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We are given two vectors $X=(X_1,X_2, . . . ,X_n)$ and $Y=(Y_1, Y_2, . . . , Y_n)$ with equal joint distributions. Do their marginal distributions $P_{X_i}$ and $ P_{Y_i}$ have to be equal?

I have no idea so far how to approach this problem. I'm sure there is a simple counterexample with a small $n$. I suppose I should look for dependent random variables.

Could you help me a bit?

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    $\begingroup$ Consider what "equal joint distributions" means. Then sum or integrate over all indices except $i$. $\endgroup$
    – Henry
    Mar 1, 2015 at 11:37
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    $\begingroup$ I see, the joint probability distribution of $n$, discrete random variables is $$P(X_1 = x_1, ..., X_n = x_n)=$$ $$ = P(X_1 = x_1)P(X_2 = x_2 | X_1 = x_1)...P(X_n = x_n|X_1=x_1,...,X_{n-1}=x_{n-1})$$ and if we sum this $\sum_i \sum_j ... \sum_k P(X_{1}=x_{1i},X_2 = x_{2j}...,X_{n}=x_{nk})=1$ but I don't see what should happen if we sum over all indices except $i$. Could you show me that? $\endgroup$
    – Sasha
    Mar 1, 2015 at 12:23

1 Answer 1

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If $$P(X_1 = z_1, \ldots, X_j = z_j,\ldots, X_n = z_n)=P(Y_1 = z_1, \ldots, Y_j = z_j,\ldots, Y_n = z_n)$$ for all $z_1,\ldots,z_j,\ldots,z_n$ is what is meant by "equal joint distributions" then you might say the marginal probabilities are $$\displaystyle P(X_i=z_i) = \sum_{z_1}\cdots \sum_{z_j, j\not = i} \cdots \sum_{z_n} P(X_1 = z_1, \ldots, X_i = z_i,\ldots, X_j = z_j,\ldots, X_n = z_n)$$ and $$\displaystyle P(Y_i=z_i) = \sum_{z_1}\cdots \sum_{z_j, j\not = i} \cdots \sum_{z_n} P(Y_1 = z_1, \ldots, Y_i = z_i, \ldots, Y_j = z_j, \ldots, Y_n = z_n)$$ and these are clearly equal to each other since they are sums of equal probabilities over the same indicies.

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  • $\begingroup$ Ok, I see that now. Thank you a lot for writing that down and explaining it to me! $\endgroup$
    – Sasha
    Mar 1, 2015 at 13:29

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