In an article I found the formula for the standard deviation of a sample size $N$

$\sigma=\frac{\overline{R}}{2.534}$

where $\overline{R}$ is the average range of subsamples (size $6$) from the main sample. How the number $2.534$ is calculated? This is the correct number?

  • 6
    References please. More importantly: 1. There can't be a "correct number" here independently of the kind of distribution you are drawing from. 2. These rules usually come from interest in short-cut methods of estimating the SD from the range. Now we have computers.... Do you want to do that and why? Why not just use the data? – Nick Cox Sep 9 '13 at 16:28
  • 3
    @Nick Sorry: you were correct. A value around $4$ works for the standard deviation when the sample size is around $15$ to $50$; $3$ works for sample sizes around $10$, etc. I will delete my previous comment so it does not confuse anyone other than myself! – whuber Sep 9 '13 at 17:03
  • 1
    @NickCox it is old russian source and I didn't see the formula before. – Andy Sep 10 '13 at 6:35
  • 3
    Giving references is rarely a bad idea. Let readers decide for themselves whether they are interesting or accessible. (There are plenty of people here who can read Russian, for example.) – Nick Cox Sep 10 '13 at 10:53
up vote 17 down vote accepted

In an a sample $x$ of $n$ independent values from a distribution $F$ with pdf $f$, the pdf of the joint distribution of the extremes $\min(x)=x_{[1]}$ and $\max(x)=x_{[n]}$ is proportional to

$$f(x_{[1]})\left(F(x_{[n]})-F(x_{[1]})\right)^{n-2}f(x_{[n]})dx_{[1]}dx_{[n]} = H_F(x_{[1]}, x_{[n]})dx_{[1]}dx_{[n]}.$$

(The constant of proportionality is the reciprocal of the multinomial coefficient $\binom{n}{1,n-2,1} = n(n-1)$. Intuitively, this joint PDF expresses the chance of finding the smallest value in the range $[x_{[1]},x_{[1]}+dx_{[1]})$, the largest value in the range $[x_{[n]},x_{[n]}+dx_{[n]})$, and the middle $n-2$ values between them within the range $[x_{[1]}+dx_{[1]}, x_{[n]})$. When $F$ is continuous, we may replace that middle range by $(x_{[1]}, x_{[n]}]$, thereby neglecting only an "infinitesimal" amount of probability. The associated probabilities, to first order in the differentials, are $f(x_{[1]})dx_{[1]},$ $f(x_{[n]})dx_{[n]},$ and $F(x_{[n]})-F(x_{[1]}),$respectively, now making it obvious where the formula comes from.)

Taking the expectation of the range $x_{[n]} - x_{[1]}$ gives $2.53441\ \sigma$ for any Normal distribution with standard deviation $\sigma$ and $n=6$. The expected range as a multiple of $\sigma$ depends on the sample size $n$:

Normal

These values were computed by numerically integrating $\binom{n}{1,n-2,1}\left(y-x\right)H_F(x,y)dxdy$ over $\{(x,y)\in\mathbb{R}^2|x\le y\}$, with $F$ set to the standard Normal CDF, and dividing by the standard deviation of $F$ (which is just $1$).

A similar multiplicative relationship between the expected range and the standard deviation will hold for any location-scale family of distributions, because it is a property of the shape of the distribution alone. For instance, here is a comparable plot for uniform distributions:

Uniform

and exponential distributions:

Exponential

The values in the preceding two plots were obtained by exact--not numerical--integration, which is possible due to the relatively simple algebraic forms of $f$ and $F$ in each case. For the uniform distributions they equal $\frac{n-1}{(n+1)}\sqrt{12}$ and for the exponential distributions they are $\gamma + \psi(n) = \gamma + \frac{\Gamma'(n)}{\Gamma(n)}$ where $\gamma$ is Euler's constant and $\psi$ is the "polygamma" function, the logarithmic derivative of Euler's Gamma function.

Although they differ (because these distributions display a wide range of shapes), the three roughly agree around $n=6$, showing that the multiplier $2.5$ does not depend heavily on the shape and therefore can serve as an omnibus, robust assessment of the standard deviation when ranges of small subsamples are known. (Indeed, the very heavy-tailed Student $t$ distribution with three degrees of freedom still has a multiplier around $2.3$ for $n=6$, not far at all from $2.5$.)

  • 6
    Wonderful exposition! You may be interested to know that this appears to have been investigated back in the 1920s. See Tippet 1925. In Tippet's tables (Table X) the expected value for the range given a sample of size 6 is $2.53441\sigma$. He shows the derivation of the complete distribution of the range for the normal distribution. This was used by David et.al. (1954) to calculate probability points of the range distribution for a test for normality (see D'Agostino & Stephens 9.3.3.4.2). – Avraham Sep 9 '13 at 18:16
  • @Avraham Thank you for the illuminating comments. What struck me when I added the graphics is that the really clever part of this whole approach is the use of subsamples of size six because that's where the multipliers all tend to be about the same regardless of distributional shape. – whuber Sep 9 '13 at 18:58
  • Thanks! Tippet's tables actually give the appropriate multiplier for all numbers between 2 and 1000. He does mention running into calculation issues; of course, this was back in 1925 a good 20 years before ENIAC. – Avraham Sep 9 '13 at 19:03
  • @whuber can you show how the number (2.534) was calculated? – Andy Sep 10 '13 at 6:51
  • I edited the answer to include explanations of the calculations. – whuber Sep 10 '13 at 14:17

That approximation is very close to the true sample standard deviation. I wrote a quick R script to illustrate it:

x = sample(1:10000,6000,replace=TRUE)

B = 100000
R = rep(NA,B)
for(i in 1:B){
    samp = sample(x,6)
    R[i] = max(samp)-min(samp)
}

mean(R)/2.534

sd(x)

which yields:

> mean(R)/2.534
[1] 2819.238
> 
> sd(x)
[1] 2880.924

Now I am not sure (yet) why this works but it at least looks like (at face value) that the approximation is a decent one.

Edit: See @Whuber's exceptional comment (above) on why this works

  • 1
    You are drawing subsamples of size $6$ from an approximately uniform distribution. For a truly uniform distribution the ratio is $10\sqrt{3}/7\approx 2.474$. Indeed, if you were to use that factor in your simulation you would obtain mean(R)/2.474 equal to $2887.6$, very close to sd(x). – whuber Sep 9 '13 at 17:48
  • Very true! > mean(R)/2.474 [1] 2887.611 – user25658 Sep 9 '13 at 17:59

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