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Consider two random variables $X_1, X_2$ on sample space $\Omega$.

Let $P_1, P_2$ are two probability distributions over $X_1, X_2$ for which only order differ. I mean $P_1 (X_1=x_1, X_2=x_2)$ and $P_2 (X_2=x_2, X_1=x_1)$ and all the values are probability values are same.

My doubt in first step is that how to order of random variables matter at all since intersection between sets is commutative? $P_1 (X_1=x_1, X_2=x_2) = P_2 (X_2=x_2, X_1=x_1) = P(\{\omega\mid X_1(\omega) = x_1\} \cap\{\omega \mid X_2(\omega) = x_2\})$

hanugm
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    The order doesn't matter. Did someone tell you it does? – Michael Hardy Nov 21 '19 at 06:28
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    There is only one probability measure $P$ on $\Omega$. So you should write $P (X_1=x_1, X_2=x_2)$ and leave out the indexes on $P$. Yes, both rvs induces probability measures $P_1,P_2$ on $\mathbb R$ but that is a different story. We have e.g. $P_1({x})=P (X_1=x)$. – drhab Nov 21 '19 at 06:41
  • It might be that my interpretation is wrong and you really meant two probability measures on $\Omega$. See for yourself. – drhab Nov 21 '19 at 06:56
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    The wording of the question seems a little odd. Can you reproduce the exact wording? Are the distributions supposed to be over ordered pairs? After all, $(1,2)$ is different from $(2,1).$ – David K Nov 21 '19 at 07:06
  • @DavidK from your comment i am in the way of getting answer – hanugm Nov 21 '19 at 08:36
  • @MichaelHardy The link i provided is saying so/// – hanugm Nov 21 '19 at 08:37
  • Possibly related: https://math.stackexchange.com/questions/614629/notation-used-for-multivariate-random-variables – David K Nov 21 '19 at 13:15
  • @hanugm : If you mean the Wikipedia article on exchangeability, where does that article say anything like that? The article in effect contrasts things like $\Pr(X=2\ &\ Y=20)$ with $\Pr(Y=2\ &\ X=20)$, but that's quite different from saying that $\Pr(X=2\ &\ Y=20)$ differs from $\Pr(Y=20\ &\ X=2). \qquad$ – Michael Hardy Nov 21 '19 at 17:18

3 Answers3

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It is worth noting explicitly that you are asking for an explanation of statements made on this Wikipedia page.

The discussion on that page mentions a possibly infinite number of random variables, but even if we consider only the case where there are only two random variables, is not merely about the random variables $X_1$ and $X_2$, it is about the sequence of random variables $(X_1,X_2).$

For $X_1$ and $X_2$ to be exchangeable, the condition is not that $P_1 (X_1=a, X_2=b) = P_2 (X_2=b, X_1=a).$ The condition is that $P(X_1=a, X_2=b) = P(X_2=a, X_1=b)$: one probability measure (not two), swapping the values of the variables.

For example, consider the random variables $X_1$ and $X_2$ with support on the set ${1,2}$ such that \begin{align} P(X_1=1,X_2=1) &= \frac13,\\ P(X_1=1,X_2=2) &= \frac13,\\ P(X_1=2,X_2=2) &= \frac13.\\ \end{align}

Then $$ P((X_1,X_2) = (1,2)) = \frac13 \neq 0 = P((X_2,X_1) = (1,2)) $$ and therefore the variables are not exchangeable.

David K
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In view of comments posted under the question it seems this may be only a case of misunderstanding something in the linked Wikipedia article. \begin{align} \text{These differ from each other: } & \begin{cases} \Pr(X_1=x_1\ \&\ X_2=x_2) \\ \Pr(X_1=x_2\ \&\ X_2=x_1) \end{cases} \\[12pt] \text{These do not differ from each other: } & \begin{cases} \Pr(X_1=x_1\ \&\ X_2=x_2) \\ \Pr(X_2=x_2\ \&\ X_1=x_1) \end{cases} \end{align}

Michael Hardy
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You are dealing with two different probability measures here. $P_1 (X_1=x_1, X_2=x_2)=P_2 (X_2=x_2, X_1=x_1)$ means $X_1$ and $X_2$ have the same joint distribution under the measures $P_1$ and $P_2$.

Kavi Rama Murthy
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