Half-normal distribution: Distribution of absolute difference between two points sampled from a normal distribution I have a normally distributed random variable $X$. I sample two points $x_1$ and $x_2$, and I am interested in the absolute difference between these two sampled points: $d=|x_2-x_1|$. 
I repeat this $N$ times collecting $N$ pairs of points, giving me absolute differences of $d_1, d_2, ... d_N$. 
Is it possible to say anything about the distribution of $d$? Specifically, I would like to construct a confidence interval on $d$, i.e. "with $1-\alpha$ confidence, $d$ falls within this interval." 
 A: A sum of two normals is normal. The variance doubles. The mean is going to be zero. The absolute difference is going to be like an absolute value of a normal, i.e. the density function will be something like: $f(d)=\frac{1}{\sqrt{\pi}\sigma}e^{-\frac{d^2}{4\sigma^2}}$
Obviously, the domain is $d\in[0,\infty)$
As @A.Donda pointed out, it's a half-normal distribution, with properly a plugged variance.
You observe X, which means that you can estimate its variance $\hat\sigma^2$ by using usual estimator such as $\hat\sigma^2=\frac{1}{N-1}\sum_{i=1}^n (x_i-\bar x)^2$, where the mean estimator is usual $\bar x =\frac{1}{N}\sum_{i=1}^n x_i$
The variance of d can be easily computed using half-normal distribution properties: $\hat\sigma_X=2\hat\sigma^2(1-\frac{2}{\pi})$
A: Comment:  Reality check, using simulation, on @Aksakal's (+1) solution.
With a million iterations we can expect
about 2 or 3 place accuracy for the SD.
Let $\mu = 100, \sigma=2.$
set.seed(2022)
d = replicate(10^6, abs(diff(rnorm(2, 100, 2))))
sd(d)
[1] 1.703695   # aprx SD of d from simulation
sqrt(8*(1 - 2/pi))
[1] 1.705005   # exact SD of d

hist(d, prob=T, col="skyblue2",  
      main="Half-normal Dist'n")
 curve(exp(-x^2/16)/(2*sqrt(pi)), 
       add=T, col="brown", lwd=2)


