Let $U_1, \cdots, U_n$ be $n$ i.i.d continuous uniform random variables on $(0,1)$ and their order statistics be $U_{(1)}, \cdots, U_{(n)}$.

Define $D_i=U_{(i)}−U_{(i−1)}$ for $i=1, \cdots, n$ with $U_0=0$.

I am trying to figure out the distribution $\lim_{n\to\infty} \#\{i = 1, \cdots, n \mid nD_i \geq z\}/n$. This must be very well known, but I cannot find a reference.


1 Answer 1


As I understand correctly: $\#\{i=1,...,n | nD_i\geq z\}$ means the number of cases which satisfy condition $nD_i\geq z$. If so:

The limiting distribution of $n\cdot min(U_1,...,U_n)$ has an expotential distribution with parameter $\lambda = 1$. See proof here.

Conditionally, given $U_1$, we can calculate distribution of second variable correcting for $U_1$, so the distance [0-1] was the same: $$((n-1)\cdot min(U_2,...,U_n)-U_1)\cdot(1-U_1) | U_1$$ The result would be the same. We have $n-1$ random variables uniformly distributed at range [0-1], but in infinity $n$ and $n-1$ are the same. And this holds for every $U_1$, so we can infer, that unconditionally, no matter the randomisation, distance between first and second variable has also expotential distribution with parameter $\lambda = 1$. Now this way we can calculate recursively, that every spacing between two consecutive variables has same distribution as the minimal random variable.

If we define variables $Z_1,...,Z_n$ which are equal to $1$ if the difference would be greater or equal than $z$ and $0$ otherwise.

From expotential distribution we know that:

$$P(nD_i\geq z) = e^{-z} = E(Z_i)$$

and i guess this is it, because when limiting with $n \rightarrow \infty$, what actually is defined is expected value.

If looking for the distribution of such cases when $n$ is not going to infinity, we would need to use Central Limit Theorem. Calculate variance and expectation of $Z_i$ which is simple, and use CLT with $n$, $Var({Z_i})$ and $E(Z_i)$.

Some R codes showing the same with RNG:

#Distribution of D_i
n = 10000000
vec = diff(sort(c(runif(n),0)))*n
hist(vec, freq=F)

#Distribution of Z_i for finite n
rep = 10000
n = 100
z = 0.5
dist = c()

for(i in 1:rep){
    vec = diff(sort(c(runif(n),0)))*n
    dist = c(dist,sum(vec>z))
hist(dist, breaks=30)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.