| Independence |
|---|
Two events A and B are said to be independent if
It is a custom to drop the sign of intersection, ∩, and write the
intersection of a collection of events as
\[
\Pr [A\cap B] = \Pr [A]\,\Pr [B] .
\]
An arbitrary collection of events is independent if, for every
finite subcollection of two or more events, say
A1, A2, ... ,Ar,
\[
\Pr\left[ A_1 \cap A_2 \cap \cdots \cap A_r \right] = \Pr [A_1]\,\Pr [A_2]
\cdots \Pr [A_r ] .
\]
Two events A and B are defined to be dependent if
they are not independent. Similar definition is valid for a collection of
events.
▣
\[
\Pr\left[ A_1 \cap A_2 \cap \cdots \cap A_r \right] =
\Pr\left[ A_1 \, A_2 \, \cdots \, A_r \right] .
\]
Two events A and B are independent when occurence of A
does nothing to change the probability of B. This concept of
independence should not be mixed with the definition
of two events A and B to be mutually exclusive:
\[
\Pr\left[ A \cap B \right] = 0 .
\]
Actually, two mutually exclusive events A and B with positive
probabilities are always dependent.
Example:
Let ∅ be the empty set and Ω be the universal set. If B is
any event, then ∅ and B are independent because
\[
\Pr\left[ \varnothing \cap B \right] = \Pr [\varnothing ] =0 \qquad\mbox{and}
\qquad \Pr [\varnothing ]\, \Pr [B] = 0\cdot \Pr [B] = 0.
\]
Similarly, Ω and B are independent because
\[
\Pr\left[ \Omega \cap B \right] = \Pr [B ] \qquad\mbox{and}
\qquad \Pr [\Omega ]\, \Pr [B] = 1\cdot \Pr [B] = \Pr [B] .
\]
Therefore, the empty set ∅ is independent of every event including
itself. So is Ω.
There are only two events that are independent of itself---the empty set and the universal set. If A were independent of itself, then \( \Pr [A\cap A ] \) would have to equal Pr[A] Pr[A] . Since A∩A = A, this says that the number Pr[A] has to equal its own square. The only numbers that equal their own squares are 0 and 1 that correspond to the empty set and to the universal set, respectively.
Note that in some continuous spaces, there could be events other than ∅ that have zero probability. Their complements would have probabilities 1. For example, the probability that a new born baby has weight of exactly 2.5 kg is zero, so such event is independent to itself. Therefore, there could exist events other than ∅ and Ω that are independent to itself. ■
Theorem: If events A and B are independent, then so are A' and B and A and B', where A' = Ac is the complement to event A. ▣
Example:
Suppose that a balanced die is rolled. Let A be the event that an odd
number is obtained, and let B be the event that one of the number 1, 2,
3, and 4 is obtained. We shall show that the events A and B are
independent.
First, we calculate probabil;ities of these events:
\[
\Pr [A] = \frac{1}{2} \qquad\mbox{and}\qquad \Pr [B] = \frac{4}{6} =
\frac{2}{3} .
\]
The intersection of these two events consists of two digits:
\( \Pr [A\cap B ] = \Pr \left[ \left\{ 1, 3 \right\} \right] = \frac{2}{6} = \frac{1}{3} . \) Hence, Pr[A B] = 1/3 =
Pr[A] Pr[B] and these events are independent.
■
Example:
Let us roll a fair die twice, and let A be the event "first die shows
an odd number," B the event "second die shows an odd number," and
C "the sum is an odd number." The probabilities of these three events
are the same:
\[
\Pr [A] = \Pr [B] = \Pr [C] = \frac{1}{2} .
\]
But their intersections are
\[
\Pr [A\cap B] = \Pr [B\cap C] = \Pr [A\cap C] = \frac{1}{4} .
\]
However, the three-event intersection is empty:
\( \Pr [A\cap B \cap C ] = \Pr [\varnothing ] = 0 \)
and these three events are dependent.
■
- Ollerton, R.L., A unifying framework for teaching probability event types, International Journal of Mathematical Education in Science and Technology, 2015, Vol. 46, No. 5, pp. 790--794; https://doi.org/10.1080/0020739X.2015.1005702