Two fair dice are thrown. We have three events:

  • A: The first die shows an odd number

  • B: The second die shows an even number

  • C: Both are odd or both are ven

Show that $A,B,C$ are piecewise independent but not independent.

My answer:

$P(A) = P(B) = P(C) = \frac{1}{2}$.

$P( A \cap B) = P( A \cap C) = P( B \cap C) = \frac{1}{4}.$

This means that all the events are pairwise independent. However:

$P(A \cap B \cap C) = 0$ while $P(A)P(B)P(C) = \frac{1}{8}$, so the events are not independent.

Is this correct (disregarding that I didn't explain how I got those probabilities)?

  • 2
    $\begingroup$ I might have extended your second line to say explicitly $P( A \cap B) = P( A \cap C) = P( B \cap C) = \frac{1}{4} = P(A)P(B) = P(A)P(C)=P(B)P(C)$ $\endgroup$ – Henry Feb 10 '16 at 8:27

Yes, this is correct. The definitions are as follow (see here).

A finite set of events $\{A_i\}$ is pairwise independent if and only if every pair of events is independent — that is, if and only if for all distinct pairs of indices $m, k$, $$ \mathrm{P}(A_m \cap A_k) = \mathrm{P}(A_m)\mathrm{P}(A_k). $$ A finite set of events is mutually independent if and only if every event is independent of any intersection of the other events — that is, if and only if for every $n$-element subset ${A_i}$, $$ \mathrm{P}\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathrm{P}(A_i). $$ You have shown that the events $A,B$ and $C$ are pairwise independent,but not mutually independent.


In this case you are using that $A$ and $B$ are independent iff $P(A\cap B)=P(A)·P(B)$

I think clarifying that is important. Beyond that, it's all correct.

They are pairwise independent because $P(A)·P(B)={1\over 4}$ too (and the same's true for $A$, $C$ and $B$,$C$).

The important part, though, is that you're correct.


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