Let $X_{1},X_{2},X_{3}$ be i.i.d samples from $N(\mu,\sigma^2)$. Let $u$ denotes $$u=\frac{X_{(3)}-X_{(1)}}{S_{3}}\text{ , where }\,X_{(i)}\text{ denotes the }i^{\text{th}}\,\text{order statistic,}\\S_{3}=\sqrt{\frac{\sum_{i=1}^3(X_i-\bar{X})^2}{2}}$$
Now I know the density function of $\,u\,$ is$$f(u)=\frac{3}{\pi}\left(1-\frac{u^2}{4}\right)^{-\frac{1}{2}}(\sqrt{3}\le u\le2)$$ But how to get that? Does anyone have any ideas? Would appreciate some help.
Note that the vector $(X_1,X_2,X_3)$ has a radially symmetric distribution since the density is a function of $X_1^2+X_2^2+X_3^2$. Therefore it’s rotationally symmetric around the vector $(1,1,1)$. Projecting onto the plane $x+y+z=0$ and normalizing preserves that symmetry, so we get that $Y_i=(X_i-\bar X)/\sqrt{\sum (X_i - \bar X)^2}$ are uniformly distributed along a unit circle that lies on the plane perpendicular to $(1,1,1)$.
Looking at $u$ we can shift, scale, and reorder the variables without affecting the value, so $u$ is a function of the $Y_{(i)}$.
We can parametrize the entire circle as $Y_1=\frac{\sqrt6}{3}\cos(z+2\pi/3)$, $Y_2=\frac{\sqrt6}{3}\cos(z+4\pi/3)$, $Y_3=\frac{\sqrt6}{3}\cos(z)$ for $0<z<2\pi$. We get the ordering $Y_1<Y_2<Y_3$ when $0<z<\pi/3$, so using symmetry, we can therefore get this to be a distribution on $Y_{(i)}$ by having $z$ be uniform on the restricted range $[0,\pi/3]$, and on that range $$u(z)=(Y_{(3)}-Y_{(1)})/\sqrt{\sum_i Y_i^2/2}=\frac{\sqrt6}{3}(\cos(z)-\cos(z+2\pi/3))/\sqrt{1/2}$$ $$=\frac{\sqrt6}{3}(2 \sin(z) \sin(2\pi/3))/\sqrt{1/2}=2\sin(z)$$
Since $\frac{du}{dz}=2\cos(z)=\sqrt{1-u^2/4}$ and z is uniformly distributed with density $3/\pi$ from 0 to $\pi/3$, we get:
$f(u)=\frac 3 \pi (\frac{du}{dz})^{-1}=\frac 3 \pi (1-u^2/4)^{-1/2}$ as desired.