Re: Random number generation using radioactivity

From: Ernst Lippe (ernstl-at-planet-dot-nl_at_ignore.this)
Date: 08/15/04 

Date: Sun, 15 Aug 2004 12:50:46 +0200

On Mon, 09 Aug 2004 08:27:33 +0200, Michael Jørgensen wrote:

>
> <juuitchan@hotmail.com> wrote in message
> news:9159d95e.0408082112.428be67b@posting.google.com...
>> I wonder if this would be a good method for random number generation
>> using radioactive decay:
>>
>> You have a radioactive source and a Geiger counter. This setup is
>> connected to a computer. The computer's internal timer is good enough to
>> split the average time between two consecutive decays into several
>> hundred parts.
>>
>> You take the intervals between successive decays modulo P (P is a
>> smallish prime number, like 41 or 59). When you get about 30 or so of
>> these numbers, you concatenate them, read it as a single base-P number,
>> convert it to decimal, throw away the first 10 or so digits, and keep
>> the rest as random digits.
>
> Well, it won't be *exactly* uniform.
Indeed, it probably won't be uniform.

> Each measured interval follows an known distribution (geometric?)
The expected distribution is more or less similar to the
exponential distribution for large time intervals. But because
detectors generally have a dead-time during which they cannot
detect a new decay, this approximation is not valid for small intervals,
because there will be too few of them.

> with an
> average value of around 200 (you wrote "several hundred parts"). Taken
> modulo 41, we have a random integer in the range [0, 40]. The distribution
> is still not uniform, there will be a substantial bias towards low
> numbers.
>
> What happens after that I'm not quite sure about. However, if we look at
> information content, then each measurement gives you approximately 5 bits
> of information. After collecting 30 values and converting to decimal and
> throwing away 10 digits we get (5*30 - 10*3) = 120 bits of information.
> That should give you approximately 40 random digits, but keeping the
> non-uniformity in mind, I would not trust them all to be independent.
With perfect instruments, the results would be independent, but
they would still be biased.

> Here's an alternate approach, that tries to achieve uniformity:
>
> [disclaimer: This is something I just thought of, while replying to this
> post. Use at your own risk!]
>
> Take the difference between pairwise measurements. This gives an integer
> in the range [-40, 40], with a peak at the value 0. Now add together 8 of
> such values (using a total of 16 measurements). This gives an integer in
> the range [-15*40, 15*40] which almost follows a normal distribution. You
> must now normalize, so that the standard deviation becomes 1.
The problem with adding these values is that you loose a lot of entropy
in this process. Also you don't know a priori the standard deviation
of the resulting values and estimating it from the sampled values is
both complicated and dangerous.
 
> Now if my memory serves me well; if X and Y are normal distributed
> independent variables, then exp(-(X^2 + Y^2)) is uniformly distributed in
> the interval ]0,1].

No, you need to use the inverse of the cumulative normal distribution.

Ernst Lippe

Relevant Pages

  • Re: Chi square for small frequencies
    ... The raw data are actually counts, it's often the case that the counts ... distributed may be a reasonable approximation because the Poisson ... approaches a normal distribution in the limiting case. ... minimizing chi-square. ...
    (sci.stat.edu)
  • Re: statistics
    ... I am looking for an approximation a formula for calculating ... random variable with normal distribution with ux & varx. ...
    (comp.soft-sys.matlab)
  • Re: Probability distribution of throwing dice
    ... each dice, k=the numbers of die, and S=the sum of both die? ... S has an approximate normal distribution. ... value of k, the better the approximation. ... thank you for helping a mathematically ...
    (sci.math)
  • Re: Applying the Poisson distribution.
    ... so a Poisson approximation for the number of no-shows is ... > where Z has a standard normal distribution. ... Is the normal distribution not appropriate for this reason as well? ...
    (sci.math)