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Suppose that $X_1, X_2$ are iid random variables with the same CDF, is it true that $P(X_2>X_1) = X_1$?

My intuition here is that if the CDF function is $F(x)$, then:

$$ P(X_2>X_1) = F(X_1) \sim \text{Uniform} $$

over $[0,1]$. Is this true?

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    What do you mean by $P(X_2>X_1)=X_1$? The left-hand side is a number, while the right-hand side is a random variable. – 2017-01-11
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    @carmichael561 Is there a correct way to find this? – 2017-01-11
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    I don't really understand what you're asking. – 2017-01-11
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    If they are continuous then the probability is just one-half. – 2017-01-11
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    Your intuition can be corrected by conditioning: $P[X_2>X_1] = \int_{-\infty}^{\infty} P[X_2>X_1|X_1=x]f_{X_1}(x)dx$. – 2017-01-11

1 Answers 1

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Let $X_1$ and $X_2$ both uniform over $[0,1]$ (and independent). Then

$$P(X_1

That is , as in general, $P(X_1

However, the conditional probability

$$P(X_1

Then, indeed, one can say that

$$P(X_1

is a random variable with a distribution. (Here $F$ is the common distribution of $X_1$ and $X_2$.) The important fact is that our random variables are i.i.d.

Finally, the distribution of $F(X_2)$ is uniform if $F$ is continuous. For continuous $F$'s:

$$P(F(X_2)

between $0$ and $1$ and is $0$ if $x<0$ and is $1$ if $x>1$.