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I think I get that if $P(B) = 0$, we need not have $P(A\mid B) = 1$ from 1 2 3 4 5 6 7 8

But what if $B = \emptyset$?

'$\omega \in \emptyset \to \omega \in A$' is a vacuously true statement right?

It seems then that $P(A\mid\emptyset) = 1$

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BCLC
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    The conditional probability (of events) is only defined when $P(B) > 0$. – Zhanxiong Nov 22 '15 at 23:08
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    I see no way to assign a value to $P(A|\emptyset)$ in a principled way - possibly only $P(\emptyset|\emptyset)=0$ and $P(\Omega|\emptyset)=1$. @Solitary We can condition on non-empty sets of measure zero but it takes some extra care. $P(B|B)=1$ certainly makes sense even if $P(B)=0$. – A.S. Nov 23 '15 at 00:30
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    @A.S. Sorry but I would not bet on P(B|B) having any meaning at all when P(B)=0. – Did Nov 23 '15 at 01:39
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    @Did You need for once to step away from formalism and follow common sense intuition of what $P(B|B)$ is representing/should represent. And it's representing a probability that $B$ happens given that $B$ happened. For non-empty $B$ it's unequivocally $1$ and there is a limiting procedure yielding that. Other assignments of $P(B_i|B_i)$ cannot hold over the collection of zero-measure sets $B_i$ that adds up to a set of positive measure. – A.S. Nov 23 '15 at 02:07
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    @A.S. No. By the way, what would be the advantages of this extension of rigorously defined notions? – Did Nov 23 '15 at 02:14
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    Because defining $P(B|B)=1$ for all non-empty $B$ doesn't contradict any of the established definitions and makes the most intuitive sense. And those definitions in the first place were created to reflect/assist our intuition. By the same notion, in general it should hold that $E(X|A)\in[\inf_{\omega\in A} X,\sup_{\omega\in A} X]$. Stop your one-word trolling. – A.S. Nov 23 '15 at 02:19
  • @A.S. To sum up, the answer to the question in my previous comment is "None". (The rest of your interventions, that is, their general hysterical/insulting tone, including some vulgar and sexist remarks fortunately now deleted after having been flagged, does not concern me. Sorry.) – Did Nov 23 '15 at 10:08

1 Answers1

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The definition of the conditional expectation (applied to the special case of the conditional probability) says: $Y=P(A\mid \mathscr B)$ is a $\mathscr B$ measurable random variable defined with probability $1$ such that

$$\int_B\chi_A(\omega)\ dP(\omega)=\int_B Y(\omega) \ dP(\omega)$$

for all $B\in\mathscr B\subset \mathscr A$ with respect to a probability space $(\Omega,\mathscr A, P).$ So, for all $\omega\in\Omega$ (except a set of measure $0$)

$$Y(\omega)=P(A\mid \mathscr B)(\omega).$$

In our case $\mathscr B=\{\Omega,\emptyset\} $ and we are looking for a $\{\Omega,\emptyset\}$ measurable function $Y$ for which

$$P(A)=\int_{\Omega}\chi_A(\omega) dP(\omega)=\int_{\Omega}Y(\omega)\ dP(\omega).$$

$Y(\omega)$ cannot change as $\omega$ is changing since it has to be $\{\Omega,\emptyset\}$ measurable. So,

$$Y(\omega)=P(A\mid \mathscr B)(\omega)=P(A),\ \, \forall \omega \in \Omega.$$

Since, $\not \exists \ \omega \in \emptyset$ $Y(\omega)=P(A\mid \mathscr B)(\omega)$ is not defined over $\emptyset.$

That is,

$$P(A\mid \{\Omega,\emptyset\})=P(A)=P(A\mid \Omega)$$

is a constant.

$P(A\mid \emptyset)$ was not defined. But this was not because of $P(\emptyset)=0$. In the theory of conditional expectation events of zero probability may appear as conditions. This and that $P(A)=P(A\mid \Omega)$ may be tempting to try to say something about $P(A\mid \emptyset)$. It seems that $P(A\mid \emptyset)$ has to remain undefined.

