Is absolute value of the difference of two vectors associated with its standard deviation (variance)? Suppose there are two random variables that are independent with each other. Is there any relationship between the absolute value of the difference of these two variables and variance of the two variables?
u1 <- rnorm(100, 0, sd=0.2)
u2 <- rnorm(100, 0, sd=0.8)
abs(u2-u1)

It will give something like

0.8553218366 0.1559001753 0.3589231965 0.1126236494
1.1591884681 1.0089059851 0.6116833748 0.1552008638...

If I look for the mean of absolute value of the difference,
mean(abs(u2-u1))

It will give me 0.66..
I am wondering, this abs(u2-u1) could have any mean association with sd (0.2 or 0.8)? In other words, how can I approximate the abs(u2-u1) given the sd(0.2 or 0.8)?
 A: Mathematically, the expectation of abs(u2-u1) should be
> sqrt(0.2**2 + 0.8**2)*sqrt(2/pi)
[1] 0.6579525

abs(u2-u1) actually follows the folded normal distribution, and you will find more information from this link
A: Yes, there is a meaning behind this combination. As you can find here, the difference of two normal random variables has the following distribution:
$$
U_1-U_2 \sim\ U_1 + a U_2\ \sim\ \mathcal{N}\big( \mu_{U_1} + a\mu_{U_2},\ \sqrt{\sigma_{U_1}^2 + a^2\sigma_{U_2}^2} \big) = \mathcal{N}\big( \mu_{U_1} - \mu_{U_2},\ \sqrt{\sigma_{U_1}^2 +  \sigma_{U_2}^2} \big)
$$
Hence, the difference of these two comes from a normal distribution $\mathcal{N}(0, \sqrt{(0.2)^2 + (0.8)^2}) \sim \mathcal{N}(0, 0.82)$.
For the difference you can use the folded normal distribution. Hence, $|U_1 - U_2|$ has the mean like following (see this link for more details which $Y = |U_1 - U_2|$ in our case):

So, expected value of the distribution of absolute difference of two normal variables $U_1$ and $U_2$ is $0.82 \times \sqrt{\frac{2}{\pi}}$ (because $\mu = 0$ in distribution of $U1 - U_2$).
Finally, as the average of $n$ instance ($n = 100$) from the same distribution (i.i.d.) is an estimation of the mean of the distribution we can say mean(abs(u1 - u2)) = 0.82 * (\sqrt(2/pi))  ~ 0.657952.
