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I have found an exercise with two variables X&Y and this joint distribution table:

X\Y -5 -4 -3 -2 -1 0 1 2 3 4 5 X
2 0 0 0 0 0 1/36 0 0 0 0 0 1/36
3 0 0 0 0 1/36 0 1/36 0 0 0 0 2/36
4 0 0 0 1/36 0 1/36 0 1/36 0 0 0 3/36
5 0 0 1/36 0 1/36 0 1/36 0 1/36 0 0 4/36
6 0 1/36 0 1/36 0 1/36 0 1/36 0 1/36 0 5/36
7 1/36 0 1/36 0 1/36 0 1/36 0 1/36 0 1/36 6/36
8 0 1/36 0 1/36 0 1/36 0 1/36 0 1/36 0 5/36
9 0 0 1/36 0 1/36 0 1/36 0 1/36 0 0 4/36
10 0 0 0 1/36 0 1/36 0 1/36 0 0 0 3/36
11 0 0 0 0 1/36 0 1/36 0 0 0 0 2/36
12 0 0 0 0 0 1/36 0 0 0 0 0 1/36
Y 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 1

And a text below the table said "X and Y are not equally distributed". I don't understand this... if X and Y have the same marginal distribution doesn't that mean they are equally distributed? Even the joint distribution is totally symmetric, so I was wondering what are the requirements for X and Y to be equally distributed? Is it mandatory that the variables have the same values too?

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    $\begingroup$ They have different distributions: for example $Y$ (which looks like one die roll minus another) has a mean of $0$ while $X$ (the sum of the same two dice) has a mean of $7$. It would be true that $X$ and $Y-7$ have the same marginal distributions $\endgroup$
    – Henry
    Sep 4, 2022 at 12:27
  • $\begingroup$ Thanks a lot @Henry, I think I understand what you are saying but I still have some doubts and I would be very grateful if you could answer them. It's the same to say "they have the same distribution" as "they are equally distributed"? Don't they have the same distribution function even if the values of the variable are different? Can two variables have the same distribution but different values? $\endgroup$
    – Ángel
    Sep 5, 2022 at 6:38
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    $\begingroup$ Two random variables can have the same distribution but different values in the sense that if A is the value of the first die and B the value of the second die, you need not have $A=B$ at each roll though they have the same distribution. So here $Y$ and $X-7$ have the same distribution (typo in my earlier comment) and for example $P(Y=2) = P(X-7=2)$. But $P(X=2)\not = P(Y=2)$ so $X$ and $Y$ have different distributions even if you might say they have the same shaped distribution. $\endgroup$
    – Henry
    Sep 5, 2022 at 7:31

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