_GRandom()

[Yashied]

I recently had the need to generate random numbers that are distributed according to the law of Gauss. The fact that the function Random() generates a uniformly distributed random numbers. Histogram for the function Random() is as follows (in the range from -1 to 1).

 

By the way, the histogram for RtlRandomEx() function from ntdll.dll is as follows (much worse than for the Random() function - Mersenne Twister).

 

I wrote a function _GRandom() similar to Random(), but returns a numbers that are distributed according to the law of Gauss. I hope that it will be useful for someone.

 

Func _GRandom($Min = 0, $Max = 1, $Variance = Default, $Mean = Default)

    Local $D, $X1, $X2, $Result

    If $Min > $Max Then
        Return SetError(1, 0, 0)
    EndIf
    If $Min = $Max Then
        Return $Min
    EndIf
    If $Mean = Default Then
        $Mean = ($Min + $Max) / 2
    EndIf
    If $Variance = Default Then
        $Variance = ($Max - $Min) / 2
    EndIf
    If ($Mean < $Min) Or ($Mean > $Max) Then
        Return SetError(1, 0, 0)
    EndIf
    If $Variance <= 0 Then
        Return SetError(1, 0, 0)
    EndIf
    Do
        Do
            $X1 = Random(-1, 1)
            $X2 = Random(-1, 1)
            $D = $X1 ^ 2 + $X2 ^ 2
        Until $D <= 1
        $Result = $Mean + $X1 * $Variance * Sqrt(-2 * Log($D) / $D)
    Until ($Result >= $Min) And (($Result <= $Max))
    Return $Result
EndFunc   ;==>_GRandom

$Min

The smallest number to be generated. The default is 0.

$Max

The largest number to be generated. The default is 1.

$Variance

Variance relative to the range that specifies by $Min and $Max parameters.

$Mean

Expected value within the range of $Mim...$Max. The default is the value in the middle of this range - 0.5.

Below shows a histogram for this function.

_GRandom(-1, 1, 0.3, 0)

All histograms were drawn by using this function.

Edited December 22, 2013 by Yashied

[andybiochem]

Sorry to be critical, but if you are selectively rejecting numbers based on an upper and lower limit (min,max), then that automatically prevents the distribution from being Gaussian. The true Gaussian integral spans from -∞ to +∞.

What you are doing is rejecting everything outside of +/- 2 SD of the mean. Whilst you may end up with a bell-curve of probability favouring the mean (which is probably useful in itself), it isn't Gaussian.

The most widely accepted true gaussian algorithms are the Box-Muller method and the Ziggurat Method.

[EDIT]

Also, what you term 'variance' is actually standard deviation. For example, this:

_GRandom(0,200,100,100)

should give me a data set with mean = 100 and SD = 10, but it actually gives an SD of 53 ish.

Edited February 7, 2010 by andybiochem

[ValeryVal]

I wrote a function _GRandom() similar to Random(), but returns a numbers that are distributed according to the law of Gauss. I hope that it will be useful for someone.

Your UDF is invalid.

See about the true solution here

[Yashied]

Your UDF is invalid.

See about the true solution here

What exactly it is invalid? The fact that I limited the range of generated values? Histogram says otherwise.

EDIT:

Yes, I know that the Gaussian function is defined at -/+ infinity, but in practice, we use a finite interval.

Edited February 8, 2010 by Yashied

[ValeryVal]

To verify your UDF's distribution try The Kolmogorov–Smirnov test from any statistical tool.

The Kolmogorov–Smirnov test can be modified to serve as a goodness of fit test. In the special case of testing for normality of the distribution, samples are standardized and compared with a standard normal distribution. This is equivalent to setting the mean and variance of the reference distribution equal to the sample estimates, and it is known that using the sample to modify the null hypothesis reduces the power of a test.

[andybiochem]

Using

$s = ""
For $i = 1 to 500
    $s &= _GRandom(50,150,100,100) & @CRLF
Next
ClipPut(StringStripWS($s,2))

...to generate 500 data points with Mean = 100, and SD=10, gives the following analyses...

TEST

Ho: The sampled population is normally distributed

alpha = 0.05

RESULTS

n = 500

Mean = 101.7177478

Standard Deviation = 27.56706523

Variance = 759.9430853

K^2 = 6.26

p = 0.04350404

VERDICT

Reject Ho: The distribution is not normal.

TEST

Ho: The sampled population is normally distributed

alpha = 0.05

RESULTS

n = 500

Mean = 100.1

Standard Deviation = 28.79

Variance = 828.8641

AD = 5.460

P = <0.005

VERDICT

Reject Ho: The distribution is not normal.

TEST

Ho: The sampled population is normally distributed

alpha = 0.05

RESULTS

n = 500

Mean = 100.1

Standard Deviation = 28.79

Variance = 828.8641

RJ = 0.979

P = <0.010

VERDICT

Reject Ho: The distribution is not normal.

TEST

Ho: The sampled population is normally distributed

alpha = 0.05

RESULTS

n = 500

Mean = 100.1

Standard Deviation = 28.79

Variance = 828.8641

KS = 0.064

P = <0.010

VERDICT

Reject Ho: The distribution is not normal.

The only way to get the UDF to pass a test for normality is to set the min and max massively below and above the mean to essentially disable the rejection of the tail-end numbers.

I don't want to put you down; it's a good effort...and can be made statistically correct very easily.