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I always understood standard deviation to be the average distance of the observations from the mean. But when I generated a standard normal distribution N(0,1) with n = 1,000,000 in Excel, and took the average of all negative observations and the average of all positive observations, I got around -0.80 and +0.80 respectively, when I would have expected to get -1 and +1. The empirical results indicate that the average distance of an observation from EV = 0 is 0.8, yet the standard deviation is 1.

How do I reconcile this conceptually for my own understanding?

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  • $\begingroup$ The number you are estimating here is $\frac{2}{\sqrt{2\pi}} \approx 0.7979$, the expectation of normal distribution conditioned on being positive. $\endgroup$
    – A. Webb
    Commented Mar 17, 2015 at 19:00

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The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean, and dividing it by $n$.

$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$

Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...

Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.

It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign when they get squared.

Hope this helps,

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  • $\begingroup$ I see - so it is really just the difference between the sum of the squares vs. the square of the sums. I didn't think the difference would be that significant (20%) - just wasn't sure if something else was at play. $\endgroup$
    – Jon
    Commented Mar 17, 2015 at 17:10
  • $\begingroup$ I just edited the post; It is important to understand that negative and positive deviations are not treated separately. Don't calculate the sd for the negative and positive values in a separate step, a large negative deviation and a large positive deviation both let increase the variance in the same manner. $\endgroup$
    – bask0
    Commented Mar 17, 2015 at 17:19
  • $\begingroup$ "It is also wrong to..." could easily be misinterpreted. The separate contributions to the variance from negative and positive residuals are meaningful. They are used in financial risk assessment, for instance: see en.wikipedia.org/wiki/Downside_risk#Examples. $\endgroup$
    – whuber ♦
    Commented Mar 17, 2015 at 17:29
  • $\begingroup$ Ok, whuber, I didn't knew this. But that's a special case, I think for a general understanding of the variance/standard deviation, we can ignore this. However, thanks for the hint. $\endgroup$
    – bask0
    Commented Mar 17, 2015 at 21:45
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Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean, which is the standard deviation This follows from the triangle inequality, for example, or from Jensen's inequality.

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

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