       Re: bimodal distribution in sign of difference of Pi digits]

• To: mathgroup at smc.vnet.net
• Subject: [mg51833] Re: bimodal distribution in sign of difference of Pi digits]
• From: Roger Bagula <tftn at earthlink.net>
• Date: Wed, 3 Nov 2004 01:24:26 -0500 (EST)
• References: <cm21dn\$gab\$1@smc.vnet.net> <cm7cv1\$lh6\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```The correct probability is {40/90,10/19,40/90) for the Sign types (1,0,-1).
The model of the probabilities becomes: ( with no zero state )

f[n_]:=f[n]=f[n-1]+Random[Integer,{1,40}]/90-Random[Integer,{1,40}]/90
f=Random[Integer,{1,40}]/90-Random[Integer,{1,40}]/90

My friend has used stepwise calculation
in Mathematica to go to a very high number of Pi digits
( 10 of millions) and the deviations from zero still remains and grows.
It appears there is no "ideal" of randomness that can be reached by our
current
methods of calculation.
Roger Bagula wrote:

>I'm learning to analyze this type of problem.
>The sum comes down to a cumlative integer probality sum ( Sign is only
>integer).
>I simulated it using the (a,b) two simple probabilities of 10 symbols to
>get
>(7/18,1/9,7/18) out 90 possible states.( 2*Binomial[10,2])
>The result behaves just as the other digits simulations did without
>using the digits:
>I also realize that "independent" probabilities may be an "ideal" myth
>as nothing comes from nowhere, but still is is the ideal from such
>probaility as a thought experiment.
>Thus, using a pseudorandom that is in any tinture Markov  or dependent
>on it's history is
>a "fault" to the simulation.
>The trouble is we actually lack an ideal probability type pseudorandom.
>No such algorithm exist as far as I know
> or have been able to search out in the last 30 years of study.
>(* simulation of 10's digits equal probabuility  (a,b) independently*)
>(* using (7/18,1/9,7/18) as probabilities that the Sign of the
>difference is (1,0,-1)*)
>digits=50000
>SeedRandom[Random[Integer,digits]]
>f[n_]:=f[n]=f[n-1]+Random[Integer,{0,7}]/18-Random[Integer,{0,7}]/18
>f=Random[Integer,{0,7}]/18-Random[Integer,{0,7}]/18
>a=Table[Floor[f[n]],{n,1,digits}];
>ListPlot[a,PlotJoined->True]
>b=Flatten@{0,Length/@Split[Sort@a], 0}
>ListPlot[b,PlotJoined->True];
>
>Roger Bagula wrote:
>
>
>
>>Dear jasonp,
>>I don't know.
>>This method is a new way to investigate Pi digits.
>>I had done some counts of base ten digits frequencies before this.
>>I have no real explaination of why the difference is higher in higher number of digits.
>>The groups of positive "Sign"s should
>>random. It is Sign[x]-> {-1,0,1} depending on the difference in consecutive
>>differences. It is the probability of a digit pair:
>>{a,b}--> Sign[a-b]
>>p=Probability [a]*Probability[b]
>>If they are equal as p0:
>>p->p0^2
>>If the Mathematica for such a probability would be:
>>p0->Random[Integer,{0,9}] as a Distribution
>>Since this is an straight type probabilty and not a Gaussian
>>the probabilies are equal and should be over a long term
>>1/10 each or a total of
>>p-->1/100
>>different for different combinations:
>>{a>b}->+1,{a-1},{a=b}->0
>>at {4/10,4/10,2/10} that gives something like
>>4/1000,4/1000,2/1000
>>I'm not seeing that kind of behavior except for the bimodal
>>which is expected as
>>(a=b) is
>>only about 2/10 of the 1/100 and I'm seeing more zeros than that.
>>It appears to be a much more complex distribution.
>>I want to try E and orther irrational numbers by this method as well!
>>until now!
>>I can simulate the probability above in Mathematica
>>and see what I get
>>and compare them.
>>jasonp at boo.net wrote:
>>
>>
>>
>>
>>
>>>Quoting Roger Bagula :
>>>
>>>
>>>
>>>
>>>
>>>>(* Sum of the sign of the differences between the first 2000 digits of Pi*)
>>>>  >>
>>>>
>>>>
>>>Shouldn't this behave like a random walk, i.e. the variance
>>>increases over time?
>>>
>>>jasonp
>>>
>>>
>>>------------------------------------------------------
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>>>
>>>
>>>
>>>
>>>
>>
>>
>
>
>

--
Respectfully, Roger L. Bagula