3

This is something I never really got in either Elementary Probability Theory or Advanced Probability Theory because my professors mainly discussed independence between 2 objects. Please tell me if my understanding is right:

  1. Events $A_1, A_2,\dots,A_n$ are independent: For any distinct indices $i_1, i_2, \dots, i_n$

$$P(A_{i_1}, A_{i_2}, \dots, A_{i_n}) = \prod_{i = i_1}^{i_n} P(A_i).\tag{A}$$

This is not the same as $P(A_1, \dots, A_n) = \prod_{i = 1}^{n} P(A_i)$.

  1. Sigma-algebras or Pi-systems $\mathscr{A}_1, \mathscr{A}_2, \dots, \mathscr{A}_n$ are independent: For any distinct indices $i_1, i_2, \dots, i_n$

$$P(A_{i_1}, A_{i_2}, \dots, A_{i_n}) = \prod_{i = i_1}^{i_n} P(A_i)\mbox{ where } A_{i_1} \in \mathscr{A}_{i_1}, A_{i_2} \in \mathscr{A}_{i_2}, \dots, A_{i_n} \in \mathscr{A}_{i_n}.\tag{B}$$

I'm guessing this is not the same as $P(A_1, \dots, A_n) = \prod_{i = 1}^{n} P(A_i)$ for similar reasons (where $A_{1} \in \mathscr{A}_{1}, A_{2} \in \mathscr{A}_{2}, \dots, A_{n} \in \mathscr{A}_{n}$).

  1. However, I saw that in Stochastic Calculus when random variables $Y_1, Y_2, ... Y_n$ are independent, we CAN say that for all Borel sets $B_i$,

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

Apparently, that is equivalent to saying for any distinct indices $i_1, i_2, \dots, i_n$ and for all Borel sets $B_i$, $$P\left(\bigcap_{i=i_1}^{i_n} (Y_i \in B_i)\right) = \prod_{i=i_1}^{i_n} P(Y_i \in B_i)\tag{$C_1$}.$$

I was surprised because I thought $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ does not establish $k$-wise independence, but apparently it does.

Is there an analogue of $C_1$ for A or B?

Note: I acknowledge the title may not be very good. Please suggest a better title if needed.

BCLC
  • 13,459
  • Not sure I understand your whole question, but note that if $P(A \cap B) = P(A)P(B)$ then $P(A | B) = P(A)$ and $P(B | A) = P(B)$. Hence it "doesn't matter" whether $B$ is true, the probability of $A$ is the same either way (and vice versa). This is the reason we call them independent events. – Alex G. Jul 12 '15 at 14:23
  • 2
    I think (provided I understood correctly the question) the difference between the first item and the last two (algebras and random variables) is that by taking some of the $A_i$'s (resp. $B_i$'s) to be the whole space $\Omega$, you can "remove" some indices (intersection with $\Omega$/multiplication by $1$ on the LHS and RHS). I.e., you can "simulate" taking only the $i_1,\dots,i_n$ by setting the sets on the other indices to be $\Omega$, thus effectively removing them. – Clement C. Jul 12 '15 at 14:27
  • What @ClementC. explained. (Unrelated: I am nearly sure this OP already asked this question and was given this answer.) – Did Jul 12 '15 at 15:04
  • 1
    Related: http://math.stackexchange.com/questions/956869/mutual-independence-definition-clarificaiton –  Jul 12 '15 at 16:14
  • @Did I think this is not a duplicate because my previous question was asking about C and C1 being equivalent while the one I am asking now is the one I later added in the comments section. – BCLC Jul 12 '15 at 20:09
  • Good luck with arguing this is not a dup (compare the answer you receive here to the answers you received there, anything familiar?). Anyway, if you brought 1% of the attention you bring to the differences between your questions, to the answers you receive, this would not happen. – Did Jul 12 '15 at 20:19

1 Answers1

1

As you said independece of events requires more than $P(A_1, ..., A_n) = \prod_{i = 1}^{n} P(A_i)$ (as one can see in http://www.engr.mun.ca/~ggeorge/MathGaz04.pdf). We just need to see why the condition

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

Is equivalent to saying for any distinct indices $i_1,i_2,\ldots,i_k$ and for all Borel sets ${B}_k$,

$$ P(\bigcap_{i=i_1}^{k} (Y_i \in B_i)) = \Pi_{i=i_1}^{k} P(Y_i \in B_i)\qquad \qquad (B)$$

1) $(A) \Rightarrow (B):$ fix $B_{i_1},\ldots ,B_{i_k} $ and for $j \not in \{i_1,i_2,\ldots,i_{k}\}$ take $B_j = \Omega$.

Note that

$$P(\bigcap_{p=1}^{k} (Y_{i_p} \in B_{i_p}))=P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i) = \Pi_{p=1}^{k} P(Y_{i_p} \in B_{i_p})$$

2) $(B) \Rightarrow (A):$ just take $i_1= 1, \ldots 1_n = n$.

Conrado Costa
  • 6,071
  • 1
    What is the difference intuitively? $Y_i in B_i$ is simply an event so what is so special about it? How come we can say probabilities of intersection is product of probabilities (intuitively). Also, I think you mean $B_j = \mathbb{R}$ ? – BCLC Jul 16 '15 at 19:54
  • 2
    The difference is whether you are dealing with fixed events or a family of events associated with a random variable. Yes, $\Omega - \Bbb{R}$. – Conrado Costa Jul 16 '15 at 20:01
  • 1
    Conrad Costa, ah, do you mean that $Y_i \in B_i$ is not fixed because it depends on the random variable? – BCLC Jul 16 '15 at 20:03
  • 2
    No, because it depends on $B_i$ – Conrado Costa Jul 16 '15 at 20:06
  • 1
    Conrad Costa, oh sorry lol that's what I meant to say. Well that does explain things intuitively. Thanks ^-^ – BCLC Jul 16 '15 at 20:15