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Topic: problem on record-breaking values in probability
Replies: 14   Last Post: Apr 14, 2013 11:36 PM

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James Waldby

Posts: 396
Registered: 1/27/11
Re: problem on record-breaking values in probability
Posted: Feb 27, 2013 4:05 PM
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On Wed, 27 Feb 2013 07:10:08 -0500, David Bernier wrote:
> On 02/27/2013 05:49 AM, David Bernier wrote:
>> On 02/27/2013 05:31 AM, David Bernier wrote:
>>> I used Marsaglia's 64-bit SUPER KISS pseudo-random number generator
>>> to simulate uniform r.v.s on [0, 1] that are independent, as
>>> X_1, X_2, X_3, ad infinitum
>>> For each go, (or sequence) I define its 1st record-breaking value
>>> as R(1) as X_1, its 2nd record-breaking value R(2) as the
>>> value taken by X_n for the smallest n with X_n > X_1, and in general

[ R(k+1) is the value taken by X_n for the smallest n with X_n > R(k)]
>>> In my first simulation I get: R(20) = 0.999999999945556
>>> or about 5.4E-11 less than 1 , a one in 18 billion event.

>>> In fact, R(20) is about 1 - (0.307)^20 ...
> I finally got another 20th record-breaking value, on my
> second sequence, but it took much longer. The values
> 1 - R(20), one per sequence, are what I call "p-values"
> from the simulations so far, a lot of variance in orders
> of magnitude:
> the corresponding p-value is 0.000000000054 // seq. 1
> the corresponding p-value is 0.000000000001 // seq. 2
> the corresponding p-value is 0.000000002463
> the corresponding p-value is 0.000000000782
> the corresponding p-value is 0.000000106993
> the corresponding p-value is 0.000000342142
> the corresponding p-value is 0.000000001978


It would be useful to report the number of trials each simulation
took to find its 20th RBV. If a simulation takes m trials, the
variance of the value X_m is approximately 1/(m^2), where X_m is
the mth smallest (ie the largest) number among m trials. (See
Your simulation data implies there is a wide variance among the
values of m required to find a 20th RBV.

In following, let L(n) = Pr(n'th item of n is lowest). (Distribution
of the lowest item should be similar to distribution of 1-(highest
item).) I suppose that L(n) = 1/n and that the expected value of the
number of record-low-values (RLV's) in m trials is sum{i=1 to m}(1/i),
or about H_m, the m'th harmonic number, which can be approximated by
log(m) + gamma, with gamma = Euler-Mascheroni constant, about 0.5772.
When H_m ~ 20, log(m) ~ 19.42 and m is about 272 million. That is,
simulations to find the 20th RLV or RBV will take about 272 million
steps, on average. (I don't know which kind of average applies, or
what the variance might be.)


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