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I am looking at the expansion of the chain rule for probability. $$ P\left(\bigcap_{k=1}^nA_k\right)=\prod_{k=1}^nP\left(A_k\middle|\bigcap_{j=1}^{k-1}A_j \right) $$ if n=2, then the terms will be: $$ P\left(A_1,A_2\right)=P\left(A_1\middle|\bigcap_{j=1}^{0}A_j\right)P\left(A_2\middle|\bigcap_{j=1}^{1}A_j\right)=P\left(A_1\right)P\left(A_2\middle|A_1\right) $$ What is the definition of the intersection where top limit is smaller than bottom limit? $$ \bigcap_{j=1}^{0}A_j=\hspace{1mm}? $$ An empty sum = 0 while an empty product = 1. Is this an empty intersection?

Thanks for the help!

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  • $\begingroup$ Yes, it is the empty set, and by convention, $P(A | \emptyset) = P(A) $ $\endgroup$ – Tryss Jul 19 '15 at 21:17
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Let's unfold the definition. $x\in\bigcap_{j=1}^0 A_j$ iff $\forall 1\leq j<0:x\in A_j$ iff $\forall j:1\leq j<0\Rightarrow x\in A_j$. Now $1\leq j<0$ is never true, so $1\leq j<0\Rightarrow x\in A_j$ is always true and thus $\forall 1\leq j<0:x\in A_j$ and $x\in\bigcap_{j=1}^0 A_j$, no matter what $x$ is. This means that empty intersection is collection of all objects under consideration (in this case, whole space of events).

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  • $\begingroup$ I never said that $x\in A_j$ is true. I've said that the implication $1\leq j<0\Rightarrow x\in A_j$ is true. It's true no matter if $x\in A_j$ or not, because the RHS of implication is false. $\endgroup$ – Wojowu Jul 19 '15 at 21:26
  • $\begingroup$ So is the answer the full space S which would be $A_1\bigcup A_2$ in my example? Which would mean $P(A_1|S)=\frac{P(A_1,S)}{P(S)}=\frac{P(A_1)}{1}=P(A_1)$. $\endgroup$ – Yuri Brovman Jul 19 '15 at 22:25

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