Suppose for proposition $A$ we have that
$$P(A) = 1$$
Then does it follow that for all $B$
$$P(A) = P(A \mid B) = P(A \mid \neg B) = 1?$$
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Not necessarily. Lets say you want to pick a number from $[0,1]$ uniformly. Let $A$ be the event that you pick an irrational number. Let $B$ be the event that you pick a rational number.
Then $P(A)=1$ but $P(A|B)=0$.
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But why is $P(A)=1$ in the first place? – Demosthene May 05 '15 at 19:28
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Because the irrationals in $[0,1]$ have Lebesgue measure $1$. – Frank May 05 '15 at 19:29
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Oh, I didn't know we were allowed to do that. +1 then :) – Demosthene May 05 '15 at 19:30
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Allowed to do what? – May 05 '15 at 19:30
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In fact, it may depend on the definition. If you use the formula, you will have to divide through zero, as P(B)=0. – wythagoras May 05 '15 at 19:32
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In case our base set is finite, does the proposition follow? – PP121 May 05 '15 at 19:33
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Yes. Because then P(A)=0 implies that A is impossible and P(A)=1 implies that not-A is impossible. – wythagoras May 05 '15 at 19:36
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No. Although it seems likely, we might consider the following probability: We throw darts at a square bord of unit size. What is the probability that the dart doesn't land on the diagonal, event A? Since the diagonal is infinitely small, it is 1. i.e. it is almost certain that it will happen. However, if we know that the dart has landed on the intsect of the diagonals, event B, we have that P(A|B)=0.
wythagoras
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TL;DR If $P(A) = 1$, then $P(A|B) = 1$ or $P(B) = 0$. So if $P(B) > 0$, $P(A|B) = 1$
Precisely, consider a sample space $\Omega$. Let $A, B_1, B_2, ...$ be events s.t. $P(A) = 1$ and the $B_i$'s partition $\Omega$ in the sense that:
- Their union is the sample space:
$$\bigcup_{i \in \mathbb{N}} B_i = \Omega$$
- They are pairwise disjoint:
$$B_q \cap B_j = \emptyset \ \forall q \ne j$$
Does this mean that
$$\forall i \in \mathbb{N}, P(A|B_i) = 1?$$
Suppose $\exists m \in \mathbb N$ s.t. $P(A | B_m) < 1$.
By Bayes' Theorem, we have
$$P(A) = 1 = \sum_{i \in \mathbb{N}} P(A | B_i)P(B_i)$$
$$\to 1 = \sum_{i \in \mathbb{N}} P(A | B_i)P(B_i)$$
$$= \sum_{i=1}^{m-1} P(A | B_i)P(B_i) + P(A | B_m)P(B_m) + \sum_{i=m+1}^{\infty} P(A | B_i)P(B_i)$$
$$= \sum_{i=1}^{m-1} P(B_i) + P(A | B_m)P(B_m) + \sum_{i=m+1}^{\infty} P(B_i)$$
$$= \sum_{i=1}^{\infty} P(B_i) - P(B_m) + P(A | B_m)P(B_m) \ (*)$$
$$= 1 - P(B_m) + P(A | B_m)P(B_m)$$
$$= 1 - P(B_m) + P(A | B_m)P(B_m)$$
$$\to 1 - 1 + P(B_m) = P(A | B_m)P(B_m)$$
$$\to P(B_m) = P(A | B_m)P(B_m)$$
$$\to 1 = P(A | B_m) \ \text{or} \ P(B_m) = 0$$
So there's no contradiction if $P(B_m) = 0$.
QED
$(*)$
$$1 = P(\Omega) = P(\bigcup_{i \in \mathbb{N}} B_i) = \sum_{i \in \mathbb{N}} P(B_i)$$
BCLC
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