Re: Generating "independent" random numbers

[Maxime Devos <maximedevos@telenet.be>]

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Op 11-10-2023 om 01:04 schreef Keith Wright:
> Maxime Devos <maximedevos@telenet.be> writes:
> 
>> Op 04-10-2023 om 18:14 schreef Keith Wright:
>>> From: Zelphir Kaltstahl <zelphirkaltstahl@posteo.de>
>>>
>>>> my goal is normal distributed floats (leaving aside the finite
>>>> nature of the computer and floats).
> 
>>> The following is either provably correct up to round-off,
>>> or totally stupid.
> 
>> It's not stupid at all, but neither is it correct (the basic approach is
>> correct, and holds more generally for other probability distributions as
>> well, but some important details are off ...).
> 
> Not surprising, I just made it up on the fly...
> 
>>> First define the cumulative distribution for the normal distribution:
>>>
>>> $$f(x)= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{x} e^{-x^2/2} dx $$
>>
>> Instead of sqrt(pi) you need sqrt(2pi).
> 
> Seems plausible.
> 
>> Also, this is the pdf, not the cdf. For the cdf, you need integrate this
>> expression from -infinity to x.
> 
> I was using TeX notation.  That's exactly what $\int_{-\infty}^{x}$
> means.

Ok, didn't notice the \int.

>>> Computing the inverse of the cumulative normal distribution is left as
>>> an exercise, because I don't know how, but it seems possible.
> 
>> While the pdf is easy to invert (multiply by constant factor, take log,
>> multiply by constant factor), resulting in an expression only involving
>> constants, multiplication, logarithms (and, depending on simplification,
>> subtraction), the cdf isn't.
> 
> I was thinking of numerical integration, but too busy to write the code.
 >
> I was feeling guilty about being so sloppy, and thinking of writing
> it more carefully, but...
> 
>> (the basic approach is correct, and holds more generally for other
>> probability distributions as well)
> 
> I just saw some pictures in my head and thought it must be a general
> theorem, but it's so simple and general that it must be "well known".
> Somebody sometime must have already written it better than I could;
> but I don't know who.  It is Theorem (X?) on page (Y?) of book (Z?).

You can probably find it in most course texts on probability and random 
number generation.  I could provide you a reference, but I don't think 
my source is publicly available.

> -- Keith
>     
> 
> PS: While we are cleaning up details, substitute "modulo" for "/" in:
> 
> From: Maxime Devos <maximedevos@telenet.be>
> 
>> So, to generate an (approximately) uniform random number on [0,1), you
>> can simply do
>>
>> (define (random-real)
>>      (exact->inexact (/ (random N) N)))
>>
>> for a suitably l

I'm pretty sure '/' is correct here.  (random N) produces integers in 
[0,N), dividing by N yields something in [0,1).

Best regards,
Maxime Devos.

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