Talk:Triangular distribution

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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, ba 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)

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[edit] 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)

[edit] 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 \le x \le c \\ & \\
                          \frac{2(b-x)}{(b-a)(b-c)} & \mathrm{for\ } c \le x \le 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)

[edit] 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. —Preceding unsigned comment added by 143.53.57.46 (talk) 09:09, 28 March 2008 (UTC)

[edit] Link

Can somebody please create the page Triangular Distribution and redirect it here. I keep searching for that page, and get nothing every time. —Preceding unsigned comment added by 192.91.171.36 (talk) 14:21, 28 May 2008 (UTC)