I have a list (lets call it $ \{L_N\} $) of N random numbers $R\in(0,1)$ (chosen from a uniform distribution). Next, I roll another random number from the same distribution (let’s call this number “b”).

Now I find the element in the list $ \{L_N\} $ that is the closest to the number “b” and find this distance.If I repeat this process, I can plot the distribution of distances that are obtained through this process.

When $N\to \infty$, what does this distribution approach?

When I simulate this in Mathematica, it appears as though it approaches an exponential function. And if the list was 1 element long, then I believe this would exactly follow an exponential distribution.

Looking at the wikipedia for exponential distributions, I can see that there is some discussion on the topic:

But I’m having trouble interpreting what they are saying here. What is “k” here? Is my case what they are describing here in the limit where $n\to \infty$?

EDIT: After a very helpful helpful intuitive answer by Bayequentist, I understand now that the behavior as $N \to \infty$ should approach a dirac delta function. But I’d still like to understand why my data (which is like the minimum of a bunch of exponential distributions), appears to also be exponential. And is there a way that I can figure out what this distribution is exactly (for large but finite N)?

Here is a picture of what the such a distribution looks like for large but finite N:

EDIT2:

Here’s some python code to simulate these distributions:`%matplotlib inline import math import numpy as np import matplotlib as mpl import matplotlib.pyplot as plt numpoints = 10000 NBINS = 1000 randarray1 = np.random.random_sample((numpoints,)) randarray2 = np.random.random_sample((numpoints,)) dtbin = [] for i in range(len(t1)): dt = 10000000 for j in range(len(t2)): delta = t1[i]-t2[j] if abs(delta) < abs(dt): dt = delta dtbin.append(dt) plt.figure() plt.hist(dtbin, bins = NBINS) plt.show()`

**Answer**

If you had been looking for the distance to the next value above, and if you inserted an extra value at $1$ so this always had an answer, then using rotational symmetry the distribution of these distances $D$ would be the same as the distribution of the minimum of $n+1$ independent uniform random variables on $[0,1]$.

That would have $P(D \le d) = 1-(1-d)^{n+1}$ and so density $f(d)=(n+1)(1-d)^n$ when $0 \le d \le 1$. For large $n$ and small $d$ this density can be approximated by $f(d) \approx n e^{-nd}$, explaining the exponential shape you have spotted.

But your question is slightly more complicated, as you are interested in the signed distance to the nearest value above *or* below. As your Wikipedia link shows, the minimum of two i.i.d. exponential random variables with rate $\lambda$ is an exponential random variable with rate $2\lambda$. So you need to change the approximation to the density to reflect both the doubled rate and the possibility of negative values of $d$. The approximation actually becomes a Laplace distribution with $$f(d) \approx n e^{-2n|d|}$$ remembering this is for large $n$ and small $d$ (in particular the true density is $0$ unless $-\frac12 \lt d \lt \frac12$). As $n$ increases, this concentrates almost all the density at $0$ as in Bayequentist’s response of the limit of a Dirac delta distribution

With $n=10^6$ the approximation to the density would look like this, matching the shape of your simulated data.

**Attribution***Source : Link , Question Author : Steven Sagona , Answer Author : Henry*