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We know if a collection of events are independent, then they are pairwise independent. In general, the converse is not true. However, I'm wondering if there's a condition under which the converse holds. I haven't been able to find anything on this. Any help is appreciated.

epsilon-delta
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    When there are only two events in the collection. – bof Oct 11 '14 at 02:14
  • it is not true in the other direction either. – Calculon Oct 11 '14 at 02:18
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    @bof . . . and if you want to fit a straight line, then obtain only two data points. – Michael Hardy Oct 11 '14 at 02:30
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    @Calculon why isn't it true in the other direction? Say we have three events $A,B,C$ if they are independent then we have $P(A \cap B) = P(A)P(B)$, $P(A \cap C) = P(A)P(C),$ $P(B\cap C) = P(B)P(C)$ and $P(A\cap B \cap C) = P(A)P(B)P(C)$. Surely, they are pairwise independent, no? This generalizes to a collection of $k$ events. – epsilon-delta Oct 11 '14 at 02:38

2 Answers2

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One situation in which this is true is when the random variables involved are jointly normally distributed. If $X_1,\ldots,X_n$ are jointly distributed in such a way that for every sequence $a_1,\ldots,a_n$ of constants (i.e. non-random), the random variable $a_1 X_1+\cdots+a_n X_n$ has a $1$-dimensional normal distribution, then that is joint normality of the distribution of $X_1,\ldots,X_n$. If these are pairwise independent, then they are independent.

At the opposite extreme, you have the case of $Y_1,Y_2, Y_3$, where $Y_1, Y_2$ are independent and identically distributed and $\Pr(Y_1=1)=p\in(0,1)$ and $\Pr(Y_1=0)=1-p$, and $Y_3$ is the mod-$2$ sum of $Y_1$ and $Y_2$. These three are pairwise independent, but the values of any two of them determine the value of the third.

Michael Hardy
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I was also wondering the same, and I think it's instructive to just restate mutual independence in an equivalent way. I will do it below for 3 real random variables but the statement is generalizable.

Let $X, Y, Z$ be pairwise independent random variables and also assume that $(X,Y)$ is independent of $Z$. Then $X,Y,Z$ are mutually independent.

To prove this, just pick Borel sets $A, B, C$, then it holds that:

$$ \begin{aligned} \Pr[ X \in A, Y \in B, Z \in C] &= \Pr[ (X,Y) \in A\times B, Z \in C] \\ &= \Pr[(X,Y) \in A\times B] \Pr[Z \in C] \\ &= \Pr[X \in A, Y \in B] \Pr[Z \in C] \\ &= \Pr[X \in A] \Pr[X \in B] \Pr[Z \in C] \end{aligned} $$

air
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