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Law of total probability

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Statement

Let $(S,\Omega,\mathbb{P})$ be a probability space, let $(U_i)_{i\eq 1}^n\subseteq\Omega$ be a finite collection of $\mathbb{P}$ -measurable sets such that:

$S\subseteq\bigcup_{i\eq 1}^n U_i$ ^{[Note 1]}, and,
the $(U_i)_{i\eq 1}^n$ are pairwise disjoint, $\forall i,j\in\{1,\ldots,n\}\subseteq\mathbb{N}\big[(i\neq j)\implies(U_i\cap U_j\eq\emptyset)\big]$ ^{[Note 2]}

then we claim:

∀A∈Ω[P[A]=n∑i=1P[A∩Bi]]
- There are a few alternate forms:
  1. $\forall A\in\Omega\Big[\mathbb{P}[A]\eq\sum_{i\eq 1}^n\mathbb{P}[A\vert B_i]\mathbb{P}[B_i]\Big]$ , or even
  2. $\forall A\in\Omega\Big[\mathbb{P}[A]\eq\sum_{i\eq 1}^n\mathbb{P}[B_i\vert A]\mathbb{P}[A]\Big]$

Notes

Jump up ↑
As:
- $\forall A\in\Omega[A\subseteq S]$ , specifically each $U_i\subseteq S$ , and,
- as a union of subsets is a subset of the union (in this case: $\bigcup_{i\eq 1}^n U_i \subseteq \bigcup_{A\in\{S\} } A$ specifically)
we automatically have:
- $\bigcup_{i\eq 1}^n U_i\subseteq S$
Combine this with the requirement that
- $S\subseteq\bigcup_{i\eq 1}^n U_i$
We see:
- $S\eq\bigcup_{i\eq 1}^n U_i$
Thus:
- $\big[S\eq\bigcup_{i\eq 1}^n U_i\big]\iff\big[S\subseteq\bigcup_{i\eq 1}^n U_i\big]$
So it doesn't matter if we use \eq or \subseteq
Jump up ↑
The property of pairwise disjointedness is often stated using the contrapositive, that is:
- $\big[(i\neq j)\implies(U_i\cap U_j\eq\emptyset)\big]\iff\big[(U_i\cap U_j\neq\emptyset)\implies(i\eq j)\big]$
giving:
- $\forall i,j\in\{1,\ldots,n\}\subseteq\mathbb{N}\big[(U_i\cap U_j\neq\emptyset)\implies(i\eq j)\big]$ as opposed to
- $\forall i,j\in\{1,\ldots,n\}\subseteq\mathbb{N}\big[(i\neq j)\implies(U_i\cap U_j\eq\emptyset)\big]$ which we gave above

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