I have four independent variables $x_i$, each of them normally distributed with $\mu = 0$ and $\sigma = 1$. What is the distribution of the range of the four variables, i.e., $\max(x_i) - \min(x_i)$?

For example, I draw a random sample of $n=10000$ of each variable and calculate the $\Delta x = \max(x_i) - \min(x_i)$, which gives $10000$ range values. I am interested in the expected distribution of this range value $\Delta x$ under the condition that each $x_i$ is normally distributed as explained above.

  • 6
    $\begingroup$ The phrase "peak-to-valley," which apparently has nothing to do with the range of the sample you describe, suggests you might have a different setting in mind: are you sure you have asked a question relevant to your underlying statistical problem? $\endgroup$
    – whuber
    Sep 16, 2019 at 10:37
  • 1
    $\begingroup$ @whuber is correct. With "Peak-to-Valley" in finance we denote the distance between the local minimum (maximum) and the next local maximum (minimum) with the term "Maximum Drawdown". Maybe you want to derive the distribution of the maximum drawdown. A quick google search will give you some paper addressing your question (if you are concerned with this measure) $\endgroup$ Sep 17, 2019 at 12:35

1 Answer 1


The range of a sample $X = x_1, x_2, \ldots, x_n$ is the difference between its maximum $x_{(n)}=\max(X)$ and minimum $x_{(1)}=\min(X):$

$$\operatorname{range}(X) = x_{(n)} - x_{(1)}.$$

When $X$ is a simple random sample of size $n\ge 2$ from a continuous distribution with distribution function $F$ and density function (PDF) $f=F^\prime,$ the joint distribution of the minimum and maximum is also continuous and, following the analysis at https://stats.stackexchange.com/a/78559/919, has a density function

$$f_{(X_{(1)}, X_{(n)})}(x,y) = n(n-1) f(x)f(y)\left(F(y)-F(x)\right)^{n-2}\, \mathcal{I}(y\ge x).$$

(The indicator $\mathcal{I}(y\ge x)$ means this function is zero when $y \lt x.$)

Upon changing variables from $(x,y)$ to $(x,r)$ with $r=y-x\ge 0$ representing possible values of the range, it follows that $$\mathrm{d}x\mathrm{d}y = \mathrm{d}x\mathrm{d}(x+r) = \mathrm{d}x\mathrm{d}r$$ and you can then integrate out the variable $x$ to obtain the PDF for the range,

$$f_{\operatorname{range}(n)}(r) = n(n-1)\mathcal{I}(r\ge 0)\, \int_{\mathbb R} f(x)f(x+r)\left(F(x+r)-F(x)\right)^{n-2}\mathrm{d}x.$$

I understand $f_{\operatorname{range}(4)}$ responds to the question's request for an "expected distribution."

In general--and particularly for Normal distributions--there is no closed form expression for the integral: it needs to be evaluated numerically. Neither is there generally a closed form expression for its expectation; that too requires numerical evaluation. But for modest values of $n$--less than $10^{36},$ approximately, when working in double precision--these integrals can be accurately evaluated.

As an example, the following plots show (from left to right)

  1. A histogram of the results of 100,000 simulated ranges of standard Normal samples of $n=4$ observations on which a plot of $f_{\operatorname{range}(4)}$ is superimposed to show how closely the numerical integral adheres to the simulated frequencies;

  2. Plots of $f_{\operatorname{range}(n)}$ for $n=3$ (red) through $n=6$ (blue), with that for $n=4$ outlined in black, to show how the range distribution changes with sample size;

  3. A plot of $f_{\operatorname{range}(10^{36})}$ to show what the range distribution looks like for large $n.$


The following R code produced the figure. It shows how to compute $f_{\operatorname{range}(n)}$ numerically and how to simulate ranges.

# Compute the density of the Normal range distribution at value `r` for a
# sample of size `n`.
f <- Vectorize(function(r, n, mu=0, sigma=1, ...) {
  q <- qnorm(1e-4/n)
  f <- function(x) dnorm(x, mu, sigma, log=TRUE)
  ff <- function(x) pnorm(x, mu, sigma, log=TRUE)
  logdiff <- function(y, x, k) {
    u <- x-y
    ifelse(2*u < log(.Machine$double.eps), y-exp(u), y+log(1-exp(u))) * k
  h <- function(x,y) {
    log(n) + log(n-1) + f(x) + f(y) + logdiff(ff(y), ff(x), n-2)
  integrate(function(x) exp(h(x,x+r)), mu+q*sigma, mu-q*sigma, ...)$value
}, "r")
# Simulate a large number of ranges of samples of size `n`.
# This takes about a second for n=10^5.
n.sim <- 1e5
n <- 4
x <- apply(matrix(rnorm(n.sim*n), ncol=n.sim), 2, function(y) diff(range(y)))
# Create the figure.
# Histogram of the simulation.
hist(x, freq=FALSE, breaks=50, cex.main=1,
     main=expression(paste("Histogram of ", 10^{5}, " Standard Normals"))
curve(f(x, n), add=TRUE, lwd=2, n=201)
# Plots of `f` for small `n`.
range.plot <- function(n, add=FALSE, n.pts=201, color="Gray", outline="Black", ...) {
  x <- seq(max(0, -2*qnorm(1/n)-3), -2*qnorm(1/n)+5, length.out=n.pts)
  y <- f(x, n, stop.on.error=FALSE, rel.tol=1e-8)
  i <- y > 1e-3
    plot(x[i], y[i], type="n", xlab="Range", ylab="Density",
         main=expression(paste("PDF for Range of ", 10^{36}, " Standard Normals")),
  polygon(x, y, col=color, border=NA)
  lines(x, y, lwd=2, col=outline)
plot(c(0,6), c(0,.5), type="n", cex.main=1,
     main=expression(paste("PDFs for Ranges of ", 3-6, " Standard Normals")),
     xlab="Range", ylab="Density")
invisible(sapply(3:6, function(i) {
  a <- sapply(c(.4, .9), function(a) rainbow(5, 0.7, 0.9, alpha=a)[i-2])
  range.plot(i, add=TRUE, color=a[1], outline=a[2])
curve(f(x, 4), add=TRUE, lwd=2, n=201)
# Plot `f` for very large `n`.
range.plot(1e36, n.pts=501, cex.main=1)
  • $\begingroup$ Why the log=TRUE in the calls to dnorm and pnorm? I mean, how did you get to the conclusion that it was better to use logarithms...? $\endgroup$
    – Vicent
    Nov 28, 2023 at 16:53
  • $\begingroup$ @Vicent This reduces loss of precision when working in finite precision arithmetic. It's part of the art of scientific computation. $\endgroup$
    – whuber
    Nov 28, 2023 at 17:08

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