Talk: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 BristowReply

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

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 BristowReply

With the formulation

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

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)Reply

Paul is right−the pdf should look like this:

${\displaystyle f(x|a,b,c)={\begin{cases}0&\mathrm {for\ } x<a,\\{\frac {2(x-a)}{(b-a)(c-a)}}&\mathrm {for\ } a\leq x<c,\\[4pt]{\frac {2}{b-a}}&\mathrm {for\ } x=c,\\[4pt]{\frac {2(b-x)}{(b-a)(b-c)}}&\mathrm {for\ } c<x\leq b,\\[4pt]0&\mathrm {for\ } b<x,\end{cases}}}$ 

For the same reason, the cdf should be fixed: F(c) = (c-a) / (b-a).

I'll make these changes if nobody objects. -- UKoch (talk) 13:14, 2 May 2012 (UTC)Reply

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)Reply

I agree that the median formula does not appear to be correct. Does anyone have a link or reference for a derivation somewhere? 207.34.120.71 (talk) 20:40, 19 August 2009 (UTC)Reply

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)Reply

The mean of two standard uniform r.v.'s is called Bates(2), apparently.146.186.250.171 (talk) 17:28, 15 April 2011 (UTC)Reply

In this edit, part of the article was changed to read as follows:

to model events which take place within an interval defined by a minimum, most often

and maximum value. The triangle distribution is better as it provide a more accurate statistics to complete the task in delay as well as early completion of a task which can be zero, unlike the normal distribution.

Is the following interpretation of the above correct?

to model events which take place within an interval defined by a minimum and maximum value. The triangle distribution is better suited than the normal distribution, as it provides a more accurate statistics if either the minimum or the maximum time needed to complete the task is equal to the most probable time.

-- UKoch (talk) 19:19, 29 April 2018 (UTC)Reply

Better yet might be:

to model events which take place within an interval defined by a minimum and maximum value. The triangle distribution is better suited than the normal distribution, as it provides a more accurate statistics if either the minimum or the maximum time needed to complete the task is close or equal to the most probable time.

However, since it's not clear what was meant, I've undone the edit in question for the time being. -- UKoch (talk) 18:21, 1 May 2018 (UTC)Reply

Is the ref in question really about the triangular distribution? I highly doubt it. -- UKoch (talk) 18:28, 16 July 2021 (UTC)Reply