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Suppose $X_1$ and $X_2$ are iid random variables. I want to determine $P(X_1=X_2)$.

If they are integer-valued random variables, then $P(X_1=X_2) = \sum_{i \in \mathbb{Z}} P_{X_1,X_2}(i,i) = \sum_{i \in \mathbb{Z}} P_{X_1}(i)^2. $

If they are continuous random variables, then $P(X_1=X_2) = \int_{x \in \mathbb{R}} f_{X_1,X_2}(x,x) dx = \int_{x \in \mathbb{R}} f_{X_1}(x)^2 dx. $ But when $X_1$ and $X_2$ are uniformly distributed over $[0,1)$, $P(X_1=X_2) = \int_{x \in \mathbb{R}} f_{X_1}(x)^2 dx = \int_{x \in [0,1)} 1 dx = 1. $ Intuitively it is not possible, since $P(X_1\neq X_2) > 0$. So is there some mistake I have made? Thanks!

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    @Srivatsan: Thanks! For some integral of a density function of a random vector, how can one tell if the integral is the density of some r.v. at some value, or the probability of some other r.v. at some value? Is it told from whether the dimension of the integral is strictly less than the dimension of the original random vector?2011-09-12

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No, your calculation for the continuous case is wrong. It should be $ P(X_1 = X_2) = \displaystyle\iint_D f_{X_1}(x) f_{X_2}(y)\ dx \ dy$, where $D = \{(x,y) \in {\mathbb R}^2: x = y\}$. But $D$ has two-dimensional measure (i.e. area) $0$, so the answer is $0$.

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    Well, @Tim, the same reason why a random chosen number from the interval $[0,1]$ has a *zero* probability of being *exactly* $0$.2011-09-12