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Edited Question: This question is about estimating the variance (or $\text{SD}_0$) of a parent distribution from which a set of random independent samples is taken. The samples are taken in sequence and the mean of absolute values of the first differences (between each pair of adjacent observations) is known, and so is the sample size, but nothing else. The sequence is assumed stationary and there is no autocorrelation.

I assume that the parent distribution is Gaussian, or at least not too far from it.

My way of trying to figure it out is that it seems that the absolute first differences should have the Half-normal distribution constructed from an underlying Normal distribution with $\text{mean}=0$ and $\text{SD}=\sqrt{2}\cdot\text{SD}_0.$ Therefore the mean of this Half-normal distribution (mean_delta) should be

$$\sqrt{2}\cdot \text{SD}_0\cdot\sqrt{2}/\sqrt{\pi} = 2\cdot\text{SD}_0/\sqrt{\pi}. $$

Therefore the $\text{SD}_0$ of the original distribution could be estimated as $\sqrt{\pi}\cdot \text{mean_delta}/2 \approx 1.128\cdot\text{mean_delta}.$

My question is: am I even on the right track?

Original Question: The absolute difference of two random independent samples from a standard Normal distribution seems to be Half-normal.

What would be the $\text{SD}$ of it if the negative half of it was also present? I've seen the $\sqrt{1-2/\pi}$ value but is that the $\text{SD}$ of the resultant Half-normal distribution or the $\text{SD}$ of the Normal distribution, right half of which being the Half-normal in question?

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  • $\begingroup$ If $X$ and $Y$ are iid $N(0,1)$ then $X-Y \sim N(0,2)$ and so $|X-Y|$ has a half-normal distribution with $\sigma=\sqrt{2}$ $\Big($not to be confused with its standard deviation of $\sqrt{2-\frac4\pi}\Big)$. $\endgroup$
    – Henry
    Commented May 12 at 19:18
  • $\begingroup$ @Henry, why does $| X -Y |$ imply that the $\sigma = \sqrt{2}$? $\endgroup$
    – nwaldo
    Commented May 12 at 19:24
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    $\begingroup$ @whuber Yes, I assume that the parent distribution is Gaussian, or at least not too far from it. $\endgroup$ Commented May 14 at 14:47
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    $\begingroup$ @whuber The question indeed evolved a bit (sorry) but it relates to the same problem of a real life measure of blood pressure variability and it's reasons and consequences. The use of the mean of absolute first differences became common in literature. Since you've mentioned it, how much better the RMSE (error being the first difference here) would have been if used instead, I wonder? $\endgroup$ Commented May 14 at 15:02
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    $\begingroup$ There are a huge number of ways to do that and the answers can vary tremendously. The basic problem is that retaining only the mean absolute difference has destroyed almost all information about the variance. $\endgroup$
    – whuber
    Commented May 17 at 14:12

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