Can two random variables $X$ and $Y$ be identically distributed while satisfying $P(XI realise this is not possible if $X$ and $Y$ are discrete with a finite support and I think it is true in general, but I am having trouble with the continuous and discrete-with-infinite-support cases.
Any help would be appreciated.
 A: Step 1
If $U \leq V$ and $U,V$ are bounded above with same  distribution then $U=V$ a.s. To prove this note that $e^{U}$ and $e^{V}$ have the same distribution. Since $e^{U}\leq e^{V}$ and $Ee^{U}=Ee^{V}$ we get $e^{U}=e^{V}$ as and hence $U=V$ a.s
Step 2
Note that for any real number $t$,  $\min \{t,X\} \leq \min \{t, Y\}$ for every real number $t$. Applying Step 1 we see that $\min \{t,X\} =\min \{t, Y\}$ a.s . Letting $t \to \infty$ we get $X=Y$ a.s.
A: The answer is negative.
Suppose that $X$ and $Y$ has the same distribution function $F$.  Put $B_n =  \{ X + \frac{1}{n} \le Y <  X + \frac{1}{n-1} \}$, $n \ge 1$, thus $\Omega = \sqcup_{n=1} B_n$.
$$\sum_n P(X \le t, B_n) = P(X \le t) = P(Y \le t) = \sum_n P(Y \le t, B_n) \text{${}$ ${}$(1)} $$
But $P(Y \le t, B_n) \le P(X \le t, B_n)$ and it follows from (1) that  $P(Y \le t, B_n) = P(X \le t, B_n)$ for all $n$.
We have
$$ P(X \le t-\frac1n, B_n) \le P(X \le t, B_n)  = P(Y \le t, B_n) =$$
$$ = P(Y \le t, X + \frac{1}{n} \le Y <  X + \frac{1}{n-1}) \le $$
$$\le P(X + \frac1n \le t, X + \frac{1}{n} \le Y <  X + \frac{1}{n-1}) =$$
$$ =P(X + \frac1n \le t, B_n) = P(X \le t-\frac1n, B_n) \text{${}$ ${}$(2)} $$
and it follows that all values in (2) are equal. Hence
$P(X \le t-\frac1n, B_n) = P(X \le t, B_n) $ and that's why
$$P(X \in (t-\frac1n, t], B_n) = 0.$$
Fix $n$. As $t$ is arbitrary we get $P(B_n) = 0$.
As  $\Omega = \sqcup_{n=1} B_n$ we have $1 = P(\Omega ) = \sum_n 0 = 0$. We got a contradiction, q.e.d.
