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:
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) {
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)
curve(dminmax(x,5), add)
produces the plot
