Re: A question on a simple algorithm to apprximate normal distribution
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Sat, 6 Jun 2009 00:23:12 -0700 (PDT)
On Jun 5, 9:05 pm, Adan Mithrillion <diablodestruct...@xxxxxxxxx>
wrote:
If we add n random variables x[i] uniformly distributed in (0,1),and
subtract the sum by n/2, the result is approximately normally
distributed. What is the relationship between the range of the random
variable (x[i]) or the number of variables added (n) and the variance
of the result? In other words, how can I apply this algorithm to a
situation where a given variance must be satisfied?
-A.M.
The variance of a sum (of independent r.v.s) is the sum of their
variances, so for n terms Var = n*Var1, where Var1 = variance of
uniform (0,1). For Uniform (a,b) the variance is (b-a)^2/12, so for
(0,1) it is 1/12. Thus, Var = n/12. For this reason, n is often chosen
as 12, since that makes the variance = 1 without the need for re-
scaling. See
http://en.wikipedia.org/wiki/Uniform_distribution_(continuous) for
material about the uniform distribution.
R.G. Vickson
.
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- From: Adan Mithrillion
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