# Empty intersection in chain rule for probability [duplicate]

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!

• Yes, it is the empty set, and by convention, $P(A | \emptyset) = P(A)$ – Tryss Jul 19 '15 at 21:17

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).
• 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. – Wojowu Jul 19 '15 at 21:26
• So is the answer the full space S which would be $A_1\bigcup A_2$ in my example? Which would mean $P(A_1|S)=\frac{P(A_1,S)}{P(S)}=\frac{P(A_1)}{1}=P(A_1)$. – Yuri Brovman Jul 19 '15 at 22:25