When random variables $Y_1, Y_2, ... Y_n$ are independent, we say that

$$P\left(\bigcap_{i=1}^{n} (Y_i \in B_i)\right) = \prod_{i=1}^{n} P(Y_i \in B_i)\tag{F1}$$

or for any distinct indices $i_1, i_2, \dots, i_n$ and for all Borel sets $B_i$, $$P\left(\bigcap_{i=i_1}^{i_n} (Y_i \in B_i)\right) = \prod_{i=i_1}^{i_n} P(Y_i \in B_i).\tag{F2}$$

If random variables $Y_1, Y_2, ...$ are independent, from Rosenthal's book I guess the definition can be stated:

For any distinct indices $i_1, i_2, \dots, i_n$ and for all Borel sets $B_i$, $$P\left(\bigcap_{i=i_1}^{i_n} (Y_i \in B_i)\right) = \prod_{i=i_1}^{i_n} P(Y_i \in B_i)\tag{I3}$$?

From which, we can infer:

For any $n \in \mathbb{N}$ and for all Borel sets $B_i$, $$P\left(\bigcap_{i=1}^{n} (Y_i \in B_i)\right) = \prod_{i=1}^{n} P(Y_i \in B_i)\tag{I2}$$?

Can we say any of the following:

For all Borel sets $B_i$, $$P\left(\bigcap_{i=1}^{\infty} (Y_i \in B_i)\right) = \prod_{i=1}^{\infty} P(Y_i \in B_i)\tag{I1}$$

For any distinct indices $i_1, i_2, \dots,$ (eg even numbers) and for all Borel sets $B_i$, $$P\left(\bigcap_{i=i_1, i_2, \dots} (Y_i \in B_i)\right) = \prod_{i=i_1, i_2, \dots} P(Y_i \in B_i)\tag{I4}$$?

I have a feeling that the answer may be related to this question, which however has to do with events rather than random variables.

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    $\begingroup$ You answer the question I2 and I3 yourself (even before posing them as a question). $\endgroup$ – drhab Jul 15 '15 at 11:24
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    $\begingroup$ As far as I remember from lectures and as far as Wikipedia can confirm, I3 is correct. en.wikipedia.org/wiki/… $\endgroup$ – Bernhard Jul 15 '15 at 11:26
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    $\begingroup$ Why do not grab a book with a definition of independence in it and read the thing? $\endgroup$ – Did Jul 15 '15 at 22:52
  • $\begingroup$ @drhab Do I? Do I2 and I3 follow from F1 and F2 or something? Anyway, I found them in Rosenthal's book. $\endgroup$ – BCLC Jul 16 '15 at 19:50
  • $\begingroup$ @Did Thanks. How lazy (or distrusting or something, since I can't seem to remember my professors in elementary probability, advanced probability or any of my statistics classes discussing or emphasizing things like pairwise vs mutual independence, independence or infinite RVs, etc) of me. XD Rosenthal takes care of I2 and I3, which I kind of suspected to be true or the definition. My concern is mainly I1 and I4: how can they be deduced from I2 or I3, if they can? $\endgroup$ – BCLC Jul 17 '15 at 10:27

If $(A_n)_n$ is a decreasing sequence of events with $A_n\downarrow A=\bigcap_{n=1}^{\infty}A_n$ then $P(A_n)\downarrow P(A)$.

You can apply this on $A_n=\bigcap_{i=1}^n\{Y_i\in B_i\}$ to find:$$P(\bigcap_{i=1}^{\infty}\{Y_i\in B_i\})=\lim_{n\rightarrow\infty}P(\bigcap_{i=1}^n\{Y_i\in B_i\})=\lim_{n\rightarrow\infty}\prod_{i=1}^nP(\{Y_i\in B_i)=\prod_{i=1}^{\infty}P(\{Y_i\in B_i)$$

  • $\begingroup$ Thanks drhab. That takes care of I1, I2, I3. What about I4? $\endgroup$ – BCLC Jul 15 '15 at 11:51
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    $\begingroup$ If the indexset is countable then there is no essential difference between I1 and I4. Just define $Z_j:=Y_{i_j}$ and $C_j:=B_{i_j}$ and apply my answer on the $Z_j\in Cj$. $\endgroup$ – drhab Jul 15 '15 at 12:37
  • $\begingroup$ Do we replace index 1, 2, ... with $j=i_1, i_2, ...$? $\endgroup$ – BCLC Jul 16 '15 at 20:01
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    $\begingroup$ Applying my answer you find $P(\bigcap_{j=1}^{\infty}\{Z_{j}\in C_{j}\})=\prod_{j=1}^{\infty}P(\{Z_{j}\in C_{j})$. This for $Z_{j}=Y_{i_{j}}$ and $C_{j}=B_{i_{j}}$. These substitutions give you (I4). $\endgroup$ – drhab Jul 16 '15 at 20:15
  • $\begingroup$ Oh applying the answer IS replacing the index. Okay thanks! ^-^ $\endgroup$ – BCLC Jul 16 '15 at 20:18

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