\operatorname{\mathsf{P}}\left({X=x}\right)=F_{X}(x)-\lim_{i\rightarrow\infty}% F_{X}(x-1/i)

\operatorname{\mathsf{P}}\left({X=x}\right)=F_{X}(x)-\lim_{i\rightarrow\infty}% F_{X}(x-1/i)

3.2 The Law of the Unconscious Statistician

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

3.1.1 Probability of a countable intersection of sets

We have $A_{n}=\{\omega\in\Omega:x-1/n<X(\omega)\leq x\}$ . These sets (or ‘events’) are nested

\displaystyle A_{1}\supseteq A_{2}\supseteq A_{3}\dots,

(C.1)

and $\cap_{n=1}^{\infty}A_{n}=\{\omega\in\Omega:X(\omega)=x\}$ . We wish to show that

\displaystyle\operatorname{\mathsf{P}}\left({\cap_{n=1}^{\infty}A_{n}}\right)=% \lim_{n\rightarrow\infty}\operatorname{\mathsf{P}}\left({A_{n}}\right).

(C.2)

We will show that C.2 holds for any sequence of nested sets C.1. However, it is impossible to prove this from our current set of axioms; as you will see, we will need another axiom.

We first transform to the complementary problem. Let $B_{n}=A_{n}^{c}$ , so

\displaystyle B_{1}\subseteq B_{2}\subseteq B_{3}\dots.

(C.3)

Equation C.2 then reads as $1-\operatorname{\mathsf{P}}\left({\cup_{n=1}^{\infty}B_{n}}\right)=\lim_{n% \rightarrow\infty}1-\operatorname{\mathsf{P}}\left({B_{n}}\right)$ , so proving C.2 is equivalent to showing that for nested sets as in C.3,

\displaystyle\operatorname{\mathsf{P}}\left({\cup_{n=1}^{\infty}B_{n}}\right)=% \lim_{n\rightarrow\infty}\operatorname{\mathsf{P}}\left({B_{n}}\right).

(C.4)

Define $C_{1}=B_{1}$ and, for $n\geq 2$ $C_{n}=B_{n}\setminus B_{n-1}$ . We have $\cup_{n=1}^{\infty}B_{n}=\cup_{n=1}^{\infty}C_{n}$ but the sets $C_{n}$ are disjoint. Hence $\operatorname{\mathsf{P}}\left({B_{n}}\right)=\sum_{i=1}^{n}\operatorname{% \mathsf{P}}\left({C_{n}}\right)$ , by the partition law. The usual abbreviation for $\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\operatorname{\mathsf{P}}\left({C_{n}}\right)$ is $\sum_{i=1}^{\infty}\operatorname{\mathsf{P}}\left({C_{n}}\right)$ . So showing C.4 subject to C.3 is equivalent to showing that for a countable sequence of disjoint sets, $C_{1},C_{2},\dots$ ,

\displaystyle\operatorname{\mathsf{P}}\left({\cup_{n=1}^{\infty}C_{n}}\right)=% \sum_{i=1}^{\infty}\operatorname{\mathsf{P}}\left({C_{i}}\right).

(C.5)

This is like the partition law, but applied to a countably infinite union of disjoint sets. Since the partition law only applies to a finite union of disjoint sets we cannot use it to prove C.5. We must take C.5 as an axiom; it is known as countable additivity.

The equivalence of the statements for $A_{n},B_{n}$ and $C_{n}$ means that the axiom of countable additivity implies C.4 and C.2 for sequences of events satisfying C.3 and C.1, respectively.