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Mutual independence

Probability and Statistics
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Probability Theory
schedule Jul 1, 2022
Last updated
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A collection of events $A_1, A_2 \ldots {A_n}$ are said to be independent if:

$$\mathbb{P}({A_i} \cap {A_j} \ldots \cap {A_m}) = \mathbb{P}(A_i)\, \mathbb{P}(A_j)\cdots\mathbb{P}(A_m)$$

for any distinct indices $i, j, m$

Because the above should hold for any number of events, we should be able to take any two events from a collection and they should also be independent (pairwise independence).

For example if we had a collection of three events this should mean that we have the following pairwise independence between any combination of two events:

$$\begin{align*} \mathbb{P}({A_1}\cap{A_2})&=\mathbb{P}(A_1)\,\cdot\,\mathbb{P}(A_2)\\ \mathbb{P}({A_1}\cap{A_3})&=\mathbb{P}(A_1)\,\cdot\,\mathbb{P}(A_3)\\ \mathbb{P}({A_2}\cap{A_3})&=\mathbb{P}(A_2)\,\cdot\,\mathbb{P}(A_3) \end{align*}$$
WARNING

Just because we have pairwise independence does not necessarily mean mutual independence between all events in the collection.

Example

Imagine we have 4 cards in a bag, numbered from 1 - 4 and each card has an equal probability of being drawn from the bag (i.e. 0.25).

Given we draw one card from the bag, let us define events $A$, $B$ and $C$ as follows:

$A$ is the event that a 1 or 2 is drawn. This can be represented as $A = \{1,2\}$.

$B$ is the event that a 1 or 3 is drawn. This can be represented as $B = \{1,3\}$.

$C$ is the event that a 1 or 4 is drawn. This can be represented as $C = \{1,4\}$.

It is clear from the above that:

$\mathbb{P}(A) = \frac{2}{4} = \frac{1}{2}$

$\mathbb{P}(B) = \frac{2}{4} =\frac{1}{2}$

$\mathbb{P}(C) = \frac{2}{4} = \frac{1}{2}$

To check whether we have pairwise independence:

$$\begin{align*} \mathbb{P}(A \cap {B}) &= \frac{1}{4} = \mathbb{P}(A)\cdot \mathbb{P}(B) = \frac{1}{2} \cdot \frac{1}{2} \,\,\,\,\,\,\,\, \text{True}\\ \mathbb{P}(A \cap {C}) &= \frac{1}{4} = \mathbb{P}(A)\cdot \mathbb{P}(C) = \frac{1}{2} \cdot \frac{1}{2} \,\,\,\,\,\,\,\, \text{True}\\ \mathbb{P}(B \cap {C}) &= \frac{1}{4} = \mathbb{P}(B)\cdot \mathbb{P}(C) = \frac{1}{2} \cdot \frac{1}{2} \,\,\,\,\,\,\,\, \text{True} \end{align*}$$
NOTE

Taking the event $(A \cap {B})$ as an example, being told that event $B$ occurred does not change the probability that event $A$ occurred. If event $B$ has occurred, then the card drawn must have been a 1 or 3. However, as event $A$ only occurs when a 1 is drawn, $\mathbb{P}(A)$ remains $\frac{1}{2}$. As the additional information on occurrence of event $B$ does not change our view on probability of event $A$, the two events are considered independent.

To check whether we have mutual independence:

$$\begin{align*} \mathbb{P}({A}\cap{B}\cap{C})=\frac{1}{4} \not&= \mathbb{P}(A)\cdot\mathbb{P}(B)\cdot\mathbb{P}(C) \\ &= \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{8} \,\,\,\,\,\,\,\,\,\,\, \text{False} \end{align*}$$
NOTE

The reason that we do not have mutual independence is that if we know that $A$ and $B$ both occurred, then we know that the card with number 1 must have been drawn from the bag as this is the only way both events can occur. This means we already know that event $C$ must have also occurred, and can see that there is a dependence here.

robocat
Published by Arthur Yanagisawa
Edited by 0 others
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