Re: Generating "independent" random numbers

[Maxime Devos]

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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