8
$\begingroup$

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)?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ In my opinion, it is still zero. Case of very large mean, the absolute value transform is not material and the expected difference remains zero. Instead, consider moving the mean to zero. This implies we have a Truncated Normal distribution (see en.wikipedia.org/wiki/…) which is truncated at the mean (now zero). As this, per Wikipedia, is "a mean preserving contraction" again no effect, the answer remains zero. $\endgroup$ – AJKOER Sep 25 at 5:18
  • 6
    $\begingroup$ @AJKOER …what? I think you've probably misread the question (and also the Wikipedia article you reference). In particular, you seem to be considering the difference of the absolute values of two i.i.d. normal random variables, whereas the OP is clearly asking about their absolute difference (i.e. $|X-Y|$, not $|X|-|Y|$). Also, as Wikipedia clearly says, "truncation is a mean-preserving contraction combined with a mean-changing rigid shift" (emphasis mine), and thus is not mean-preserving as a whole. $\endgroup$ – Ilmari Karonen Sep 25 at 17:56
12
$\begingroup$

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
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Is there an immense gratitude button anywhere on the internet? $\endgroup$ – Ritesh Singh Sep 25 at 5:20
  • 5
    $\begingroup$ Yes, it's the little checkmark you already clicked - thank you! $\endgroup$ – Stephan Kolassa Sep 25 at 5:21
  • 1
    $\begingroup$ It's the formula for the mean "$\mu_Y$" in the sidebar at the Wikipedia page, where we substitute $\mu=0$ for the mean of $X-Y$ and use $\sqrt{2}\sigma$ in the place of $\sigma$ for the standard deviation of $X-Y$. $\endgroup$ – Stephan Kolassa Sep 25 at 5:35
  • 1
    $\begingroup$ This is a general result on sums of independent normal variables: if $X\sim N(\mu_X, \sigma^2_X)$ and $Y\sim N(\mu_Y, \sigma^2_Y)$ are independent, then $-Y\sim N(-\mu_Y,\sigma^2_Y)$ and $X-Y=X+(-Y)\sim N(\mu_X-\mu_Y, \sigma^2_X+\sigma^2_Y)$. $\endgroup$ – Stephan Kolassa Sep 25 at 17:37
  • 1
    $\begingroup$ @RiteshSingh You can start a bounty for the question, then assign it to Stephan Kolassa if you want. $\endgroup$ – Acccumulation Sep 25 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.