Let $X_{1},X_{2}$ be a random sample of $N(0,1)$, $n=2$.
$Y_{1}= X_{(1)} \quad Y_{2}= X_{(2)}$
In that case $Y_{1}=min \{ X_{1},X_{2}\}$ and $Y_{2}=max \{ X_{1},X_{2}\}$
I know that the density of $(Y_{1},Y_{2})$ is $ \ f_{Y_{1},Y_{2}}(y_{1},y_2)=n(n-1)(F(y_n)-F(y_1))^{(n-2)} f(y1)f(y_n) \implies f_{Y_{1},Y_{2}}(y_{1},y_2)=2f(y1)f(y_2)$
Let $R=Y_{2}-Y_{1}$ and $T=Y_{2} \implies Y_1 = T -R \quad Y_2=T$
We can use the Change of Variables Theorem: $f_{R,T}(r,t)= f_{Y_{1},Y_{2}}(t-r,t) * |J| \ $ where $|J|$ is Jacobian matrix and determinant.
$|J|=1$
We have $f_{R,T}(r,t)=2f(t-r)f(t) \implies f_R = \int _{ - \infty}^{\infty} 2f(t-r)f(t) dt =\int _{ - \infty}^{\infty} 2 \frac{1}{\sqrt{2 \pi} } e^{- \frac{ (t-r)^{2}}{2}} \frac{1}{\sqrt{2 \pi} } e^{ -\frac{ (t)^{2}}{2}} dt $
Are the calculations correct?
Well $f_{Y_1,Y_2}(y_1,y_2) = n~(n-1)~f(y_1)~f(y_2)~(1-F(y_2))^{n-2}~\mathbf 1_{y_1<y_2}$.
But indeed when $n=2$, we have $f_{Y_1,Y_2}(y_1,y_2)=2~f(y_1)~f(y_2)~\mathbf 1_{y_1<y_2}$
When $R=Y_2-Y_1$ and $T=Y_2$, then $f_{R,T}(r,t)=f_{Y_1,Y_2}(t-r, t)$ , and the Jacobian determinant is $1$.
$$\begin{align}f_{R,T}(r,t) &= 2~f(t-r)~f(t)~\mathbf 1_{0<r}\\[2ex]f_R(r) &= 2~\mathbf 1_{0<r} \int_\Bbb R \dfrac{\mathrm e^{-(t-r)^2/2}\mathrm e^{-t^2/2}}{2\pi}\,\mathrm d t\\[1ex]&= \mathrm e^{-r^2/2}\sqrt{~\dfrac{2}{\pi}~}~\mathbf 1_{0<r}\end{align}$$
So, it checks out $\huge\color{green}\checkmark$.