Talk:Triangular distribution

The article says: a (location), b (scale) and c (shape) are the triangular distribution parameters. I would have said that c was a more natural location parameter as the mode, b−a the scale (or range and something like ${\frac {a+b-2c}{2(b-a)}}$ best for the shape, being related to the idea of skewness. --Henrygb 03:31, 25 Mar 2005 (UTC)

Kurtosis and kurtosis of Triangular distribution?

The kurtosis excess given 12/5 appears to be the 'true' kurtosis?

[@http://mathworld.wolfram.com/TriangularDistribution.html Wolfram ]give the kurtosis excess as -3/5.

which suggests that the value given as excess is in fact the 'true' excess 12/5 (kurtosis excess + 3 = 15/5 -3/5 = 12/5)

Paul A Bristow 18:34, 7 December 2006 (UTC) Paul A Bristow

Right, thanks. Its fixed. PAR 19:31, 7 December 2006 (UTC)

Formulae for pdf cause divide by zero if a or b = mode

Formulae for pdf cause divide by zero if a or b = c (the mode) (c-a = 0 if c == a). This are the two right angle triagle cases.

In these cases the value is the apex value of 2/(b-a)?

Should this specified separately?

Paul A Bristow 10:27, 11 December 2006 (UTC) Paul A Bristow

With the formulation

$f(x|a,b,c)=\left\{{\begin{matrix}{\frac {2(x-a)}{(b-a)(c-a)}}&\mathrm {for\ } a\leq x\leq c\\&\\{\frac {2(b-x)}{(b-a)(b-c)}}&\mathrm {for\ } c\leq x\leq b\end{matrix}}\right.$

when c=a the first form produces your problem, but the second is fine, even for x=c. I wouldn't bother adding more. --Henrygb 11:26, 11 December 2006 (UTC)

Is the median correct?

I have a feeling that the two cases for the median should be separtated by c = (b+a)/2 rather than c = (b-a)/2 as given on the page.