2
$\begingroup$

When we speak about the 3xSD rule for the normal distribution, or the Tschebyshev inequality, saying how many data are covered by the appropriate multiplicity of SDs, do we mean the square root of the second central moment, or the distribution-characteristic measure?

For example, for the normal distribution, we have mean +- 3xSD, but correspondingly, for the log-normal distribution we have the geom. mean */ geom. SD^3. The two are equivalent, if the "mean +- 3SD" is on the log scale.

So now my question is: are all the theorems expressed in means and SDs based on the classic 1st raw moment and the sqrt(2nd central moment), or on their distriution-related measures?

I am confused - the mean+-3SD doesn't make any sense for multiplicative distribution on the untransformed scale, so it seems it's on the log scale, which means we use now the log-normal specific geometric counterparts of the mean and SD.

Does the Tschebyshev inequality also use the "distribution specific" (here: geometric mean and SD for the Log-normal) measures?

But in every textbook I saw, the mean was the sum of the data divided by its amount, and the SD was the square root of the sum of squared differences of the data from mean. There was no "logarithms" or anything else "distribution specific" mentioned at all!

$\endgroup$
1
  • 1
    $\begingroup$ The three sigma rule is intended to apply only to distributions which can be approximated sufficiently well by a normal distribution. E.g. it does not apply to a Bernoulli distribution. Chebyshev's inequality, using the usual definition of the standard deviation, is true for any distribution with finite mean and variance. $\endgroup$
    – fblundun
    Mar 21 '21 at 15:18
1
$\begingroup$

Comment: The Empirical Rule states that for a 'mound shaped' (presumably that means approximately normal) sample, about 95% of the observations lie within $\bar X \pm 2S.$ Although some elementary texts seem to 'cherry-pick' particular datasets for which results are very near 95%, the ER does work pretty well in practice.

x = rnorm(100)
a = mean(x);  s = sd(x)
mean(x > a-2*s & x < a+2*s)
[1] 0.98

x = rnorm(100)
a = mean(x);  s = sd(x)
mean(x > a-2*s & x < a+2*s)
[1] 0.92

However, I was surprised to discover that this part of the ER also seems to apply to a variety of distinctly non-normal continuous distributions. I illustrate this for standard exponential and for $\mathsf{Beta}(2,3).$

x = rexp(1000)
a = mean(x); s = sd(x)
mean(x > a-2*s & x < a+2*s)
[1] 0.952

x = rbeta(1000, 2, 3)
a = mean(x); s = sd(x)
mean(x >= a-2*s & x <= a+2*s)
[1] 0.974

Also, this part of the ER 'works' for some discrete distributions, that converge to normal, even when parameters do not provide a roughly normal shape.

x = rbinom(1000, 20, .1)
a = mean(x); s = sd(x)
mean(x > a-2*s & x < a+2*s)
[1] 0.961

x = rpois(100, 10)
a = mean(x); s = sd(x)
mean(x >= a-2*s & x <= a+2*s)
[1] 0.94

If anyone has an explanation for this, I would be interested to see it.


The three-SD version of the ER states that 'all or almost all' of the sample values lie within $\bar X \pm 3S.$ Because version suggests a proportion near $1$ and because Chebyshev's inequality guarantees $8/9,$ perhaps it is less surprising that it also works well for some distinctly non-normal samples.

x = rnorm(100)
a = mean(x);  s = sd(x)
mean(x > a-3*s & x < a+3*s)
[1] 1

x = rexp(100)
a = mean(x);  s = sd(x)
mean(x > a-3*s & x < a+3*s)
[1] 0.99

x = rbeta(100, 2, 5)
a = mean(x);  s = sd(x)
mean(x > a-3*s & x < a+3*s)
[1] 0.99

x = rpois(100, 7)
a = mean(x);  s = sd(x)
mean(x > a-3*s & x < a+3*s)
[1] 0.99
$\endgroup$
1
  • $\begingroup$ Thank you very much! Indeed, I made a small simulation and the equivalency of the m-+3s for the normal vs. gm/*s^3 for the lognormal is immediate, but also if the measures are used oppositely, it works well, as you noticed - close to the Chebyshev,s inequality. > set.seed(1000) > x <- rlnorm(1000, 0, 1) "Amount of observations of the log-normal for m-+3s: 97.5%" "Amount of observations of the log-normal for gm/*s^3: 99.8%" > x <- rnorm(1000, 10, 1) #as above "Amount of observations of the normal for m-+3s: 99.8%" "Amount of observations of the normal for gm/*s^3: 99.5%" $\endgroup$ Mar 21 '21 at 20:31

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.