Suppose that $X_1,X_2$ are IID with continuous CDF $F$. I would like to show that $$ \Pr(X_1>X_2)=\frac{1}{2}. $$ By symmetry, we have $\Pr(X_1>X_2)=\Pr(X_2>X_1)$ but how do I show rigorously that $\Pr(X_1=X_2)=0$?
Probability of ties under for IID random variables with continuous CDF
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probability
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2 Answers
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@chandu1729 has dealt with the absolutely continuous case, and the continuous case is not much different. Let $\mu$ be the probability measure on $\Bbb R$ with cdf $F$. The continuity of $F$ means that $\mu(\{x\})=0$ for each $x\in\Bbb R$. The joint distribution of $X_1$ and $X_2$ is the product measure $\mu\otimes\mu$; in particular, $$ P(X_1=X_2)=P((X_1,X_2)\in D)=\mu\otimes\mu(D), $$ where $D:=\{(x,x)\in\Bbb R^2:x\in\Bbb R\}$. Now use Fubini to write $\mu\otimes\mu(D)$ as an iterated integral, and evaluate.
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As $F$ is continuous, $X_1$ and $X_2$ will have density function $f$. If $A$ is Borel set of $\mathbb{R}^2$, we have the following :
$$P((X_1,X_2) \in A) = \int_A f(x) f(y) dxdy$$
Hence, $P(X_1=X_2) = \int_{y=x}f(x)f(y)dxdy$
Since, the set $A=\{(x,y):y=x\}$ is of Lebesgue measure $0$ in $\mathbb{R}^2$, $P(X_1=X_2)=0$.