Let $x$ be a number within (0,1) and let $Y_{i}$ be a uniform RV $Y_{i}\sim U(0,1)$. We define function $f_{i}(x)$ such that:
$$ f_{i}(x) =\left\{ \begin{array}{ll} 1, & \mbox{if $x>Y_{i}$}\\ 0, & \mbox{otherwise} \end{array} \right. $$
Are $f_{1}(x)$ and $f_{2}(x)$ IID?
My intuition is that they are not because $x$ is obviously correlated with itself. However, I am having trouble coming up with a proof given that $Y_{1}$ and $Y_{2}$ are clearly IID.
Assuming $Y_i$ are generated independently, then $f(x_1)$ and $f(x_2)$ with $x_1=x_2$ are IID. For example, in the first trial you generate a number from the uniform distribution ($Y_1$) and then using that number you calculate $f(\frac{1}{2})$. In the second trial you generate another random number from the uniform distribution ($Y_2$), independently from the first random number in the first trial, and then again you calculate $f(\frac{1}{2})$. In both triales, we don't care $x_1=x_2=\frac{1}{2}$. Generating $Y_1$ and $Y_2$ doesn't depend on $x$, because first you generate them and then you compare with $x$.
First we show they are identically distributed. It is sufficient to show that they have the same cumulative distribution functions (CDF).
Let $F$ and $G$ be CDFs of $f(x_1)$ and $f(x_2)$ respectively and let $0 \leq z < 1 $. We have,
$$F(z) = P(f(x_1) \leq z) = P(f(x_1) = 0) = P(Y \geq x_1) = 1-x_1.$$
so $$F(z) = \begin{cases} 0 & z < 0 \\ 1-x_1 & 0\leq z < 1 \\ 1 & z \geq 1 \end{cases}$$
Similarly
$$G(z) = \begin{cases} 0 & z < 0 \\ 1-x_2 & 0\leq z < 1 \\ 1 & z \geq 1 \end{cases}$$ if $x_1=x_2$, then $F(z) = G(z)$ for all $z$ and this means they have the same distributions.
It is left to show they are independent. Let $0 \leq y,z <1 $,
$$P(f(x_1) \leq y, f(x_2) \leq z) = P(Y_1 \geq x_1, Y_2 \geq x_2) = P(Y_1 \geq x_1)P(Y_2 \geq x_2) = P(f(x_1) \leq y) P(f(x_2) \leq z).$$
Other cases for $y$ and $z$ can be also checked.