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A family $\mathcal A=(A_i:i\in I)$ of events is called independent, if for all finite subsets $J$ of $I$, $$ \mathbb P\left(\bigcap_{i\in J}A_i\right)=\prod_{i\in J}\mathbb P(A_i). $$

I'm wondering why we need $J$ to be a finite subset? Infinite products are defined after all, and we also have the following: $$ \mathbb P\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty \mathbb P(A_i), $$ where $A_1,A_2,\dots$ are disjoint events in $\mathcal F$. So I don't see why the definition requires $J$ to be finite; where would it go wrong otherwise?

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    $\begingroup$ Good question (+1). I do not know the exact reason, but then, we need to define infinite multiplications. We are putting more requirements to the probability and the events. Also there could be uncountable events/sets gets multiplied $\endgroup$
    – Jay Zha
    Apr 16, 2017 at 18:29
  • $\begingroup$ @YujieZha Thank you. In the end, it is of course a definition, though I can't help but wonder what extra requirements we would get, if we were to define it with infinite sets. $\endgroup$
    – Sha Vuklia
    Apr 16, 2017 at 18:35
  • $\begingroup$ Where did you get the definition? Maybe it's an elementary probability textbook where countably infinite families are not dealt with. $\endgroup$
    – BCLC
    Apr 15, 2018 at 20:04

1 Answer 1

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In short, the two definitions are equivalent, and the one with finite sets is more convenient.

Claim. A family $\mathcal A=(A_i:i\in I)$ of events is independent, if and only if for all finite or countably infinite subsets $J$ of $I$, $$ \mathbb P\left(\bigcap_{i\in J}A_i\right)=\prod_{i\in J}\mathbb P(A_i). $$

Proof. Clearly this implies independence, since finite sets are 'finite or countably infinite'.

Conversely, for any countably infinite set $J$, let $\{J_n\}_{n=1}^{\infty}$ be an increasing sequence of finite sets such that $J=\bigcup_{n=1}^{\infty}J_n$. Then by continuity of measure, $$ \mathbb P\left(\bigcap_{i\in J}A_i\right)=\lim_{n\to\infty}\mathbb P\left(\bigcap_{i\in J_n}A_i\right)=\lim_{n\to\infty}\prod_{i\in J_n}\mathbb P(A_i)=\prod_{i\in J}\mathbb P(A_i). $$

So one may define independence using either finite or countably infinite sets. In practice, it is more convenient to use the definition with finite sets to verify that a given family of random variables is independent. Thus people stick with the first definition, since using countably infinite sets doesn't gain anything more.

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  • $\begingroup$ Shouldn't your sequence $\{J_n\}_{n=1}^\infty$ be decreasing, so that we have $J=\cap_{n=1}^\infty J_n$? $\endgroup$
    – Sha Vuklia
    Apr 16, 2017 at 22:16
  • $\begingroup$ And how do you justify: $\lim_{n\to\infty}\prod_{i\in J_n}\mathbb P(A_i)=\prod_{i\in J}\mathbb P(A_i)$? I can see that in the first equality, you're using continuity of measure, in the second equality you're using independence, but what are you using in the last equality? Is it a definition, or is it trivial? If it is trivial, would you mind to elaborating a little bit? $\endgroup$
    – Sha Vuklia
    Apr 16, 2017 at 22:25
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    $\begingroup$ Response to first comment: no. Think of $I=J=\mathbb N$ then $J_n$ could be $\{1,\ldots,n\}$. The sets $\{J_n\}$ form an exhaustion of $J$. The term "increasing" means, in this context, that $J_n\subset J_m$ for $n<m$. $$$$ Response to second comment: This is the definition of the infinite product. By analogy, thing of what the definition of an infinite sum is: the limit of the partial sums. Again think of $J_n=\{1,\ldots,n\}$ here, in which case the product over $J_n$ is a "partial product" of the infinite product. $\endgroup$
    – pre-kidney
    Apr 16, 2017 at 23:39
  • $\begingroup$ Thanks for the reply! I get it now. $\endgroup$
    – Sha Vuklia
    Apr 16, 2017 at 23:56

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