# What is the distribution of max-min for a Gaussian distribution

For a process N(t), where at any instance of t=T0, the distribution of N(T0) is Gaussain with mu=0:

What is the distribution of max(N(t))-min(N(t))?

From my simulation, it has some non-zero positive mean value and a waveform that looks like Gaussian but has a longer tail on the right side:

• Asymptotically, the maximum and the negative of the minimum follow independent Gumbel distributions with different location parameters but the same scale parameter. Asymptotically, the difference is therefore a sum of two Gumbels with the same scale and therefore also Gumbel, see en.wikipedia.org/wiki/Gumbel_distribution May 19 '20 at 15:11
• @JarleTufto that looks like an answer in itself. May 19 '20 at 15:25
• Does this answer your question? Independence of Sample mean and Sample range of Normal Distribution May 19 '20 at 18:23
• @mdewey Actually, my previous comment is wrong in that the sum of independent Gumbels is not Gumbel (which is obvious when looking at the mgfs). But there are lot of papers on the distribution of linear combinations of Gumbel distributed variables, see google.com/… May 19 '20 at 21:46
• This seems quite similar to the studentized-range distribution (see en.wikipedia.org/wiki/…), but I'm not quite confident enough to put this in an answer ... May 20 '20 at 3:26

Working with the standard normal case for simplicity, the joint density of the minimum and maximum is $$f_{X_{(1)},X_{(2)}}(x_1,x_2)=\frac{n!}{(n-2)!}\phi(x_1)\phi(x_2)[\Phi(x_2)-\Phi(x_1)]^{n-2},$$ for $$x_2>x_1$$. The joint density of the linear transformation \begin{align} Y_1&=X_{(2)}-X_{(1)}, \\ Y_2&=X_{(2)} \end{align} becomes \begin{align} f_{Y_1,Y_2}(y_1,y_2) &=f_{X_{(1)},X_{(2)}}(y_2-y_1,y_2) \\&=\frac{n!}{(n-2)!}\phi(y_2-y_1)\phi(y_2)[\Phi(y_2)-\Phi(y_2-y_1)]^{n-2} \end{align} for $$y_1>0$$. Hence, the marginal density of $$Y_1$$ is \begin{align} f_{Y_1}(y_1) &=\int_{-\infty}^\infty f_{Y_1,Y_2}(y_1,y_2)dy_2 \\&=\frac{n!}{(n-2)!}\int_{-\infty}^\infty\phi(y_2-y_1)\phi(y_2)[\Phi(y_2)-\Phi(y_2-y_1)]^{n-2}dy_2. \end{align} At least for $$n=2$$ and $$n=3$$ but perhaps also for larger $$n$$, this integral has an analytic solution. Resorting to numerical integrations using the R code

dminmax <- function(y1, n) {
g <- function(y2)
dnorm(y2-y1)*dnorm(y2)*(pnorm(y2)-pnorm(y2-y1))^(n-2)
res <- integrate(g, -Inf, Inf)
n*(n-1)*res\$value
}
dminmax <- Vectorize(dminmax)