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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:

enter image description here

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

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  • $\begingroup$ How does p. 17 in your link seem to suggest otherwise? $\endgroup$ – Stefan Hansen Oct 6 '14 at 7:15
  • $\begingroup$ @stefan hansen sorry unclear I meant it goes against me and supports you guys $\endgroup$ – BCLC Oct 6 '14 at 8:24
  • $\begingroup$ @StefanHansen added pictures. $\endgroup$ – BCLC Oct 7 '14 at 18:07
  • $\begingroup$ math.stackexchange.com/questions/924865/… $\endgroup$ – BCLC Jul 12 '15 at 13:48
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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.

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  • $\begingroup$ OMG SORRY. (C) is totally what I meant. Thanks Did. Anyway, what? Why does C' imply C? This seems to suggest otherwise. engr.mun.ca/~ggeorge/MathGaz04.pdf I mean, isn't the example what my prof was asserting? $\endgroup$ – BCLC Oct 3 '14 at 17:35
  • $\begingroup$ 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). $\endgroup$ – Did Oct 3 '14 at 20:05
  • $\begingroup$ 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... $\endgroup$ – BCLC Oct 3 '14 at 23:31
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    $\begingroup$ 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 $$P(Y_1\in B_1,Y_3\in B_3,Y_4\in B_4)=P(Y_1\in B_1,Y_2\in\mathbb R,Y_3\in B_3,Y_4\in B_4,Y_5\in\mathbb R).$$ $\endgroup$ – Did Oct 3 '14 at 23:48
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    $\begingroup$ READ MY COMMENT and THINK about it. $\endgroup$ – Did Oct 7 '14 at 18:22

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