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.
Zen
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20This 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 $1/2$. – 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.
cardinal
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Stéphane Laurent
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3(+1) Your second point, in particular, is an important one and, once understand, helps elucidate the differences in the two concepts involved. – 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 $x=1$.
Siong Thye Goh
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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](https://stats.stackexchange.com/questions/24938/can-two-random-variables-have-the-same-distribution-yet-be-almost-surely-differ#comment45504_24939))? – whuber Jan 07 '18 at 15:39