If $X$ and $Y$ are two random variables with the exact same distribution, do we then say $$X = Y?$$ Or do we say $$X = Y \ \text{almost everywhere}?$$ And if so, why? $X$ and $Y$ are maps. Why are tiwo maps equal just because they happen to have the same distribution?

And if not, when do we say they are equal?

EDIT with context:

I have two marginal distributions ($X_1, X_2)$ which are normally distributed with same mean and variance. I need to show that their joint distribution lies in a 1-dimensional space in $\mathbb{R}^2$, since it is a singular normal distribution. I would like to conclude that $X_1 = X_2$ and thus the 1D space is just the diagonal. But can I go from "$X_1$ and $X_2$ have the same distribution to saying $X_1 = X_2$?

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    $\begingroup$ This question points to the weakness of the standard definition of a random variable solely as a function from measurable sets to real numbers. In the sense that they are solely such a function, you can say $X=Y$, and you don't need to add a.e. But no one would say that $X=Y$ unless you are almost surely guaranteed the same outcome. $\endgroup$ – Paul Feb 22 '16 at 16:39
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    $\begingroup$ In the answer to your edited question, no, you cannot make that jump unless you know something specific about the covariances. $\endgroup$ – Paul Feb 22 '16 at 16:41
  • $\begingroup$ Paul. I do actually have knowledge of the covariances. What am I supposed to use and how? $\endgroup$ – Simp Feb 22 '16 at 16:43
  • $\begingroup$ If the correlation is 1, you do know that they are equal. Otherwise, you cannot assume that. $\endgroup$ – Paul Feb 22 '16 at 16:45
  • $\begingroup$ Yes, of course. If the correlation is $1$, then we know that $X_1 = \alpha + \beta X_2$ and, since they have the same distribution $\alpha = 0, \beta = 1$. However, the theorem I have at hand establishes the equality only almost everywhere. Will this influence by desired conclusion? Will I have to change my conclusion so that it says "the joint distribution lies on the diagonal $\textbf{with probability 1}$"? $\endgroup$ – Simp Feb 22 '16 at 16:52

Having the same distribution is very weak, much weaker than equality almost surely. In fact it does not even depend on the random variables being defined on the same probability space.

If I saw $X=Y$ I would probably assume it means either literal equality of functions or equality almost surely.


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