# 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:

• @JarleTufto that looks like an answer in itself. Commented May 19, 2020 at 15:25
• Does this answer your question? Independence of Sample mean and Sample range of Normal Distribution Commented May 19, 2020 at 18:23
• 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 ... Commented May 20, 2020 at 3:26

Working with the standard normal case for simplicity, the joint density of the minimum and maximum is $$f_{X_{(1)},X_{(n)}}(x_1,x_2)=\frac{n!}{(n-2)!}\phi(x_1)\phi(x_2)[\Phi(x_2)-\Phi(x_1)]^{n-2},$$ for $$x_2>x_1$$. The joint density of the linear transformation \begin{align} Y_1&=X_{(n)}-X_{(1)}, \\ Y_2&=X_{(n)} \end{align} becomes \begin{align} f_{Y_1,Y_2}(y_1,y_2) &=f_{X_{(1)},X_{(n)}}(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) {
n*(n-1)*res$value } dminmax <- Vectorize(dminmax) curve(dminmax(x,5), add)  produces the plot • I'm a bit confused in places notationally if you are trying to provide a general solution or if you are treating an$n=2\$ special case. Commented Mar 28 at 19:35