$X,Y$ are iid from distribution $F$, which is a continuous function, then is $P(X=Y)>0$? Suppose $X,Y$ are iid random variables from a distribution function $F$, which is a continuous function. Then is it always true that $P(X=Y)=0$?
For me, the answer is trivially YES. We have $\int_y P(X=y)dF(y)=0$, and as $P(X=y)=0$, hence $P(X=Y)=0$.
It has however been claimed that it is false in general and counterexample exists, and a total of 10 points have been allotted to this problem. This makes me doubt: does there really exist such $F$?
 A: More generally, if $X$ and $Y$ are independent, with cdfs $F_X$ and $F_Y$, then
$$
\eqalign{\Bbb P[X=Y]
&=\sum_{x\in\Bbb R}\Bbb P[X=x]\cdot\Bbb P[Y=x]\cr
&=\sum_{x\in\Bbb R}[F_X(x)-F_X(x-)]\cdot[F_Y(x)-F_Y(x-)],\cr
}
$$
so $\Bbb P[X=Y]=0$ provided one or the other has a continuous cdf. (Or simply if their cdfs have no common discontinuities.) For a proof see my answer to this question: Calculating $P(X+Y=0)$ for independent random variables (Problem in Durrett)
A: Assuming they are real valued: $P(X=Y)=\int_{-\infty}^\infty P(Y=x|X=x) \ P(dx)=\int_{-\infty}^\infty P(Y=x) \ P(dx)=0$, because $P(Y=x)=0$ for any $x\in\mathbb{R}$ since $Y$ has a continuous distribution (pdf). If you meant for the cdf to be continuous, then the answer changes.
In general, though $P(X=Y)>0$ is possible, but you have to relax some of your conditions: independence, identically distributed, and/or continuous distribution.
$\textbf{Example $1$:}$ Take $X$ and $Y$ to be i.i.d. fair coin flips, then $P(X=Y)=0.5$ (discrete distributions).
$\textbf{Example $2$:}$ Let $\Omega=\Omega_1\cup\Omega_2$ (disjoint), and let $Y(\omega)=X(\omega)$ for $\omega\in\Omega_1$ and $Y(\omega)=X(\omega)+1$ for $\omega\in\Omega_2$. Then $P(X=Y)=P(\Omega_1)>0$ is possible. 
If $X$ is uniform on $[0,2]$, and $\Omega_1=\{\omega \mid X(\omega)\leq 1\}$, then $Y$ is uniform on $[0,1]\cup (2,3]$. Then $Y$'s pdf is discontinuous, but it's cdf is continuous (both the pdf and cdf of $X$ are continuous). Of course, they are neither independent, nor identically distributed.
$\textbf{Example $3$:}$ Keep everything the same as Example $2$ above, except let $Y=\frac{1}{2}(X^2+1)$ on $\Omega_2$. Note that $Y=g(X)$ with both $g$ and $g^{-1}$ are continuous and differentiable (a fun calculus exercise). Then $Y$'s pdf and cdf are both continuous. Of course, $X$ and $Y$ are still neither independent, nor identically distributed.
Side Note: The last 2 examples could be described as couplings of $X$ and $Y$, since they are constructed on the same probability space. We could think of them as being "in principle independent" if we were to "simulate them separately". E.g. if I sample an $\omega$ and evaluate $X$, and you sample another $\omega$ and evaluate $Y$, then our experimental outcomes can be independent. But, this is equivalent to there being two distinct $\Omega$ spaces with identical structure -- i.e. one copy of $\Omega$ per experiment.
