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?

  • 2
    $\begingroup$ The discussion about $X_i$ and $R_n$ is actually unnecessary. Note that $R_n$ is some random variable and $F_{R_n}$ is its cumulative distribution function. In this case, the random variable $Z:= F_{R_n}(R_n)$ is always uniform. See stats.stackexchange.com/questions/161635/… $\endgroup$
    – Levent
    Commented Jun 6, 2023 at 19:12
  • $\begingroup$ @Levent But why is $Z$ not constantly equal to $1$ $\endgroup$
    – Summerday
    Commented Jun 6, 2023 at 19:14
  • $\begingroup$ That is a confusion about notation. $Z_n=F_{R_n}(R_n)=1$ is not correct because $F_{R_n}$ takes numbers as inputs, not random variables. What is meant there is the following: Sample $R_n$, say you got the number $r$. Feed it to, $F_{R_n}$, i.e., $F_{R_n}(r)=\mathbb{P}(R_n\leq r)$. This is a number between $[0,1]$ and you consider it as a sample of $Z_n$. $\endgroup$
    – Levent
    Commented Jun 6, 2023 at 19:17
  • $\begingroup$ @Levent Ah so it only means that $Z_n$ takes values in $[0,1]$. I think I got it thanks a lot! $\endgroup$
    – Summerday
    Commented Jun 6, 2023 at 19:20
  • 1
    $\begingroup$ It has to: The function $F_{R_n}$ only takes values in $[0,1]$. $\endgroup$
    – Levent
    Commented Jun 6, 2023 at 19:21

1 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.

enter image description here

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 $Z_n \sim Unif(0,1)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .