Fitting general pareto distribution to histogram

I am trying to get an understanding of how the example is done at this link: http://www.mathworks.com/help/stats/generalized-extreme-value-distribution.html Can someone explain why the gppdf is being multiplied by .5. It makes sense that you would multiply the gppdf by the number of excesses in order to get a number instead of a probability so that it fits to the histogram, but why are they are multiplying by .5 in the last two lines?

rng default % For reproducibility t = trnd(5,5000,1); y = t(t > 2) - 2; paramEsts = gpfit(y) hist(y+2,2.25:.5:11.75); xgrid = linspace(2,12,1000); line(xgrid,.5*length(y)*... gppdf(xgrid,paramEsts(1),paramEsts(2),2));

Answer 1

[Brendan Hamm]

The 0.5 has to do with the bin width. So the area under the curve of the pdf is equal to one and needs to equal the area under the curve of the histogram, which is binWidth*numberOfObservations.

[Josh Wagner]

Ahh, duh! Thanks!