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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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David Bernier

Posts: 3,892
Registered: 12/13/04
Re: problem on record-breaking values in probability
Posted: Mar 1, 2013 8:41 AM
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On 02/27/2013 10:24 PM, David Bernier wrote:
> On 02/27/2013 04:05 PM, James Waldby wrote:
>> 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

>> [etc]
>> 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.)

> Yes, I hadn't thought of using expectations and expected values,
> leading to partial sums of the harmonic series.
> You wrote:
> "simulations to find the 20th RLV or RBV will take about 272 million
> steps".
> Right, that looks like an interesting way of looking at
> record values.
> Since it's like a stopping time, the number of the simulation,
> I suppose it could be denoted S({X_i}_{i=1, ... oo}, 20)
> or S(X_{BOLD_overligned}, 20).
> Electricity should go off tomorrow morning for repair.
> I think your estimate
> E(log(S_20)) ~= 20 - gamma
> should be very good.

In the literature, a remarkable article, which may have
appeared in the Am. Math. Monthly, can be found by
searching for:
Breaking Records and Breaking Boards. Ned Glick

The author is Ned Glick, at some time at UBC in Canada.

David Bernier
dracut:/# lvm vgcfgrestore
File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID
993: sh
Please specify a *single* volume group to restore.

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