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When we speak of "joint probabilities" in a general sense, do they take the order in which the events occur into account? Or alternatively, is it -by defnition- true that

$P(A\cap B) = P(B\cap A)$

? My statistics book is not explicit on this and I can't deduce it from the examples given.

A reference that's explicit on this would be really appreciated.

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1 Answer 1

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Yes you have from ensemble theory that $A \cap B = B \cap A$ (for instance from the intersection definition there Intersection), which is intuitive : $A \cap B$ is the set of elements that are both in A and B. Therefore $\mathbb{P} (B \cap A)= \mathbb{P}(A \cap B) $. And indeed the joint distribution is just for a pair of values to happen, no matter how they are related.

If you are interested in the order of elements, you are probably more interested in conditional probability (there should be a chapter on that in your book), that is

$$ \mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} $$

which is the probability that $A$ occurs given that B has occured

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  • $\begingroup$ I'm familiar with that rule but there's something I don't get with regard to this situation: if my sample space S = {AAABBB} and I draw twice from it without replacement, then the rule states that $P(A)$ = 0.5 and $P(B|A)$ = 0.6 so $P(A\cap B)$ = 0.3. However, this doesn't take into account the 0.3 probability of drawing A and B in the reverse order so $P(A \cap B)$ really is 0.6 and not 0.3. What am I missing here? $\endgroup$
    – RubenGeert
    Nov 17, 2014 at 8:34
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    $\begingroup$ In the probability nmodel you are using, there is NO concept of "time"! Everything 'happens' simultaneously, the order you see in formulas like $P(A|B)$ is not a time order. $\endgroup$ Nov 17, 2014 at 11:09
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    $\begingroup$ You are not stating the problem correctly. In the example you give you have to separate the two draws : you have one random variable, let us name it $X_1$, that is the first drax, and and another one for the second draw, be $X_2$. Then, you want to calculate $\mathbb{P}(X_2=B|X_1=A)$, which is equal to $\frac{\mathbb{P}(X_2 = B \cap X_1 = A)}{\mathbb{P}(X_1=A)} = \frac{1/2*3/5}{1/2} = \frac{3}{5}$ $\endgroup$ Nov 17, 2014 at 17:23
  • $\begingroup$ @Aerandal: you're totally right. Indeed, adding the subscripts to indicate two separate the different probability experiments really solves the riddle. $\endgroup$
    – RubenGeert
    Nov 17, 2014 at 17:59
  • $\begingroup$ @kjetilbhalvorsen: I see how that works for events based on a single experiment: if I draw 1 person from a database, what's the probability she owns a car -given the person is female. But in the case of sampling 2 out of 6 colored balls, the conditional probability for the color of the second ball necessarily does involve a time order, right? $\endgroup$
    – RubenGeert
    Nov 17, 2014 at 18:01

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