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In a probability theory script, I read a definition of independence where I don't understand one detail (Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space)

"A family of events $(A_i)_{i \in I}$, $A_i \in \mathcal{F}$ is called independent if $\{A_i, \Omega\}$ [*] with $i\in I$ is independent." (def. 7.2.1c)

Just before there is definied: "Subsets $\mathcal{E}_i$, $i \in I$ of $\mathcal{F}$ are called independent if all finite combinations of them are independent." (which comes down to the usual formula $\mathbb{P}(A \cap B) = \mathbb{P}(A) \mathbb{P}(B)$)

Why do I have to introduce the $\Omega$ in [*] ?

(The source is this script: http://www.wias-berlin.de/people/koenig/www/WTSkript.pdf)

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Beware that the parenthesis which comes down to the usual formula $\mathbb{P}(A \cap B) = \mathbb{P}(A) \mathbb{P}(B)$ is wrong (this is pairwise independence, not independence). –  Did Jul 23 '12 at 8:45
    
Yes, I wrote it too short, actually I meant this definition also Stefan Hansen gave.."The sets (Ai)i∈I is said to be independent if that for every 1≤n≤|I| and choice of indices i1,…,in∈I we have that P(Ai1∩⋯∩Ain)=∏k=1nP(Aik)(∗)." –  Suedklee Jul 23 '12 at 13:00

1 Answer 1

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The usual definition of independence of sets is the following.

Definition: The sets $(A_i)_{i\in I}$ is said to be independent if that for every $1\leq n\leq |I|$ and choice of indices $i_1,\ldots,i_n \in I $ we have that $$ P(A_{i_1}\cap\cdots\cap A_{i_n})=\prod_{k=1}^n P(A_{i_k}) \qquad (*). $$

If one were to check the definition in $(*)$, this would involve computing products with $2,3,\ldots, |I|$ terms. Therefore it is sometimes more convenient to introduce an equivalent characterization.

The sets $(A_i)_{i\in I}$ are independent if and only if $$ P(B_1\cap\cdots \cap B_{N})=\prod_{k=1}^N P(B_k), \qquad (\text{here } N=|I|), $$ for every choice $B_1,\ldots,B_N$ where $B_i\in \{A_i,\Omega\}$.

Note that this is again equivalent to saying that the sets $(\{A_i,\Omega\})_{i\in I}$ are independent.

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While I agree that the two definitions are equivalent, I wonder about the comment about checking whether $N$ events are independent. In one case, there are $$\sum_{n=2}^N \binom{N}{n} = 2^N - 1 - N$$ cases to be checked, while in the other case, there are $2^N$ cases to be checked of which precisely $N+1$ are trivial since they involve exactly one of the $B_i$ being $A_i$, or none of the $B_i$ being $A_i$. –  Dilip Sarwate Jul 23 '12 at 19:05
    
@DilipSarwate: You are absolutely right. It is in no way more ''optimal'' in terms of how many cases there is to be checked. But sometimes the equivalent definition is more handy to work with (i've seen it be used in proofs). –  Stefan Hansen Jul 23 '12 at 19:29
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There is also another version that comes in handy at times, though it is slightly more work to check. In the $2^N$ equations you state, change "$B_i \in \{A_i, \Omega\}$" to "$B_i \in \{A_i, A_i^c\}$" –  Dilip Sarwate Jul 23 '12 at 19:38

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