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I see mixed messages. heropup commented $(C \mid B) \mid (A \mid B)$, but Michael Hardy chided that "$\color{Red}{\text{There's no such thing as A∣B.}}$ When one writes Pr(A∣B), one is NOT writing about the probability of something that's called A∣B". Tom Loredo answered

Now suppose our information about the problem tells us that $A$ and $B$ are conditionally independent given $C$. A conventional notation for this is: $$ A \perp\!\!\!\perp B \,|\, C, $$ which means (among other implications), $ P(A|B,C) = P(A|C).$

But why is there $\color{Red}{\text{"no such thing as A∣B"}}$? A|B makes perfect sense to me. $A \mid B$ would simply mean the event A, given that and after the event B happened and anteceded A. A|B would meaningfully and gainfully differ from $A \cap B$, because $A \cap B$ reveals nothing about the echelon/order/tier of A and B! But A|B and B|A do!

I retort Bey's rebuttal. $(A|B)|(C|D)$ can be construed meaningfully! $(A|B)|(C|D)$ would simply mean A, given B, given C, given D. So D happened first. Then C happened after D. Then B happened after C, D. Finally A happened after B, C, D.

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    $\begingroup$ Flip a coin twice and record the results. The set of outcomes may be written $\Omega = \{HH, HT, TH, TT\}$ and the set of events is the power set. Now assign $A = \{HT, HH\}$ and $B = \{HH, HT, TH\}$. Which event is $A\mid B$? $\endgroup$ Dec 25, 2021 at 1:31
  • $\begingroup$ $\mu(A \cap B) / \mu(B)$ is not $\mu($anything $)$ $\endgroup$
    – David Lui
    Dec 25, 2021 at 1:42
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    $\begingroup$ Probability is concerned with finding the likelihood of events happening; if an event is known to have happened, then you use conditional probability, which, as expressed by Feller and others, is unfortunately written with $P(A\mid B).$ Some statisticians write $P_B(A)$ and emphasise that $P_B$ is just another probability measure and $A$ is still the event. $A \mid B$ isn't an event and talking about it as such is just plain nonesense. $\endgroup$
    – William M.
    Dec 25, 2021 at 2:57
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    $\begingroup$ Conditioning does not tell you ordering -- it is simply telling you the relevant sub-region of the sample space that should be given measure 1. $\endgroup$
    – Annika
    Dec 25, 2021 at 7:41
  • $\begingroup$ You could say that $A|B$ is the set $A$ considered in the probability space $B$ with probability measure $P_B$ normalized so that $P_B(B) = 1$. I don't know if anyone defines it this way, though. $\endgroup$ Jan 1, 2022 at 3:13

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There's nothing about timing in the notation $A\mid B.$

For example: an urn contain five red marbles and five blue marbles. Two marbles are to be chosen without replacement.

Let $A$ be the event that the first one chosen is red.

Let $B$ be the event that the second is red.

Then $\Pr(B\mid A) = \dfrac 4 9 = \Pr(A\mid B).$

One can choose the two marbles at the same time, and the "first" may then be, not the first one chosen, but the first one whose color is revealed. That does not alter the statement about the two conditional probabilities above.

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  • $\begingroup$ Thanks. You are correct that $A|B$ doesn't always stipulate timing, but you didn't address "order" in your answer? Your last paragraph features echelon/order/tier because you wrote "the first one whose color is revealed". So there is a first and second! $\endgroup$
    – user53259
    Dec 25, 2021 at 7:26
  • $\begingroup$ @PGTK if we extended this to three marbles and let C be the event "the third marble is red" then $A|B|C =A|C|B = A|C\cap B$ -- in either case, you are formulating a probability in lights of everything you know, regardless of whether I told you B or C first. $\endgroup$
    – Annika
    Dec 25, 2021 at 7:45
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The pipe cannot really be “chained” in a coherent way — the right hand side fixes the “universe” we are considering when thinking about events.

Another way to think about this is how would you represent the event $A|B|C$ using a Venm diagram? This notation is hopelessly ambiguous.

A more precise version of this is $ A|(B\cap C)$. However, what if we had $(A|B) | (C|D)$?

Note that $C|D$ is effectively telling us the same thing as $C \cap D$ in terms of events. The only difference is that the conditional expressions is telling us what serves as $\Omega$ in any probability calculation.

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  • $\begingroup$ 1. Why must we be able to represent $A|B|C$ using a Venn diagram"? We cannot picture higher dimensions, which mathematics still accept. 2. $(A|B) | (C|D)$ would simply mean A, given B, given C, given D. So D happened first. Then C happened after D. Then B happened after C, D. Finally A happened after B, C, D. 3. "C|D is effectively telling us the same thing as C∩D in terms of events." No it doesn't. $ C∩D$ reveals nothing about the echelon/order/tier, as I explained in my post. $\endgroup$
    – user53259
    Dec 25, 2021 at 7:29
  • $\begingroup$ @PGTK I suggested this to help show that $A|B$ is not an event. Conditioning restricts the sample space, it doesn't tell you about the order. I think you are misunderstanding Michael Hardy's answer -- he is saying that you can calculate conditional probabilities that differ from the order in which the events happen. $\endgroup$
    – Annika
    Dec 25, 2021 at 7:37
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I think there is a sense in which A|B can be thought of as an event, but nothing to do with timing. Rather, it relies on the construction of a new kind of algebra: https://en.wikipedia.org/wiki/Goodman%E2%80%93Nguyen%E2%80%93Van_Fraassen_algebra

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  • $\begingroup$ I have seen another sense of $A\mid B$ that I think may have been introduced by Bruno de Finetti. de Finetti sets a thought experiment, as follows: You sell someone a "ticket" (thought of as like a lottery ticket), which is a promise to pay the bearer of the ticket $\$1$ if the event $A$ is shown to happen, or the proposition $A$ is shown to be true. You may set the price of the ticket, but then the other party to the transaction may choose either to buy the ticket from you at that price, or to make you buy the same ticket from them at the same price. Then$\,\ldots\ldots\qquad$ $\endgroup$ Dec 25, 2021 at 20:30
  • $\begingroup$ $\ldots\ldots\,$a more complicated ticket is called $A\mid B,$ and is subject to the same rules about price-setting and who buys and who sells. If you buy $A\mid B,$ you get $\$1$ if $A$ and $B$ both occur, and $\$0$ of $B$ occurs and $A$ does not, and your ticket price is refunded if $B$ does not occur. For example, $A$ may be the event that the Yankees beat the Red Sox tomorrow, and $B$ may be the event that the game actually takes place as opposed to being rained out. The rules of this game do not require$\,\ldots\ldots\qquad$ $\endgroup$ Dec 25, 2021 at 20:33
  • $\begingroup$ The rules of this game do not require the person who sets prices to set them so that $\operatorname{price}(A\mid B)\times\operatorname{price}(B) = \operatorname{price}(A\cap B),$ but it is demonstrably bad strategy not to do so, as follows: If you set the price of $A\mid B$ higher than what that rule prescribes, then your opponent can make you buy that ticket, while your opponent buys the other two. In that case, you lose regardless of which of the three outcomes eventuates. And similarly if you set the price too low. $\qquad$ $\endgroup$ Dec 25, 2021 at 20:35

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