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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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up vote 24 down vote

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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10  
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 $1/2$. – cardinal Mar 20 '12 at 17:07
Really cool example, @cardinal! – Zen Mar 20 '12 at 19:06
up vote 12 down vote

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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(+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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