Equivalent Characterizations of Independence

Why Conditional Probabilities Characterize Independence

Independence means that learning one event occurred does not change the probability assigned to another event. Conditional probability makes that idea precise: if $B$ has positive probability, compare $P(A\mid B)$ with $P(A)$. Equality means that conditioning on $B$ leaves the probability of $A$ unchanged. This tutorial proves the equivalence between that comparison and the probability-factorization definition of independence.

Definition: Events $A$ and $B$ are independent if $$ P(A\cap B)=P(A)P(B). $$ If $P(B)>0$, then $P(A\mid B)=P(A\cap B)/P(B)$. Consequently, for a positive-probability conditioning event $B$, independence is equivalent to $P(A\mid B)=P(A)$.

The condition $P(B)>0$ is essential for the conditional probability $P(A\mid B)$ to be defined. The event $A$ itself does not need to have positive probability for this direction of the characterization.

The Equivalence and Its Justification

Assume first that $P(B)>0$. Independence can be tested in either of two equivalent ways: factor the joint probability, or check whether conditioning on $B$ changes $P(A)$. The equivalence follows directly from the definition of conditional probability.

Start with independence.
If $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)$.

Use the positive-probability condition.
Since $P(B)>0$, divide the factorization by $P(B)$.

Recognize conditional probability.
The quotient $P(A\cap B)/P(B)$ is $P(A\mid B)$, so it equals $P(A)$.

\begin{aligned} P(A\cap B)=P(A)P(B) &\quad\Longrightarrow\quad \frac{P(A\cap B)}{P(B)}=P(A),\\ &\quad\Longrightarrow\quad P(A\mid B)=P(A). \end{aligned}

For the converse, suppose $P(B)>0$ and $P(A\mid B)=P(A)$. Multiplying both sides by $P(B)$, then applying the definition of conditional probability, gives the factorization:

P(A\mid B)=P(A) \quad\Longrightarrow\quad \frac{P(A\cap B)}{P(B)}=P(A) \quad\Longrightarrow\quad P(A\cap B)=P(A)P(B).

Theorem: For events $A$ and $B$ with $P(B)>0$, $$ A\text{ and }B\text{ are independent} \quad\Longleftrightarrow\quad P(A\mid B)=P(A). $$ If $P(A)>0$, the corresponding reverse-direction statement is also valid: $$ A\text{ and }B\text{ are independent} \quad\Longleftrightarrow\quad P(B\mid A)=P(B). $$

The reverse-direction statement follows by the same argument, now dividing by $P(A)$. The two conditional statements are not both required when using the theorem: either one is sufficient if its conditioning event has positive probability. If the conditioning event has probability zero, the conditional probability is undefined, so it cannot be used in this test. The factorization definition remains the appropriate statement in that case.

Worked Examples

Worked Example: A Coin Toss and a Die Roll

A fair coin is tossed once and a fair six-sided die is rolled once. The experiment has 12 equally likely outcomes, each with probability $1/12$. Let $A$ be the event that the coin shows heads, and let $B$ be the event that the die shows an even number. Use conditional probability to check whether $A$ and $B$ are independent.

Method. Find $P(A)$, $P(B)$, and $P(A\mid B)$, then compare the conditional and unconditional probabilities of $A$.

There are six outcomes with heads, so $P(A)=6/12=1/2$. There are six outcomes with an even die result (either coin result paired with 2, 4, or 6), so $P(B)=6/12=1/2>0$. Of those six outcomes in $B$, three have heads. Therefore,

P(A\mid B)=\frac{3}{6}=\frac{1}{2}.

Since $P(A\mid B)=1/2=P(A)$, the theorem shows that $A$ and $B$ are independent. As a check using factorization, $P(A\cap B)=3/12=1/4$, while $P(A)P(B)=(1/2)(1/2)=1/4$. The two checks agree.

Common mistake avoided: Do not compare $P(A\mid B)$ with $P(B)$. The question is whether conditioning on $B$ changes the probability of $A$, so the comparison is with $P(A)$.

Worked Example: Claim Probability Across Two Risk Groups

Consider a hypothetical portfolio in which a randomly selected policy is in group $B$ with probability $0.40$. Let $A$ denote a claim during the coming year. The model specifies $P(A\mid B)=0.12$ and $P(A\mid B^c)=0.12$, where $B^c$ means the policy is not in group $B$. Find $P(A)$, determine whether $A$ and $B$ are independent, and calculate $P(A\cap B)$.

Method. First combine the two groups using the law of total probability, then compare $P(A\mid B)$ with $P(A)$. The group probabilities are $P(B)=0.40$ and $P(B^c)=1-0.40=0.60$.

The overall claim probability is

\begin{aligned} P(A) &=P(A\mid B)P(B)+P(A\mid B^c)P(B^c)\\ &=(0.12)(0.40)+(0.12)(0.60)\\ &=0.048+0.072\\ &=0.120. \end{aligned}

Now $P(A\mid B)=0.12=P(A)$, and $P(B)=0.40>0$. Thus $A$ and $B$ are independent. Their intersection probability is

P(A\cap B)=P(A\mid B)P(B)=(0.12)(0.40)=0.048.

The factorization check gives $P(A)P(B)=(0.120)(0.40)=0.048$, confirming the result. The equal claim probabilities in the two groups are what make learning the group membership leave the claim probability unchanged.

