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# Can two random variables have the same distribution, yet be almost surely different?

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

### 2 Answers

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.

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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# Can two random variables have the same distribution, yet be almost surely different?

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.

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

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