# [pstricks] pst-funct: strange psGaussI behavior

Andrzej Pacut A.Pacut at elka.pw.edu.pl
Fri May 4 20:53:05 CEST 2012

There is a strange behavior of \psGaussI function (part of pst-func
package). I understood this function was the integral of Gaussian density.
Running a simple test below I began to doubt:

\psset{xunit=2,yunit=20mm}

\begin{pspicture}(-2,-.5)(2,2)

\psaxes[Dy=0.25]{->}(0,0)(-2,0)(2,1.25)

\uput[-90](6,0){x} \uput[0](0,1){y}

\psGaussI[mue=-1.5,sigma=0.2]{-1.7}{1.7}

\psGaussI[mue=-1.0,sigma=0.2]{-1.7}{1.7}

\psGaussI[mue=-0.5,sigma=0.2]{-1.7}{1.7}

\psGaussI[mue=-0.0,sigma=0.2]{-1.7}{1.7}

\psGaussI[mue= 0.5,sigma=0.2]{-1.7}{1.7}

\psGaussI[mue= 1.0,sigma=0.2]{-1.7}{1.7}

\psGaussI[mue= 1.5,sigma=0.2]{-1.7}{1.7}

\end{pspicture}

Only the plots for mue=-1 and mue=-0.5 look according to my expectation
(which might be erroneous). Now,

1)      For mue =-1.5, the plot has a horizontal asymptote at about 0.8.
This may show that psGaussI is not the Gaussian distribution function, but
rather a running integral of Gausian density, with the left integral
bound=-1.7 in my example. Am I right? If so, is it a way to plot a real
Gaussian distribution?

2)      All plots for mue greater than -0.5 fluctuate, showing a behavior
far from the one I would expect from a distribution function, or even from
the integral of a density function. Changing Simpson parameter does not
change the story.

I will be grateful for any help

Andrzej Pacut

Faculty of Electronics and Information Technology

Warsaw University of Technology

Warsaw, Poland

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