In less mathematical terms (but only slightly): Assume the chance of x falling between x_i and x_i+dx is as close to P(x_i)*dx as you can get, with dx very small. Then y_i = D(x_i), with formula 10 connection the two functions — meaning P(x)*dx ≈ D(x+dx)-D(x) for all x, is the chance that x < x_i. This chance is between 0 (for x_i = -inf) to 1 (for x_i = +inf).

Note that Perl's rand function returns an approximately uniform distribution between 0 and 1. Let's say you pick one of these random numbers, use them as the current value for y_i, and via the inverse function of D(x) get back a value for x_i. Well: those values for x that you get back this way, are distributed with probability function P(x). Et voilą.

So all you need is sub implementing approximately the inverse function for D(x).

Update2: I don't think this is exactly how I did it before, but it does appear to work.

#! perl -slw
use strict;
use List::Util qw[ sum min max ];

our $N ||= 10_000;
our $MEAN ||= 75;

sub skewedRnd {
    my( $start, $end, $skewedMean ) = @_;
    my $low = $skewedMean - $start;
    my $high = $end - $skewedMean;
    return ( rand() < 1 - ( $skewedMean / ( $end - $start ) ) )
        ? $start      + rand( $low )
        : $skewedMean + rand( $high );
}

my @values = map skewedRnd( 0, 100, $MEAN ), 1 .. $N;

printf "range 0 .. 100; Min: %f Max:%f Mean:%f \n",
    min( @values ), max( @values ), sum( @values ) / $N;

__END__
[23:52:06.41] P:\test>425761 -N=1000000 -MEAN=99
range 0 .. 100; Min: 0.003021 Max:99.999969 Mean:98.999126

[23:52:28.58] P:\test>425761 -N=1000 -MEAN=99
range 0 .. 100; Min: 5.154236 Max:99.997009 Mean:98.727916

[23:52:41.92] P:\test>425761 -N=1000 -MEAN=20
range 0 .. 100; Min: 0.029907 Max:99.995117 Mean:19.924415

[23:52:51.27] P:\test>425761 -N=1000 -MEAN=10
range 0 .. 100; Min: 0.008850 Max:97.561035 Mean:10.563889

[23:52:55.64] P:\test>425761 -N=1000 -MEAN=75
range 0 .. 100; Min: 0.144196 Max:99.998474 Mean:74.406415

[23:53:01.49] P:\test>425761 -N=1000 -MEAN=90
range 0 .. 100; Min: 1.606750 Max:99.976501 Mean:90.610480

[23:53:08.42] P:\test>425761 -N=1000 -MEAN=51
range 0 .. 100; Min: 0.049805 Max:99.974579 Mean:50.391814

[23:53:33.88] P:\test>425761 -N=1000 -MEAN=51
range 0 .. 100; Min: 0.221008 Max:99.923737 Mean:50.118810

[23:53:36.24] P:\test>425761 -N=10000 -MEAN=51
range 0 .. 100; Min: 0.001556 Max:99.998505 Mean:51.276380

[23:53:40.10] P:\test>425761 -N=100000 -MEAN=51
range 0 .. 100; Min: 0.000000 Max:99.997009 Mean:51.018693

[23:53:44.44] P:\test>425761 -N=1000000 -MEAN=51
range 0 .. 100; Min: 0.000000 Max:99.998505 Mean:50.992506
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UPDATE: Below here is wrong--

I'll let the math guys pick it apart, and offer it on the basis that it is simple, understandable and served my purpose. Maybe it will serve yours too.

This really isn't a Perl problem, and it's not a really well-defined statistics/math problem, either.

If I understand the question, you want to change the mean value of a random function to a given value, right? This is a linear transformation, and it is very simple to implement:

or, create a new function:

The idea is to produce random numbers within a specified range, but with a skewed average. What you describing wil not only transform the average, but also the range.

Eg. random numbers in the 0 .. 100 will, over time, tend towards an average of 50. If the desired average was say 75, then your math will produce a set of values in the range:

( 0 - 50 + 75 ) .. ( 100 - 50 + 75 ) := 25 .. 125

Which will tend towards the desired average, but is the wrong range. (I did a similar thing with my first attempt above).

The transformation required is not linear, but statistical.

---

Examine what is said, not who speaks.

Silence betokens consent.

Love the truth but pardon error.

Actually, the AM says they want random numbers distributed normally, and that the range is specified as (mean - delta, mean + delta).

I believe they want to change the mean, and keep the delta.