# What is the mean absolute difference between values in a normal distribution?

I understand that variance is mean of squared differences and that standard deviation is square root of the mean.

What, however, is the average difference between values in a normal distribution (without considering the sign, of course, since if we consider the sign, it would be 0)?

Assume that $$X, Y\sim N(\mu,\sigma^2)$$ are iid.

Then their difference is $$X-Y\sim N(0,2\sigma^2)$$. As you write, the expectation of this difference is zero.

And the absolute value of this difference $$|X-Y|$$ follows a folded normal distribution. Its mean can be found by plugging the mean $$0$$ and variance $$2\sigma^2$$ of $$X-Y$$ into the formula at the Wikipedia page:

$$\sqrt{2}\sigma\sqrt{\frac{2}{\pi}} = \frac{2\sigma}{\sqrt{\pi}}.$$

A quick simulation in R is consistent with this:

> nn <- 1e6
> sigma <- 2
> set.seed(1)
> XX <- rnorm(nn,0,sigma)
> YY <- rnorm(nn,0,sigma)
> mean(abs(XX-YY))
[1] 2.257667
> sqrt(2)*sigma*sqrt(2/pi)
[1] 2.256758