Talk:Triangular distribution: Difference between revisions
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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 (statistics)|mode]], ''b''−''a'' the scale (or [[range (statistics)|range]] and something like <math>\frac{a+b-2c}{2(b-a)}</math> best for the shape, being related to the idea of [[skewness]]. --[[User:Henrygb|Henrygb]] 03:31, 25 Mar 2005 (UTC) |
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 (statistics)|mode]], ''b''−''a'' the scale (or [[range (statistics)|range]] and something like <math>\frac{a+b-2c}{2(b-a)}</math> best for the shape, being related to the idea of [[skewness]]. --[[User:Henrygb|Henrygb]] 03:31, 25 Mar 2005 (UTC) |
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The mean of two standard uniform r.v.'s is called [[Bates_distribution|Bates(2)]], apparently.[[Special:Contributions/146.186.250.171|146.186.250.171]] ([[User talk:146.186.250.171|talk]]) 17:28, 15 April 2011 (UTC) |
The mean of two standard uniform r.v.'s is called [[Bates_distribution|Bates(2)]], apparently.[[Special:Contributions/146.186.250.171|146.186.250.171]] ([[User talk:146.186.250.171|talk]]) 17:28, 15 April 2011 (UTC) |
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== Unclear edit == |
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In [https://en.wikipedia.org/enwiki/w/index.php?title=Triangular_distribution&type=revision&diff=838784148&oldid=832293480 this edit], part of the article was changed to read as follows: |
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:to model events which take place within an interval defined by a minimum, most often |
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: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. |
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Is the following interpretation of the above correct? |
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: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. |
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-- [[User:UKoch|UKoch]] ([[User talk:UKoch|talk]]) 19:19, 29 April 2018 (UTC) |
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:Better yet might be: |
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::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. |
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:However, since it's not clear what was meant, I've undone the edit in question for the time being. -- [[User:UKoch|UKoch]] ([[User talk:UKoch|talk]]) 18:21, 1 May 2018 (UTC) |
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== Beamforming == |
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Is the ref in question really about the triangular distribution? I highly doubt it. -- [[User:UKoch|UKoch]] ([[User talk:UKoch|talk]]) 18:28, 16 July 2021 (UTC) |
Latest revision as of 10:26, 10 February 2024
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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, b−a the scale (or range and something like best for the shape, being related to the idea of skewness. --Henrygb 03:31, 25 Mar 2005 (UTC)
Kurtosis and kurtosis of Triangular distribution?
[edit]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
[edit]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
- 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)
- Paul is right−the pdf should look like this:
- 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)
Is the median correct?
[edit]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)
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)
Link
[edit]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)
example should cite Bates distribution?
[edit]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)
Unclear edit
[edit]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)
- 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)
Beamforming
[edit]Is the ref in question really about the triangular distribution? I highly doubt it. -- UKoch (talk) 18:28, 16 July 2021 (UTC)