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  1. Is it correct to say that $P(A) = 1 \to P(A|B) = 1$?

Precisely, consider a probability space $(\Omega, \mathscr{F}, \mathbb{P})$. Let $A \in \mathscr{F}$ and let $B_1, B_2, ... \in \mathscr{F}$ s.t. $\bigcup_{i \in \mathbb{N}} B_i = \Omega$ and $B_q \cap B_j = \emptyset \ \forall q \ne j$. Suppose $P(A) = 1$. Does this mean that $\forall i \in \mathbb{N}, P(A|B_i) = 1$?

What I tried:

Suppose $\exists m \in \mathbb N$ s.t. $P(A | B_m) < 1$.

By Bayes' Theorem, we have

$$P(A) = 1 = \sum_{i \in \mathbb{N}} P(A | B_i)P(B_i)$$

$$\to 1 = \sum_{i \in \mathbb{N}} P(A | B_i)P(B_i)$$

$$= \sum_{i=1}^{m-1} P(A | B_i)P(B_i) + P(A | B_m)P(B_m) + \sum_{i=m+1}^{\infty} P(A | B_i)P(B_i)$$

$$= \sum_{i=1}^{m-1} P(B_i) + P(A | B_m)P(B_m) + \sum_{i=m+1}^{\infty} P(B_i)$$

$$= \sum_{i=1}^{\infty} P(B_i) - P(B_m) + P(A | B_m)P(B_m) \ (*)$$

$$= 1 - P(B_m) + P(A | B_m)P(B_m)$$

$$= 1 - P(B_m) + P(A | B_m)P(B_m)$$

$$\to 1 - 1 + P(B_m) = P(A | B_m)P(B_m)$$

$$\to P(B_m) = P(A | B_m)P(B_m)$$

$$\to 1 = P(A | B_m) ↯ \ QED?$$

$(*)$

$$1 = P(\Omega) = P(\bigcup_{i \in \mathbb{N}} B_i) = \sum_{i \in \mathbb{N}} P(B_i)$$

Edit: The final step is supposed to be $1 = P(A | B_m)$ or $P(B_m) = 0$. Hence, it is incorrect to say $P(A) = 1 \to P(A|B) = 1$. I guess it holds if $P(B) > 0$.

  1. $P(A) < 1 \to P(A|B) = P(A)$ ?

Let's say we are given the unconditional probability $P(A) = 1/2$. Does this mean that $\forall i \in \mathbb{N}, P(A|B_i) = 1/2$? That seems to be what is suggested here. Might be interpreting incorrectly.

Community
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BCLC
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1 Answers1

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Let $X$ be a random variable uniformly distributed in $[0,1]$ and let $A$ denote the event that $X>0$ and $B$ the event that $X=0$. Although $P(A)=1$, we have $P(A|B)=0$, contradicting 1.

Update: A limiting process that defines $B$ as the limit of sets with positive measure is defining $A_\epsilon=(\epsilon,1]$ and $B=[0,\epsilon]$. In this case, since $P(A_\epsilon|B_\epsilon)=0$ for all $\epsilon$, we have $$ \lim_{\epsilon\to0}P(A_\epsilon|B_\epsilon)=0 $$ and $A_\epsilon\to A$, $B_\epsilon\to B$. Also see the comment by @A.S.

DirkGently
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    "the concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible" Kolmogorov. If you take a limiting procedure to define $P(A|B)$ you'll still get $1$. – A.S. Oct 24 '15 at 03:40
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    @A.S., Dose he give any reason? We can define it as a limit and it is useful for continuous RVs, http://math.stackexchange.com/a/450004/88378. – DirkGently Oct 24 '15 at 03:51
  • I took it from https://en.wikipedia.org/wiki/Conditioning_(probability)#The_limiting_procedure. Basically you need some additional structure to define $P(A|B)$ if $P(B)=0$. A limiting procedure in your example would give $P(A|B)=1$. On the other hand, $P(A|X)=I([0,1])=I((0,1])$ (since we equate functions that differ on the set of measure zero) hence $P(A|X=0)=0$ and $P(A|X=0)=1$ are both satisfactory definitions. Frankly speaking, I don't fully get. $P(A|B)=0$ in your example certainly makes "conventional" sense, but the more I think about this, the more I agree with the quote. – A.S. Oct 24 '15 at 04:28
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    @DirkGently Basically, we can condition on a (continuous) family of events of zero measure (conditional expectation, for example). Then arbitrarily changing any slice of measure zero (as I've done above with $P(A|X=0)$) won't make a difference. Conditioning on a single event $P(A|X=0)$ runs into problems. One limiting procedure (described in wikipedia article where we approximate sets ${X=0}$ by sets of positive measure), leads to $P(A|X=0)=1$ while approximation of measure (described in a answer you linked) would lead to $P(A|X=0)=0$. The latter approximation is more intuitive to me though. – A.S. Oct 24 '15 at 04:56
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    @DirkGEntly Here are two more links: http://stats.stackexchange.com/a/136545 https://en.wikipedia.org/wiki/Borel%E2%80%93Kolmogorov_paradox I finally understand conditioning on sets of measure zero! – A.S. Oct 24 '15 at 05:22
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    Why P(A|B)=0 here? – Did Oct 24 '15 at 20:26
  • @Did If we take a sequence of probability measures on the unit interval $m_{\epsilon}(S)=\frac 1 \epsilon \mu([0,\epsilon]\cap S)$ as $\epsilon\to 0$, won't they converge to $\delta_0$ yielding $P(A|B)=0$? – A.S. Oct 24 '15 at 20:53
  • @A.S. If this is the construction the OP has in mind (there are others), they should make this precise. As is it now, this post does not answer the question. – Did Oct 24 '15 at 21:02
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    @A.S. Thanks for the very interesting links. – DirkGently Oct 25 '15 at 01:23
  • What I did: $x(x-1) = 0 \to x(x-1)/x = 0/x \to (x-1) = 0 \to x = 1$. Thanks DirkGently! – BCLC Nov 04 '15 at 01:47