How can I show that this random variable $Z_n\sim \operatorname{Uni}(0,1)$?

Question

Let us consider $X_1,...,X_n$ iid random variables which are distributed with respect to some density $f(x)$. Let $R_n:=R(X_1,...X_n)$ be some continuous random variable which depends on $X_1,...,X_n$ and define $Z_n:=F_{R_n}(R_n)$ where $F_{R_n}(t)=\Bbb{P}(R_n\leq t)$. I want to show that $Z_n\sim \operatorname{Uni}(0,1)$.

Im a bit confused about $Z_n$ because if I replace it with the definition I would get $Z_n=F_{R_n}(R_n)=\Bbb{P}(R_n\leq R_n)=1$. This is my first confusion. But the second one is that I don't see a way to show that this random variable is uniformly distributed. I thought about showing that for all $q\in (0,1)$ $$\Bbb{P}(Z_n\leq q)=q$$ but this does not went well.

Can maybe someone help me?

Answer 1

To better illustrate the following explanation, consider that the CDF of $R_n$ is represented by the figure below.

One important detail is that the definition of $X_n$ does not matter at all in this problem, so we will just ignore all information about the RV $X$.

To solve this problem, it is important to understand the definition of the random variable $Z_n$.

$Z_n$ is a (non-linear) transformation of $R_n$, so we will obtain the CDF of $Z_n$ from $R_n$ itself.

Let $F_{Z_n}(t) = P(Z_n \le t)$. denotes the CDF of $Z_n$ calculated in a point $t \in (0,1)$.

Now, take a fixed $t \in (0,1)$. Since $R_n$ is a continuous RV, exists a $r_t \in \mathbb{R}$ such $F_{R_n}(r_t) = t$.

Note that $Z_n = t$ is equivalent $R_n = r$ for an $r$ such to $F_{R_n}(r) = t$. In that way, $Z_n = t \iff R = r_t$.

For this reason:

$$F_{Z_n}(t) = P(Z_n \le t) = P(R_n \le r_t) = F_{R_n}(r_t) = t$$

Therefore, $F_{Z_n}(t) = t$, which implies that $Z_n \sim Unif(0,1)$.