Pairwise Independence Is Not Mutual Independence

A wrong example

$$\{\begin{array}{l}A:\text{red exists},\\ B:\text{blue exists},\\ C:\text{both red and blue exist}.\end{array}$$

Apparently, $C=A\cap B$ and $P(A|C)=1\ne P(A)$. Therefore, these three events are not pairwise independent nor mutually independent.

However, in light of this, we can have a correct example using a 4-sided die. Roll a fair 4-sided die whose outcome is in $\{1,2,3,4\}$. Define three events

$$\{\begin{array}{l}A:\{1,2\},\\ B:\{1,3\},\\ C:\{2,3\}.\end{array}$$

Then we have

$$\begin{array}{r}P(AB)=P(\{1\})=\frac{1}{4}=P(A)P(B),\\ P(AC)=P(\{2\})=\frac{1}{4}=P(A)P(C),\\ P(BC)=P(\{3\})=\frac{1}{4}=P(B)P(C).\end{array}$$

But

$$P(ABC)=P(\varnothing )=0\ne P(A)P(B)P(C).$$

Therefore, pairwise independence is not mutual independence.

Simplestness

Define What’s Simpler

To show the above example is the “simplest,” we need first to define the partial order to compare simpleness: we say an example $E$ is simpler than $F$, if $E$ involves fewer events than $F$, or the cardinality of $E$‘s sample space is smaller when $E$ and $F$ involve the same number of events. Since the number of events cannot be less than 3, otherwise mutual independence will reduce to pairwise independence, we need to find a smaller sample space for a simpler example.

Select the Events

Consider the sample space $\{1,2,3\}$. The nonempty events are

$$\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}.$$

We call the first three events size-1 events, the following three events size-2 events, and the last one size-3 event. To compose the example, we need to select three events $A$, $B$, $C$. The size-3 event cannot be selected, because it is the superset of all other events, and hence is not independent of any of them. We can select up to one size-1 event; otherwise, the intersection of two size-1 events is empty, indicating that these two events are not pairwise independent (here we assume nonzero probability). However, when we select one size-1 event, the other two size-2 events either are the supersets of it, or have no intersection with it; in either situation, they are not pairwise independent. Therefore, $A,B,C$ can only be the three size-2 events.

Calculate the Probabilities

Let the probability mass function be

$$P(\{1\})=p,P(\{2\})=q,P(\{3\})=1-p-q.$$

Then we have

$$P(BC)=P(B)P(C)\Rightarrow 1-p-q=(1-q)(1-p)\Rightarrow pq=0.$$

This also indicates mutual independence, because

$$P(ABC)=P(\varnothing )=0=pq(1-p-q)=P(A)P(B)P(C).$$

In summary, such an example with a sample space of cardinality 3 does not exist.