Problem
Chapter 1 problem 1: fill in the details of the proof of Theorem 1.8.
Also prove the monotone decreasing case.
Theorem 1.8 is "Continuity of Probabilities". If $A_n \rightarrow A$ then
$$ P(A_n) \rightarrow P(A) $$
as $n \rightarrow \infty$
Proof
Suppose that $A_n$ is monotone increasing so that $A_1 \subset A_2 \subset ... $.
Let $A = lim_{n \to \infty} A_n = \bigcup_{i=1}^{\infty} A_i$
Define
- $B_1 = A_1$
- $B_2 = \{ \omega \in \Omega : \omega \in A_2, \omega \notin A_1 \} $
- $B_3 = \{ \omega \in \Omega : \omega \in A_3, \omega \notin A_2, \omega \notin A_1 \} $
- etc
It can be shown that:
- $B_1, B_2, ...$ are disjoint
- $A_n = \bigcup_{i=1}^{n} A_i = \bigcup_{i=1}^{n} B_i$ for each $n$
- $\bigcup_{i=1}^{\infty} B_i = \bigcup_{i=1}^{\infty} A_i.$
this is what we need to do for this exercise
From Axiom 3 of probability measures: if $A_1, A_2, ...$ are disjoint then $P(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} P(A_i)$
we can show:
$lim_{n \to \infty} P(A_n)$
= $ lim_{n \to \infty} \sum_{i=1}^{n} P(B_i)$
= $ \sum_{i=1}^{\infty} P(B_i)$
= $ P(\bigcup_{i=1}^{\infty} B_i) $
= $ P(A) $
Solution
1. Showing $B_1, B_2, ...$ are disjoint
Attempt 1
Disjoint means that there is null intersection.
$B_1 \cap B_2$
= $\{\omega \in \Omega: \omega \in B_1 \text{ and } \omega \in B_2\}$
= $\{\omega \in \Omega: \omega \in A_1 \text{ and } \omega \in \{ \omega' \in \Omega : \omega' \in A_2, \omega' \notin A_1 \} \}$
=$\emptyset$
The last step is clear since $\omega$ can't both be in $A_1$ and not be in $A_1$ at the same time.
This doesn't prove it for all cases though. Peeking at the solution, I got a hint.
Attempt 2
Hint: it suffices to prove $B_j \cap B_k = \emptyset, \forall j < k$ without loss of generality.
$B_k$ can be written as $A_k \cap B_{k-1}^{c}$
$B_k \cap B_{k-1}$
= $(A_k \cap B_{k-1}^{c}) \cap B_{k-1}$
= $A_k \cap (B_{k-1}^{c} \cap B_{k-1})$
= $A_k \cap \emptyset$
= $\emptyset$
Damnit, this doesn't really show it for all j and k, had to look at the solution again:
Attempt 3: solution walkthrough
Note that I was mistaken, it's not that $B_k$ can be written as $A_k \cap B_{k-1}^{c}$
It's that $B_k$ can be written as $A_k \cap A_{k-1}^{c}$
Using this insight we can proceed:
$B_j \cap B_k$
= $(A_j \cap A_{j-1}^{c}) \cap (A_k \cap A_{k-1}^{c})$
If we just look at $A_j \cap A_{k-1}^c$
we know that $A_j \subseteq A_{k-1}$ because $j \leq k$ so $A_j \cap A_{k-1}^c = \emptyset$
and if you throw an $\emptyset$ into a series of intersections you always have $\emptyset$.
2. Showing $A_n = \bigcup_{i=1}^{n} A_i = \bigcup_{i=1}^{n} B_i$ for each $n$
Since $A$ is an increasing sequence, $\bigcup_{i=1}^{n} A_i = A_n$.
To show $\bigcup_{i=1}^{n} A_i = \bigcup_{i=1}^{n} B_i$ let's use induction.
Base case
For n = 1 $\bigcup_{i=1}^{n} A_i = A_1 = B_1 = \bigcup_{i=1}^{n} B_i$
Inductive step
Given
$\bigcup_{i=1}^{n} B_i = \bigcup_{i=1}^{n} A_i$
show
$\bigcup_{i=1}^{n+1} B_i = \bigcup_{i=1}^{n+1} A_i$
$\bigcup_{i=1}^{n+1} B_i$
=$B_{n+1} \cup (\bigcup_{i=1}^{n} B_i)$
=$B_{n+1} \cup (\bigcup_{i=1}^{n} A_i)$
=$(A_{n+1} \cap A_{n}^{c}) \cup (\bigcup_{i=1}^{n} A_i)$
=$(A_{n+1} \cap A_{n}^{c}) \cup A_n$
=$(A_{n+1} \cap A_{n}^{c}) \cup (A_n \cap A_{n+1})$
=$A_{n+1} \cap (A_{n}^{c} \cup A_n)$
=$A_{n+1} \cap \Omega$
=$A_{n+1}$
=$\bigcup_{i=1}^{n+1} A_i$
3. Showing $\bigcup_{i=1}^{\infty} B_i = \bigcup_{i=1}^{\infty} A_i.$
$\bigcup_{i=1}^{\infty} B_i$
= $\bigcup_{n=1}^{\infty} (\bigcup_{i=1}^{n} B_i)$
= $\bigcup_{n=1}^{\infty} (\bigcup_{i=1}^{n} A_i)$
= $\bigcup_{n=1}^{\infty} A_n$
= $\bigcup_{i=1}^{\infty} A_i$
Also prove the monotone decreasing case.
Need to show that If $A_n$ is a decreasing sequence converging to $A$ that $P(A_n) \rightarrow P(A)$
The key trick here is to know that "convergence of set sequences is preserved under complement", so we can instead prove that:
$ A_n^c \rightarrow A^c$ as n increases implies that $P(A_n^c) \rightarrow P(A^c)$
$lim_{n \to \infty} P(A_n^c)$
= $ lim_{n \to \infty} \sum_{i=1}^{n} P(B_i^c)$
= $ \sum_{i=1}^{\infty} P(B_i^c)$
= $ P(\bigcup_{i=1}^{\infty} B_i^c) $
= $ P(A^c) $
and thus
$ P(A_n) = 1 - P(A_n^c) \rightarrow 1 - P(A^c) \rightarrow P(A)$
Sources
- 36-705 hw1 problem 1