Did
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zoli
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    @A.S.: $P(A\mid \emptyset)=P(A\mid{\Omega,\emptyset})(\omega)$ if $\omega\in\emptyset$. ($\omega\in\emptyset\Rightarrow P(A\mid \emptyset)=P(A\mid{\Omega,\emptyset})(\omega)$. Is this implication true or or false?) – zoli Nov 23 '15 at 01:16
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    @A.S.: We have "$P(A\mid{\Omega,\emptyset})(\omega)$ and $\omega\in\emptyset$" and not "$P(A\mid{\Omega,\emptyset})(\emptyset)$". The first statement is "vacuous" the second one is meaningless. – zoli Nov 23 '15 at 01:25
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    "$P(A\mid \emptyset)=a$ for any reals $0\le a\le 1$. " Dunno what that even means. // The conclusion should be that $P(A\mid {\Omega,\emptyset})(\omega)=P(A)$ for every $\omega\in\Omega$, period (note that $\omega\in\emptyset$ never happens, by definition of $\emptyset$). – Did Nov 23 '15 at 01:37
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    So you got $P(A|{\Omega,\emptyset})=P(A)$ for all $\omega$. How does it yield $P(A|\emptyset)$? – A.S. Nov 23 '15 at 01:59
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    It yields that within the framework of the theory of conditional expectation $P(A\mid\emptyset)$ can be endowed with any value. The value does not matter. Not like in the case of the traditional theory of conditional expectations where $P(A\mid\emptyset)$ is a forbidden quantity. – zoli Nov 23 '15 at 07:33
  • Starting at "So, we have to examine the following two integrals" I fail to understand the identities $$\int_{\Omega}\chi_A dP=\int_{\Omega}P(A\mid \Omega)\ dP=P(A)$$ and $$\int_{\emptyset}\chi_A dP=\int_{\emptyset}P(A\mid \emptyset)\ dP=\int_{\emptyset}a\ dP=0.$$ What the definition of conditional expectation that you partially recalled before says in this case is that $Y=P(A\mid \mathscr B)$ is a random variable such that $$\int_{\Omega}Y\ dP=\int_{\Omega}\chi_A dP=P(A)$$ and $$\int_{\emptyset}Y\ dP=\int_{\emptyset}\chi_A dP=0.$$ This, together with the condition ... – Did Nov 23 '15 at 10:01
  • ... that $Y$ is $\mathscr B$-measurable, forces $Y$ to be the constant random variable $Y=P(A)$. Thus, the last assertion of your answer should read $$P(A\mid\mathscr B)(\omega)=P(A)$$ for every $\omega$ in $\Omega$. None of this being related to the well-defined quantity $P(A\mid\Omega)$ or to the non-existent quantity $P(A\mid\emptyset)$. – Did Nov 23 '15 at 10:03
  • zoli, I like the idea that $$P(A\mid {\Omega,\emptyset})= \begin{cases} P(A),& \text{ if } \omega\in\Omega\ a,& \text{ if } \omega\in\emptyset \end{cases}$$ Was sort of thinking the same thing but couldn't seem to reconcile it w/ vacuous truths. Would that imply that since $P(A | ∅)$ can have more than one value, the set function $f(B) = P(A | B)$ is undefined for $\emptyset$ much like $g(x) = 1/x$ for 0? Or merely that $f(\emptyset)$ can take only any value in $[0,1]$ like here? – BCLC Nov 23 '15 at 22:18
  • So it can be a hundred? The same framework yields that $P(A|B)$ can be anything for any non-empty zero-measure set $B$ (including $P(B|B)$ being anything). Not very useful. – A.S. Nov 23 '15 at 22:19
  • zoli, I was thinking that ∅ is not measurable to some 'measure' $P(A|B)$, but it's not really a measure. Is that what you were trying to do w/ $P(A| {Ω,∅})$ ? I don't think that's a measure though, is it? – BCLC Nov 23 '15 at 22:23
  • Somebody should probably mention that $\omega\in\emptyset$ never happens. Kind of the definition of $\emptyset$, you know... – Did Nov 23 '15 at 23:02