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We have the definition of independence of events $A_1,A_2,\dots,A_n$:

$$\mathbf P\left(\bigcap_{i\in S}A_i\right)=\prod_{i\in S}\mathbf P\left(A_i\right), \forall S\subset\{1, 2, \dots, n\}$$

This means that, for three events $A_1, A_2, A_3$, $\mathbf P(A_1\cap A_2\cap A_3)=\mathbf P(A_1)\cdot \mathbf P(A_2)\cdot \mathbf P(A_3)$ alone is not sufficient for them to be independent because it does not imply the other three criteria:

  • $\mathbf P(A_1\cap A_2)=\mathbf P(A_1)\cdot \mathbf P(A_2)$
  • $\mathbf P(A_2\cap A_3)=\mathbf P(A_2)\cdot \mathbf P(A_3)$
  • $\mathbf P(A_3\cap A_1)=\mathbf P(A_3)\cdot \mathbf P(A_1)$

However, we can say that three random variables are independent if $p_{X,Y,Z}(x,y,z)=p_X(x)\cdot p_Y(y)\cdot p_Z(z),\forall x,y,z$.

Once this equation is satisfied, we have

$$\begin{align} p_{X,Y}(x,y)&=\sum_zp_{X,Y,Z}(x,y,z)\\ &=\sum_zp_X(x)\cdot p_Y(y)\cdot p_Z(z)\\ &=p_X(x)\cdot p_Y(y)\sum_zp_Z(z)\\ &=p_X(x)\cdot p_Y(y) \end{align}$$

Similarly, we have $p_{Y,Z}(y,z)=p_Y(y)\cdot p_Z(z)$ and $p_{Z,X}(z,x)=p_Z(z)\cdot p_X(x)$.

Why is there such a difference in the definitions of indepedence? Is there an intuitive explanation for this?

2 Answers 2

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Looking at events and random variables together we have:

$A_1,\dots, A_n$ are independent events if and only if $1_{A_1},\dots,1_{A_n}$ are independent random variables.

That's quite fair isn't it? You could say that events and random variables are treated the same way when it comes to independence.

But this induces the following definition for independence of events $A_1,\dots,A_n$:$$\Pr(E_1\cap\cdots\cap E_n)=\Pr(E_1)\times\cdots\times\Pr(E_n)\text{ where }E_i\in\{A_i,A_i^c\}\text{ for }i=1,\dots,n$$

So actually it is a cluster of $2^n$ identities.

It is equivalent to the definition you mention in your question and is a stronger condition than the (insufficient) condition $\Pr(A_1\cap\cdots\cap A_n)=\Pr(A_1)\times\cdots\times\Pr(A_n)$.

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    Your answer provides a good way to relate independence of events and random variables. But your definition has a slightly different form. It has $2^n$ identities, while that I mentioned has essentially $\sum_{k=2}^n\pmatrix{n\\k}$ identities, because intersections and products require at least two operands. – 2017-02-21
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    There are indeed redundancies in the $2^n$ equations. If e.g. $n=2$ then one of the $4$ is enough and enables you to prove the other $3$. Essential is that the equalities mentioned by you are true **if and only if** the equalities mentioned by me are true. – 2017-02-21
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We have the definition of independence of events $A_1,A_2,\dots,A_n$: $$ \mathbf P\left(\bigcap_{i\in S}A_i\right)=\prod_{i\in S}\mathbf P\left(A_i\right), \forall S\subset\{1, 2, \dots, n\}. $$

The above definition of independence is actually the definition of mutually independent events.

There's something called as pairwise independence which is defined as (as the name suggests) that every pair of events in $A_1,A_2,\dots,A_n$ satisfy $$ \mathbf P(A_i\cap A_j)=\mathbf P(A_i)\cdot \mathbf P(A_j), ~\forall i \neq j. $$

In the question details you have just shown that pairwise independence does not imply mutual independence. But mutual independence implying pairwise independence is trivial.

So in some sense mutual independence is stronger than pairwise independence.

In problems that we see usually, the term independence is viewed as mutual independence.

If a problem demands just the use of pairwise independence then use this rather than the stronger version of the independence i.e. the mutual independence.

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    In the context of "independence" the OP is especially interested in the *difference* that appears between events and random variables. But your answer only deals with events. – 2017-02-20