Hi,
I have an honours degree in mathematics with a fair amount of
statistics content; unfortunately I haven't used the statistics
much since I graduated in 1975, so feel free to shoot down what
I have to say if you think I'm wrong.
Firstly, what we are measuring is the *maximum* value from a
sequence of observations; therefore we would not expect the
observed distribution to be normal, even if the underlying
distribution were. I would expect the distribution to have a
positive skew (i.e. to be "squeezed" on the side below the
mean, and "stretched" on the side above the mean).
As an example, I'm going to stick to numbers small enough to
check easily. The probabilities come from a standard set of
tables of the normal distribution, to four significant figures.
For _any_ distribution, if the probability that a random variable
is less than x is p, then the probability that all of a set of
N random variables from the same distribution are all less than
x is p^N, i.e. the probability that at least one of the N random
variables will be greater than or equal to x is 1 - p^N
Suppose that the random variables come from a normal distribution
with mean 0 and standard deviation 1. However we take the absolute
value of the variables - the distribution is symmetric about zero,
so this means that we are dealing with twice the integral from 0
to +infinity instead of the integral from -infinity to +infinity.
We take the maximum of 100 random variables drawn from this "half-
normal" distribution and wish to look at the distribution of the
new variable represented by this maximum function.
The following tabulated values should illustrate the shape of
the distribution. The left hand column is the value of x, the second
column is the value of the cumulative distribution function of the
normal distribution (from standard stats tables), the third value
is the corrected CDF for the upper half of the distribution, and the
fourth value is the CDF of the distribution of the maxima of 100
observations.
1.0 0.8413 0.6826 0.000000
1.1 0.8643 0.7286 0.000000
1.2 0.8849 0.7698 0.000000
1.3 0.9032 0.8064 0.000000
1.4 0.9192 0.8384 0.000000
1.5 0.9332 0.8664 0.000001
1.6 0.9452 0.8904 0.000009
1.7 0.9554 0.9108 0.000088
1.8 0.9641 0.9282 0.000581
1.9 0.9713 0.9426 0.002709
2.0 0.9772 0.9544 0.009398
2.1 0.9821 0.9642 0.026104
2.2 0.9861 0.9722 0.059643
2.3 0.9893 0.9786 0.114953
2.4 0.9918 0.9836 0.191360
2.5 0.9938 0.9876 0.287150
2.6 0.9953 0.9906 0.388895
2.7 0.9965 0.9930 0.495364
2.8 0.9974 0.9948 0.593715
2.9 0.9981 0.9962 0.683367
3.0 0.9986 0.9972 0.755487
3.1 0.9990 0.9980 0.818567
3.2 0.9993 0.9986 0.869273
3.3 0.9995 0.9990 0.904792
3.4 0.9997 0.9994 0.941748
3.5 0.9998 0.9996 0.960782
.. at which point the four digit precision of the tables becomes
insufficient to make it worthwhile tabulating more data.
But I think it's clear that the CDF rises more slowly on the side
above the median (just above 2.7) than it does on the lower side.
(Sorry I chose 100 now - there is an unfortunate coincidence between
"just above 2.7" and e, which is accidental - in fact the median is
more accurately 2.705)
What does all this mean? Unfortunately I can't see much, except that
given a value which is the maximum of a set of 100 data values drawn
from a population of positive real values, we can (swallowing hard
and trusting in the central limit theorem) deduce that the standard
deviation of the underlying distribution is about 1/2.7 of the given
value.
Perhaps this is therefore the wrong angle to attack the problem from.
....................................................................
Could I suggest that the maximum convolution error is likely to occur
when whole blocks of bits are 1, thus forcing large integers into the
transform so that rounding/truncation errors are most likely to occur.
Consider the LL test algorithm x' = (x^2 - 2) (mod 2^p - 1)
If x = 1, then x' = -1, x'' = +1, x''' = -1, etc, etc, for all values
of p. And -1 (mod 2^p - 1) = (2^p - 2) (mod 2^p -1). Furthermore
2^p - 2 = 2 * (2^(p-1) - 1), a number which contains (p-1) consecutive
1 bits.
So running a couple of iterations of the LL test algorithm starting
at 1 (instead of 4), looking at the max convolution error and checking
that the value of the residual after 2 iterations is 1, might be
instructive ... ?
Regards
Brian Beesley
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