Common mistake avoided: Do not treat the conditional claim rate $P(A\mid B)$ as automatically equal to the overall rate $P(A)$. Here they agree only after the two group rates and group probabilities are combined.

Worked Example: Two Draws Without Replacement

A container holds seven distinct tokens: three blue and four yellow. Two tokens are drawn in order without replacement, with each token in the container equally likely to be selected at each draw. Let $A$ be the event that the first token is blue and $B$ the event that the second token is blue. Determine whether $A$ and $B$ are independent.

Method. Find $P(A)$ and $P(A\mid B)$. Since $P(B)>0$, equality would establish independence; inequality establishes dependence.

There are three blue tokens among seven on the first draw, so

P(A)=\frac{3}{7}.

By symmetry, the second draw is also blue with probability $3/7$, so $P(B)=3/7>0$. To calculate the conditional probability, first find the probability that both draws are blue. The probability the first is blue is $3/7$; after that, two blue tokens remain among six, so the probability the second is blue is $2/6$. Hence

P(A\cap B)=\frac{3}{7}\cdot\frac{2}{6}=\frac{6}{42}=\frac{1}{7}.

Therefore,

P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{1/7}{3/7} =\frac{1}{3}.

Because $1/3\ne 3/7=P(A)$, the events are not independent. The factorization check agrees: $P(A)P(B)=(3/7)(3/7)=9/49$, whereas $P(A\cap B)=1/7=7/49$.

Common mistake avoided: The two draws have the same marginal probability of blue, but that does not imply independence. Independence requires the conditional probability to remain unchanged, and here it changes from $3/7$ to $1/3$.

Interpreting the Test Correctly

The conditional-probability characterization is useful because it gives independence a direct interpretation: an event $B$ provides no probability information about $A$ precisely when $P(A\mid B)=P(A)$, provided $P(B)>0$. It is often faster to use this equality than to compute every probability in a larger model. The factorization form is also useful, especially when the joint probability and both marginal probabilities are given directly.

Information available	Appropriate check	Condition
$P(A\cap B)$, $P(A)$, and $P(B)$	Check whether $P(A\cap B)=P(A)P(B)$	No division is needed
$P(A\mid B)$ and $P(A)$	Check whether $P(A\mid B)=P(A)$	$P(B)>0$
$P(B\mid A)$ and $P(B)$	Check whether $P(B\mid A)=P(B)$	$P(A)>0$

A conditional probability is a probability within the restricted sample space specified by the conditioning event. Its denominator is the probability of that event. Consequently, the comparison must use the same target event on both sides: $P(A\mid B)$ is compared with $P(A)$, not with $P(B)$. Also, equality for one particular conditioning event establishes independence of that pair of events; it does not by itself assert that every other event in the model is independent of $A$.

Key takeaway: When $P(B)>0$, $A$ and $B$ are independent exactly when conditioning on $B$ leaves the probability of $A$ unchanged. If the conditioning event has probability zero, its conditional probability is undefined, and the factorization definition must be used instead.

Check Your Understanding

For each problem, use the stated model and show whether the conditional-probability test supports independence.

A fair eight-sided die, numbered 1 through 8, is rolled once. Let $A$ be the event of rolling a number greater than 4 and $B$ the event of rolling an even number. Find $P(A)$, $P(A\mid B)$, and decide whether the events are independent. (Answer: $P(A)=1/2$, $P(A\mid B)=2/4=1/2$; independent.)
For events $A$ and $B$, suppose $P(B)=0.25$, $P(A)=0.16$, and $P(A\cap B)=0.04$. Calculate $P(A\mid B)$, then test independence both by conditional probability and by factorization. (Answer: $P(A\mid B)=0.16$; independent since $0.04=(0.16)(0.25)$.)
In a hypothetical inspection model, a randomly selected component has probability $0.30$ of being inspected. Let $B$ be the event that it is inspected and $A$ the event that it has a flaw. Suppose $P(A\mid B)=0.08$ and $P(A\mid B^c)=0.05$. Find the overall flaw probability and determine whether $A$ and $B$ are independent. (Answer: $P(A)=(0.08)(0.30)+(0.05)(0.70)=0.059$; not independent because $0.08\ne0.059$.)
A box contains five green and three orange counters. Two counters are drawn in order without replacement. Let $A$ mean the first counter is green and $B$ mean the second is green. Find $P(A)$, $P(A\cap B)$, and $P(A\mid B)$, then determine independence. (Answer: $P(A)=5/8$, $P(A\cap B)=(5/8)(4/7)=5/14$, and $P(A\mid B)=(5/14)/(5/8)=4/7$; not independent because $4/7\ne5/8$.)
Suppose $P(A)=0.20$, $P(B)=0.50$, and $P(A\mid B)=0.20$. Calculate the intersection probability and use both characterizations to decide independence. (Answer: $P(A\cap B)=(0.20)(0.50)=0.10$; independent since $P(A\mid B)=P(A)$ and $P(A\cap B)=P(A)P(B)$.)

Information available	Appropriate check	Condition
\(P(A\cap B)\), \(P(A)\), and \(P(B)\)	Check whether \(P(A\cap B)=P(A)P(B)\)	No division is needed
\(P(A\mid B)\) and \(P(A)\)	Check whether \(P(A\mid B)=P(A)\)	\(P(B)>0\)
\(P(B\mid A)\) and \(P(B)\)	Check whether \(P(B\mid A)=P(B)\)	\(P(A)>0\)

What You'll Learn