From: Ted Harding <Ted.Harding_at_nessie.mcc.ac.uk>

Date: Fri 13 May 2005 - 23:00:00 EST

E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 094 0861

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Fri May 13 23:11:05 2005

Date: Fri 13 May 2005 - 23:00:00 EST

On 13-May-05 Bliese, Paul D LTC USAMH wrote:

> Interesting thread. The graphics are great, the only thing that

*> might be worth doing for teaching purposes would be to illustrate
**> the original distribution that is being averaged 1000 times.
**>
**> Below is one option based on Bill Venables code. Note that to do
**> this I had to start with a k of 2.
**>
**> N <- 10000
**> for(k in 2:20) {
**> graphics.off()
**> par(mfrow = c(2,2), pty = "s")
**> hist(((runif(k))-0.5)*sqrt(12*k),main="Example Distribution 1")
**> hist(((runif(k))-0.5)*sqrt(12*k),main="Example Distribution 2")
**> m <- replicate(N, (mean(runif(k))-0.5)*sqrt(12*k))
**> hist(m, breaks = "FD", xlim = c(-4,4), main = k,
**> prob = TRUE, ylim = c(0,0.5), col = "lemonchiffon")
**> pu <- par("usr")[1:2]
**> x <- seq(pu[1], pu[2], len = 500)
**> lines(x, dnorm(x), col = "red")
**> qqnorm(m, ylim = c(-4,4), xlim = c(-4,4), pch = ".", col = "blue")
**> abline(0, 1, col = "red")
**> Sys.sleep(3)
**> }
**>
**> By the way, I should probably know this but what is the logic of
**> the "sqrt(12*k)" part of the example? Obviously as k increases
**> the mean will approach .5 in a uniform distribution, so
**> runif(k)-.5 will be close to zero, and sqrt(12*k) increases as
**> k increases. Why 12, though?
*

The reason is indeed simple! In demonstrating the convergence of the distribution of mean(k X's) to a Normal distribution, the reference (i.e. the limiting distribution) is N(0,1), which has mean 0 and variance 1. Therefore, in comparing the distribution of mean(k X's) with N(0,1) it needs to be standardised to itself have mean 0 and variance 1. As you've already spotted, you standardise for the mean by subtracting 0.5; to standardise for the variance you need to divide by sqrt(variance(mean(k X's))).

This is sqrt(variance(X)/k). Finally (and this is where the "12" comes in), the variance of an X uniformly distributed on (0,1) is 1/12 (left as an exercise for the reader ... ). Hence 12*k.

Best wishes,

Ted.

E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 094 0861

Date: 13-May-05 Time: 13:43:54 ------------------------------ XFMail ------------------------------ ______________________________________________R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Fri May 13 23:11:05 2005

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