every cell factors, so the indicators are independent.

Conversely, if the indicators are independent, their \((1,1)\) cell must factor. Therefore

P(A\cap B)=P(I=1,J=1)=P(I=1)P(J=1)=P(A)P(B).

This is exactly event independence. The argument also covers events with probability 0 or 1; no division by an event probability is required. \(\square\)

Worked Probability Checks

Worked Example: Multiples of Two and Three

Let the sample space be \(\{1,2,\ldots,12\}\), with every outcome equally likely. Let \(A\) be the event that the selected integer is even, and \(B\) the event that it is divisible by 3. Determine whether the events, and thus their indicators, are independent.

There are 6 even integers, so \(P(A)=6/12=1/2\). There are 4 multiples of 3, so \(P(B)=4/12=1/3\). The outcomes in both events are 6 and 12, giving \(P(A\cap B)=2/12=1/6\). Compare:

P(A)P(B)=\frac12\cdot\frac13=\frac16=P(A\cap B).

The events are independent. The indicator table confirms that all four cells factor. The outcome counts in the cells \((1,1),(1,0),(0,1),(0,0)\) are respectively \(2,4,2,4\), out of 12. These probabilities are \(1/6,1/3,1/6,1/3\). The products of the marginals for those cells are respectively \((1/2)(1/3)=1/6\), \((1/2)(2/3)=1/3\), \((1/2)(1/3)=1/6\), and \((1/2)(2/3)=1/3\).

Worked Example: A Shared Outcome Does Not Guarantee Independence

A single fair six-sided die is rolled. Let \(A\) be the event that the result is at most 3, and \(B\) the event that the result is even. Check whether the events are independent using their indicators.

The events are \(A=\{1,2,3\}\) and \(B=\{2,4,6\}\). Thus \(P(A)=3/6=1/2\), \(P(B)=3/6=1/2\), and \(A\cap B=\{2\}\), so \(P(A\cap B)=1/6\). But

P(A)P(B)=\frac12\cdot\frac12=\frac14\ne\frac16=P(A\cap B).

Therefore the events are not independent, and neither are their indicators. The four indicator-cell probabilities are \(P(1,1)=1/6\), \(P(1,0)=2/6=1/3\), \(P(0,1)=2/6=1/3\), and \(P(0,0)=1/6\). If the indicators were independent, the \((1,1)\) cell would have probability \(1/4\), not \(1/6\).

The full table also checks out: each marginal indicator has probability \(1/2\) of either value, so independence would require all four cells to have probability \(1/4\). The calculated cells do not.

Worked Example: Drawing Without Replacement

A box contains two red cards and two black cards. Two cards are drawn in order without replacement, with all possible ordered draws treated as equally likely. Let \(A\) be the event that the first card is red and \(B\) the event that the second card is red. Are the events independent?

There are \(4\cdot3=12\) equally likely ordered draws of distinct cards. Each of the two red cards can be first, followed by any of the other three cards, so \(P(A)=6/12=1/2\). By the same count for the second position, \(P(B)=1/2\).

For both cards to be red, there are \(2\cdot1=2\) ordered draws, so \(P(A\cap B)=2/12=1/6\). The product of the marginal probabilities is \(1/2\cdot1/2=1/4\), which differs from \(1/6\). The events and their indicators are therefore dependent.

The other indicator cells are \(P(1,0)=P(A)-P(A\cap B)=1/2-1/6=1/3\), \(P(0,1)=P(B)-P(A\cap B)=1/3\), and \(P(0,0)=1-1/2-1/2+1/6=1/6\). They sum with the both-red cell to \(1/6+1/3+1/3+1/6=1\), checking the complete table. The dependence has a direct probability explanation: after a red first draw, only one red card remains among three cards, so the probability of a red second draw is \(1/3\), not the unconditional \(1/2\).

Using the Four Cells to Check Calculations

Indicator tables are useful even when the goal is not to decide independence. Suppose a calculation proposes values for \(P(A)\), \(P(B)\), and \(P(A\cap B)\). These values imply all four indicator probabilities through the displayed formulas. A valid joint distribution requires every cell to be nonnegative and their sum to be 1.

Probability check: Any proposed values \(p=P(A)\), \(q=P(B)\), and \(r=P(A\cap B)\) must satisfy \(0\le p,q\le1\) and \( \max(0,p+q-1)\le r\le\min(p,q) \). These bounds are exactly the requirements that the four indicator-cell probabilities be nonnegative.

For example, an alleged intersection probability cannot exceed either marginal probability, since \(A\cap B\) is contained in each event. Nor can it be too small: if \(P(A)+P(B)>1\), the two events must overlap by at least \(P(A)+P(B)-1\), because their union has probability at most 1. Independence is a more specific condition, \(r=pq\), not merely a requirement that the cells form a valid probability table.

The pointwise identity \(\mathbf{1}_A\mathbf{1}_B=\mathbf{1}_{A\cap B}\) is another way to remember why the \((1,1)\) cell records the intersection. The product is 1 exactly when both indicators are 1, and is 0 otherwise. For checking independence, however, the decisive comparison remains \(P(A\cap B)\) against \(P(A)P(B)\).

Common Mistakes

Checking only whether the events overlap. A nonempty intersection does not establish dependence or independence. Compare \(P(A\cap B)\) with \(P(A)P(B)\); the die events in the second example overlap but are dependent.
Assuming that a plausible-looking table proves independence. Independence requires each indicator cell to factor, or equivalently, by the theorem, that the intersection probability equal the product of the marginals. A table can be a valid probability table without satisfying that condition.
Forgetting the complements. The cell \((0,0)\) is \(P(A^c\cap B^c)\), not \(1-P(A\cap B)\). The latter includes outcomes in either event. Use \(1-P(A)-P(B)+P(A\cap B)\) for the both-fail cell.
Dividing by a zero probability. Independence is defined by the product equation \(P(A\cap B)=P(A)P(B)\), which remains meaningful when an event has probability zero. A conditional-probability argument may require a positive conditioning probability, but the indicator equivalence does not.

Key takeaway: The joint distribution of \((\mathbf{1}_A,\mathbf{1}_B)\) is a two-by-two table whose \((1,1)\) probability is \(P(A\cap B)\). Its indicators are independent exactly when this probability equals \(P(A)P(B)\); the remaining cells then factor automatically.

Check Your Understanding

Use the indicator table and the intersection-product comparison where appropriate.

For events \(A\) and \(B\), write the four probabilities \(P(\mathbf{1}_A=i,\mathbf{1}_B=j)\) for \((i,j)\in\{0,1\}^2\) in terms of \(P(A)\), \(P(B)\), and \(P(A\cap B)\).
If \(P(A)=0.4\), \(P(B)=0.5\), and \(P(A\cap B)=0.2\), are the events independent? Give all four indicator-cell probabilities.
Can two events with probability zero be independent under the product definition? Explain using their indicators.
A calculation gives \(P(A)=0.7\), \(P(B)=0.6\), and \(P(A\cap B)=0.1\). Identify the probability constraint that this calculation violates.
Why does independence of the two indicators imply the factorization for every pair of Borel sets, even though the indicators take only two possible values?