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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: Feb 27, 2013 10:24 PM
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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

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.

The higher moments (and the variance) of
log(S_20) are still something of a mystery to me.

I propose to change 20 to 15 or 16 because it takes
a long time with 20th RBV simulations.

David Bernier
$ pwd

$ ls -l
david 9203 Feb 27 04:48 a.out
david 1866 Feb 27 04:48 test17a.c

gcc -std=c89 test17a.c
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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