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P(ABCD)=P(A|BCD)P(B|CD)P(C|D)P(D)

I really do not get this to be true, the reason is that the conditional statement say: Given this has happened already (If I am correct). So it tells something about the TIME when certain events occur in relation to other events.

Example: P(A|B) = probability for A to occur given B already has occurred or (B occure BEFORE A).

P(ABCD) = The same as the intersection of all the events happening, but tells nothing about certain events happening before others.

Any comments about this that may clarify, thank you for reading?

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  • $\begingroup$ Conditioning has nothing to do with time. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 4, 2017 at 6:46
  • $\begingroup$ It's more one trial, no time sequence, but there is information that $B$ has ocurrred. The question is, given that information, what is the probability that $A$ also occurred. $\endgroup$
    – quasi
    Commented Oct 4, 2017 at 6:54
  • $\begingroup$ For example, if I roll a die, and tell you that the result is an even number, what is the (conditional) probability, based on that information, that the result is, $4$, say? There's no time sequence -- it's just one roll, but with (partial) information provided about the actual outcome. $\endgroup$
    – quasi
    Commented Oct 4, 2017 at 6:57
  • $\begingroup$ So basically conditioning says in P(A|B), A and B happen at once, and completely dismiss the time aspect of it, only what to be selected from a "set"? $\endgroup$
    – Rolf Eilert Johansen
    Commented Oct 4, 2017 at 7:26
  • $\begingroup$ I don't fully understand your last comment. But as I said, it's only one trial. For that trial either $A$ occurred or not, and either $B$ occurred or not. Each of those $4$ events has an absolute probability. The conditional probability, $P(A|B)$, is the probability that $A$ occurred, assuming knowledge that $B$ has occurred. That effectively cuts down the sample space. $\endgroup$
    – quasi
    Commented Oct 4, 2017 at 7:41

2 Answers 2

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So for the equation $$P(ABCD) = P(A|BCD)P(B|CD)P(C|D)P(D)$$ the RHS, reading from right-to-left, is

The probability that $D$ occurred,

times the probability that $C$ occurred, assuming knowledge (i.e., it's given that) $D\;$occurred,

times the probability that $B$ occurred, assuming knowledge that $C,D\;$both occurred,

times the probability that $A$ occurred, assuming knowledge that $B,C,D\;$all occurred.

But there is only one trial, no time sequence.

Each conditional probability represents a probability which takes into account the information provided (i.e., assuming the known occurrence of the events to the right of the vertical bar).

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  • $\begingroup$ Thank you for the effort quasi $\endgroup$
    – Rolf Eilert Johansen
    Commented Oct 6, 2017 at 7:25
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Yes, you are right. For two events $A$ and $B$: $$P(A\cap B)=P(A)\cdot P(B|A) \ \ (A, \ then \ B)$$ $$P(B\cap A)=P(B)\cdot P(A|B) \ \ (B, \ then \ A)$$ However, in the given formula, the information about the sequence of events is also indicated. Start interpreting from the end: $$P(C|D) \ (D, \ then \ C)$$ $$P(B|C\cap D) \ (C\cap D, \ then \ B)=(D, \ then \ C, \ then \ B)$$ $$P(A|B\cap C\cap D) \ (B\cap C\cap D, \ then \ A)=\cdots=(D, \ then \ C, \ then \ B, \ then \ A).$$

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  • $\begingroup$ Thanks, as a start, do I from this assume the "events" A, B, C ... etc are ordered in time? $\endgroup$
    – Rolf Eilert Johansen
    Commented Oct 4, 2017 at 7:43
  • $\begingroup$ Really not. Each condition is just additional knowledge about the same trial. In that trial, each of the $4$ events $A,B,C,D$ either occurred or not. $\endgroup$
    – quasi
    Commented Oct 4, 2017 at 7:53
  • $\begingroup$ I think it dawns, it is not about time, but selecting the right amount of information from a set. However when you mention (A then B), should it be A and B occur at the same time ? $\endgroup$
    – Rolf Eilert Johansen
    Commented Oct 4, 2017 at 8:17
  • $\begingroup$ I didn't say (A then B). That was farruhota's phrasing. It's only one trial, so if they both occur, then of course it's at the same time. The concept of $P(A|B)$ is the probability (for that single trial) that $A$ occurred if you you have information that $B$ occurred, and no other information (other than the (absolute) probabilities for events in the original sample space). $\endgroup$
    – quasi
    Commented Oct 4, 2017 at 8:24
  • $\begingroup$ I see, sorry. So this potential answer got assumptions about time that is not properly defined? $\endgroup$
    – Rolf Eilert Johansen
    Commented Oct 4, 2017 at 8:26

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