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

• To: mathgroup at smc.vnet.net
• Subject: [mg51855] Re: bimodal distribution in sign of difference of Pi digits]
• Date: Wed, 3 Nov 2004 01:26:18 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```On 11/2/04 at 2:06 AM, tftn at earthlink.net (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.

This is an error. If you have 10 distinct digits then there are 100 possible pairs of thoose digits. It should be obvious the integers from 0 t0 99 reflect this fact, i.e.,

Table[IntegerDigits[n, 10, 2], {n, 0, 99}] ==
Union[Table[IntegerDigits[n, 10, 2], {n, 0, 99}]]
True

showing as should be expected there are no duplicates

From this, it should be equally obvious there are exacly 10 pairs where both digits are equal. Further a moments thought should convince you the number of cases where the first digit is greater than the second must equalt the number of cases where the first digit is less than the second. So, Sign[x-y] will be one of {-1,0,1} with probabilities {9/20,1/10,9/20}

If you wanted, you could use this to compute the probability of finding various values in the cumulative sums you formed in your earlier posts.

For example, the first digit of the cumulative sum must be one of {-1,0,1} with the probabilities given above.

and

Union[Apply[Plus, Flatten[Outer[List, {-1, 0, 1},
{-1, 0, 1}], 1], {1}]]

gives

{-2, -1, 0, 1, 2}

the only possible values for the second sum. And the probabilities will be

{81/400,9/100,83/200,9/100,81/400}

This could be continued to compute the exact probability for the cumulative sums of any desired number of random digits but clearly the compuational difficulty rapidly increases with the number of random digits chosen. But even without actually computing the probabilities, the first few make it clear the probabilites of each value are not constant. That is the distribution of cumulative sums will not be uniform.

>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.

is true, the deviations you are seeing from a true uniform random number generator have a lot more to your analysis than problems with Random[Integer, {0,9}]
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```

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