[From: http://en.wikipedia.org/wiki/Skewness ]
If Y is the sum of n independent and identically distributed random variables, all
with the distribution of X, then the third cumulant of Y is n times that of X, and
the second cumulant of Y is n times that of X, so Skew[Y] = Skew[X]/sqrt(n).
This shows that the skewness of the sum is smaller, as it approaches a Gaussian
distribution in accordance with the central limit theorem.
[From: http://en.wikipedia.org/wiki/Kurtosis ]
Suppose that Y is the sum of n identically distributed independent random
variables all with the same distribution as X. Then
Kurt[Y] =1/n Kurt[X] (where Kurt[] is excess kurtosis)
More generally, if X1, ..., Xn are independent random variables, not necessarily
identically distributed, but all having the same variance, then
Kurt[Sum(X_i)) = 1/n^2 Sum(Kurt[X_i])
summed from i = 1 to n
These supply rules of thumb for estimating how many terms must be summed in order to produce
a Gaussian to some degree of approximation; the skewness and excess kurtosis must both be below
some limits, respectively. These are necessary but possibly not sufficient conditions; it is not clear
how both requirements could be met without also implying a good Gaussian approximation, but
there may be some clever artificially rigged mathematical construction that could have the right
moments but the wrong shape (e.g., analogous to the piecewise joint density function for two
Gaussians that are uncorrelated but not joint-Gaussian).
Example: p(x) = 5/x^6 , x > 1
This is the inverse-power law with the smallest integer exponent in the denominator that admits a
finite kurtosis. It has mean = 5/4, sigma = %(15)/5 . 0.32275, skewness = 6%(15)/5 . 4.647580015,
and kurtosis = 369/5, so excess kurtosis of about 70.8. It would take 10,000 random variables
summed to get the skewness down to about 0.05; excess kurtosis would have dropped to about
0.007. The skewness is the more difficult criterion.
Monte Carlo: average 10,000 draws, 100 thousand trials.
mean(Avg):
sigma(Avg):
skew(Avg):
kurt(Avg):
1.2499988323022309
0.0031741394047122254
0.057876526852386284
2.9970156109572583
The histogram, with abscissa n = 1000*Avg - 1000, is:
The underlying density function p(x) is: