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Is it possible that two random variables have the same distribution and yet they are almost surely different?

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Let $X\sim N(0,1)$ and define $Y=-X$. It is easy to prove that $Y\sim N(0,1)$.

But $$ P\{\omega : X(\omega)=Y(\omega)\} = P\{\omega : X(\omega)=0,Y(\omega)=0\} \leq P\{\omega : X(\omega)=0\} = 0 \, . $$

Hence, $X$ and $Y$ are different with probability one.

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    $\begingroup$ This same trick works much more generally and even in cases that might "appear" simpler to someone first encountering the subject. For example, consider $X$ and $1-X$ where $X$ is a Bernoulli random variable with probability of success being $1/2$. $\endgroup$ – cardinal Mar 20 '12 at 17:07
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Any pair of independent random variables $X$ and $Y$ having the same continuous distribution provides a counterexample.

In fact, two random variables having the same distribution are not even necessarily defined on the same probability space, hence the question makes no sense in general.

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    $\begingroup$ (+1) Your second point, in particular, is an important one and, once understand, helps elucidate the differences in the two concepts involved. $\endgroup$ – cardinal Mar 20 '12 at 20:35
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Just consider $X(x)=x$ and $Y(x)=1-x$ with $x \in [0,1]$ with Borel or Lebesgue measure. For both the accumulated probability is $F(x)=x$ and the probability distibution is $f(x)=1$. For the sum $X+Y$ the distribution is a Dirac unit mass at $x=1$.

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  • $\begingroup$ Welcome to our site. Could you clarify the sense in which your post answers the question in this thread and show how it differs from the answer given by Zen (and the comment by @Cardinal to that answer)? $\endgroup$ – whuber Jan 7 '18 at 15:39

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