Problem
For a fixed $B$ with $P(B) > 0$ , show that $P(\cdot | B)$ is a probability.
Solution
We need to show that $P(\cdot | B)$ satisfies the axioms of probability
- $P(A) \geq 0$ for every $A \subseteq \Omega$
- $P(\Omega) = 1$
- If $P_1, P_2, ... P_n$ are disjoint, then $P(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} P(A_i) $
1. $P(\cdot | B) \geq 0$
For a given $A$, $P(A | B) = \dfrac{P(A \cap B)}{P(B)}$.
We know that $P(B) > 0$ and that $P(A) \geq 0$
Consider $P(A \cap B)$. Either $A \cap B = \emptyset$ in which case $P(A \cap B) = 0$ or $A \cap B \in \Omega $, in which case $P(A \cap B) \geq 0$.
Therefore $P(A \cap B) \geq 0$ and $\dfrac{P(A \cap B)}{P(B)} \geq 0$
2. $P(\Omega | B) = 1$
$P(\Omega | B) = \dfrac{P(\Omega \cap B)}{P(B)} = \dfrac{P(B)}{P(B)} = 1$
3. $P(\bigcup_{i=1}^{\infty} A_i | B) = \sum_{i=1}^{\infty} P(A_i | B) $
$P(\bigcup_{i=1}^{\infty} A_i | B)$
=$\frac{P((\bigcup_{i=1}^{\infty} A_i) \cap B)}{P(B)}$
So this boils down to showing that, for disjoint $A_i$, $[\bigcup_{i=1}^{\infty} A_i] \cap B = \bigcup_{i=1}^{\infty} A_i \cap B$
because given that
$\frac{P((\bigcup_{i=1}^{\infty} A_i) \cap B)}{P(B)}$
=$\frac{P(\bigcup_{i=1}^{\infty} A_i \cap B)}{P(B)}$
=$\frac{\sum_{i=1}^{\infty} P(A_i \cap B))}{P(B)}$
=$\sum_{i=1}^{\infty} \frac{P(A_i \cap B)}{P(B)}$
=$\sum_{i=1}^{\infty} P(A_i | B)$
Note that there is an assumption in there that if $A_is$ are disjoint, then $A_is \cap B$ are disjoint, and this makes sense because $A_i \cap A_k = \emptyset$ then $(A_i \cap B) \cap (A_k \cap B) = A_i \cap A_k \cap B \cap B = \emptyset \cap B = \emptyset$
Ok, we still need to show that, for disjoint $A_i$, $[\bigcup_{i=1}^{\infty} A_i] \cap B = \bigcup_{i=1}^{\infty} A_i \cap B$. To do this, we will show that each contains the other.
Step 1
$[\bigcup_{i=1}^{\infty} A_i] \cap B \subseteq \bigcup_{i=1}^{\infty} A_i \cap B$
Suppose an outcome $w \in [\bigcup_{i=1}^{\infty} A_i] \cap B$. That means there exists an $i$ so that $w \in A_i$ and $w \in B$. For that $i$, $w \in A_i \cap B$ and must also be in $\bigcup_{i=1}^{\infty} A_i \cap B$.
Step 2
$ \bigcup_{i=1}^{\infty} A_i \cap B \subseteq [\bigcup_{i=1}^{\infty} A_i] \cap B $
Suppose $w \in \bigcup_{i=1}^{\infty} A_i \cap B$. That means there exists an $i$ such that $w \in A_i \cap B$, and therefore $w \in A_i$ and $w \in B$. If $w \in A_i$, then $w \in \bigcup_{i=1}^{\infty} A_i$, so $w \in [\bigcup_{i=1}^{\infty} A_i] \cap B$.