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Let $A_1,\ldots,A_n$ be $n$ events in a discrete probability space. Can someone give me an example such that $$P(A_1 \cap \cdots \cap A_n)=P(A_1)\cdot \ldots \cdot P(A_n)$$ holds, but such that there is a subset $A_{i_1},\ldots,A_{i_k}$of the $A_1,\ldots,A_n$, such that $$P(A_{i_1}\cap\ldots \cap A_{i_k})\neq P(A_{i_1})\cdot\ldots \cdot P(A_{i_k}) \ \ ?$$

If that weren't such an example that would mean that the usual definition of mutual independence - as here, for example - is too strong, since the independence of the intersection of all events already implies the independence of all partial intersections of events. In that case I would like a proof of this implication.

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The subset $A_{i_1}, \ldots, A_{i_k}$ can be pretty far from independent — they can be the same event! All you need is for the remaining events to be such that the net intersection is small enough.

For example, consider the probability space of three fair coin tosses and let $A_1=A_2=\{\text{first toss is heads}\}$. For your condition to hold, we can just take an $A_3$ such that $P(A_3)=1/2$ but $P(A_1\cap A_3)=1/8$; in particular, $A_3=\{\text{at least two tosses are tails}\}$ will do. Then $$P(A_1\cap A_2\cap A_3) = P(\{\text{heads-tails-tails}\}) = \frac18 = P(A_1)\cdot P(A_2)\cdot P(A_3),$$ but $A_1$ and $A_2$ are obviously not independent.

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  • $\begingroup$ Your edit got me thinking: Are there also examples, where we don't take the same event but give it different names ? $\endgroup$ – MyCatsHat Dec 30 '12 at 17:39
  • $\begingroup$ Just write out the 8 elements of the sample space here, and I'm sure you can find a different four-element subset $A_2$ that still forms a counterexample. $\endgroup$ – Rahul Dec 30 '12 at 17:47
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My favorite answer to this question is as follows.

Let $\Omega = \{2,3,5,30\}$ and define events $A_2 = \{2,30\}, ~ A_3 = \{3,30\}, ~ A_5 = \{5,30\}$ so that $$A_2\cap A_3 = A_2\cap A_5 = A_3\cap A_5 = A_2\cap A_3\cap A_5 = \{30\}.$$

  • If the outcomes are equally likely, then $P(A_2) = P(A_3) = P(A_5) = \frac{1}{2}$ and so $A_2, A_3$, and $A_5$ are pairwise independent events since $P(A_i \cap A_j) = \frac{1}{4} = P(A_i)P(A_j)$ but are not mutually independent events since $P(A_2\cap A_3\cap A_5) = \frac{1}{4} \neq P(A_2)P(A_3)P(A_5) = \frac{1}{8}$.

  • On the other hand, if the outcomes have probabilities $$P(\{2\})=\frac{11}{24}, ~ P(\{3\})=\frac{7}{24},~ P(\{5\})=\frac{5}{24}, ~ P(\{30\})=\frac{1}{24},$$ then $P(A_2) = \frac{1}{2}$, $P(A_3) = \frac{1}{3}$, and $P(A_5)=\frac{1}{4}$ so that $$P(A_2\cap A_3\cap A_5) = P(\{30\}) = \frac{1}{24} = P(A_2)P(A_3)P(A_5)$$ but $$\begin{align*} P(A_2\cap A_3) &= \frac{1}{24} \neq P(A_2)P(A_3) = \frac{1}{6},\\ P(A_2\cap A_5) &= \frac{1}{24} \neq P(A_2)P(A_3) = \frac{1}{8},\\ P(A_3\cap A_5) &= \frac{1}{24} \neq P(A_2)P(A_3) = \frac{1}{12}. \end{align*}$$ Thus, the usual definition of mutual independence is not too strong, and we do need to insist that not only should the $n > 2$ events satisfy $$P(B_1\cap B_2\cap \cdots \cap B_n) = P(B_1)P(B_2)\cdots P(B_n)$$ but also that a similar equality (shown below) must hold for all subsets $I$ of $\{1,2,\ldots, n\}$ of cardinality $2$ or more: $$P\left(\bigcap_{i \in I} B_i\right) = \prod_{i \in I}P(B_i)$$

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