Re: standard error: bootstrap vs standard normal
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Tue, 12 Dec 2006 16:36:55 EST
1) The number of different Bootstraps is
(n elements taken WITH REPETITION in groups of k)
(n+k-1)Ck = (9+9-1)C9 = 17C9 = 24´310
2) Let be X1 , X2 , …, Xn a random sample of size n
obtained from X~N(m,s). I denote by Xbar it mean.
______V = 1/(n-1)* sum (Xj –Xbar)^2 the estimate of
f the variance s^2.
It is well known that (n-1)*V/s has a Chi-squared
Distribution with n-1 degrees of freedom.
This result is not well known. It is not well known because it is not true when the population variance, which you've denoted by s^2, is not equal to one.
In fact,
(n-1)*V/s ~ s*Chisq(n-1).
With s=1 and n=9 we obtain for a 0.34 tail
probability the value V=1.13 which gives s= 1.06
approximately.
3) By simulation (Monte Carlo) I obtained the
fraction 52.5% of N(0,1):9 showng s>1. In detail:
0.525, 0.527, 0.520, 0.528.
When s=1, P(V>1) = P(Chisq(8)>8) = 0.433. Your simulation results are way off.
Jack
This shows that the real Distribution of V is less.
concentrate than the one obtained throughout the
Bootstrap pseudo-samples which values, naturally, are
more close together because the items are drawn WITH
REPLACEMENT. Therefore are there repeated values
which naturally decreases the *span*.
_____licas (Luis A. Afonso)
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