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Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that

$P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?

If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?), but I was under the impression that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.

So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.

p.17 here seems to suggest otherwise. idk

Help please?

Also this:

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or

enter image description here

So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?

Community
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BCLC
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1 Answers1

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I was [under][2] the [impression][3] that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_n} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_n} P(Y_j \in B_j)$.

You misread: independence of $n$ random variables $(Y_1,\ldots,Y_n)$ is equivalent to the following condition:

(C) For every distinct indices $i_1, i_2, ..., i_k$ and every $B_j$, $P(\bigcap\limits_{j=i_1}^{i_k} (Y_j \in B_j)) = \prod\limits_{j=i_1}^{i_k} P(Y_j \in B_j)$.

Indeed, choosing $k=n$ and $i_j=j$, (C) implies condition (C'):

(C') For every $B_j$, $P(\bigcap\limits_{i=1}^{n} (Y_i \in B_i)) = \prod\limits_{i=1}^{n} P(Y_i \in B_i)$.

In the other direction, if (C') holds, then, for every distinct indices $i_1, i_2, ..., i_k$ and every $B_j$, one can complete the collection of events by $(Y_s\in\mathbb R)$ for the $n-k$ missing indices $s$, then (C) follows.

Did
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  • OMG SORRY. (C) is totally what I meant. Thanks Did. Anyway, what? Why does C' imply C? This seems to suggest otherwise. http://www.engr.mun.ca/~ggeorge/MathGaz04.pdf I mean, isn't the example what my prof was asserting? – BCLC Oct 03 '14 at 17:35
  • As I said, (C) and (C') are equivalent. Note that (C) and (C') assert that some property holds for every Borel subsets B_i, not for some specific collection (B_i). – Did Oct 03 '14 at 20:05
  • My prof did say something about for every $B_i$ but how is that relevant? I honestly don't get how C follows from your splitting up of k and n-k... – BCLC Oct 03 '14 at 23:31
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    If $P(Y_1\in B_1,Y_2\in B_2,Y_3\in B_3,Y_4\in B_4,Y_5\in B_5)$ is what it should be for every $(B_1,B_2,B_3,B_4,B_5)$ then $P(Y_1\in B_1,Y_3\in B_3,Y_4\in B_4)$ is what it should be for every $(B_1,B_3,B_4)$ since $$P(Y_1\in B_1,Y_3\in B_3,Y_4\in B_4)=P(Y_1\in B_1,Y_2\in\mathbb R,Y_3\in B_3,Y_4\in B_4,Y_5\in\mathbb R).$$ – Did Oct 03 '14 at 23:48
  • Sorta getting it. How is that different from the link earlier though? I got it from Wiki: http://en.wikipedia.org/wiki/Independence_(probability_theory)#cite_ref-3 – BCLC Oct 04 '14 at 08:05
  • Remember that question of mine (now deleted but temporarily deleted) from before? It went: In [David Williams' Probability with Martingales][1]

    Given a probability space. Let $\mathfrak{I}_1, \mathfrak{I}_2, \mathfrak{I}_3$ be $\pi$-systems on $\Omega$ such that for k = 1, 2, 3,

    $\mathfrak{I}_k \subseteq \mathfrak{F}$ and $\Omega \in \mathfrak{I}_k$

    Prove that if $P( \bigcap_{i=1}^{3} I_i) = \Pi_{i=1}^{3} P(I_i)$ whenever $I_k \in \mathfrak{I}_k$ for k = 1, 2, 3, then $\sigma(\mathfrak{I}_k)$ for k = 1, 2, 3 are independent. Explain why $\Omega \in \mathfrak{I}_k$ is needed.

     – BCLC Oct 04 '14 at 08:07
  • Yeah so I figured it out. The Omega thing is needed to establish pairwise independence and to ultimately conclude independence of the $\pi$-systems. Without it, we cannot say that the $\pi$-systems are independent right? – BCLC Oct 04 '14 at 08:09
  • Did, added pictures. – BCLC Oct 07 '14 at 18:07
  • Yeah, and all this is completely offtopic with respect to the question you started with, to which I fully replied. – Did Oct 07 '14 at 18:12
  • Did, I mean, how come I can't conclude independence of the Pi-Systems right away but I can for the random variables? – BCLC Oct 07 '14 at 18:15
  • Where do you see pi-systems in the definition of independence for random variables? I see only collections of sets $(Y\in B)$ with $B$ in the Borel sigma-algebra. These collections ARE SIGMA-ALGEBRAS. So, what are you talking about exactly? – Did Oct 07 '14 at 18:17
  • Did In the Williams exercise, the probabilities splitting up does not imply independence. But for random variables it does? – BCLC Oct 07 '14 at 18:21
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    READ MY COMMENT and THINK about it. – Did Oct 07 '14 at 18:22
  • You're brilliant AND funny, Did :)) Okay fiiiiiine – BCLC Oct 07 '14 at 18:28