Is there a reason why Triangular is not an option in fitdist?

One can make triangular dists via makedist, but why not automate the mu and sigma calc. That way a user can toggle between options such as 'Normal' and 'Triangular' in the fitdist command more easily.

[dpb]

Submit enhancement request...

Answer 1

[John D'Errico]

Is there a GOOD reason why not? Probably not, but then I don't expect a triangular distribution is in high demand either. Could you use maximum likelihood techniques to estimate the parameters of a triangular distribution? I don't see why not, though there may be some numerical issues I don't immediately see. There are always things that sneak up and bite you in the ... More likely, I asssume it is not in there out of parsimony. Why write code nobody will need?

IF you truly, REALLY need it, then you could send in a feature request. Of course, don't expect it to appear next week. More likely a year or several years down the line, as this certainly would not be a high priority thing for them to do, and there are only so many resources they have to write code, so many person-hours they have available.

So, IF you desperately need it now, you would need to write the code yourself to estimate those parameters. Sorry 'bout that. Are you talking about a simple symmetric triangular distribution? That would be easy enough. A skewed one would take a little more effort, but not that much.

[Matt Weber]

Thanks for the reply. I mainly wanted to make sure I wasn't misunderstanding how to use fitdist. Triangular might be more used that 'Birnbaum-Saunders', but nothing against the developers!

[John D'Errico]

A quick check for myself, and I do not see triangular in there.

fitdist

I imagine the coding squad for the stats TB saw no real need for it. I'd have probably made the same choice. Triangular distributions always seemed to be one of those things that were good for student homework assignments, but not much more. :-) I can't say I'm not wrong in that perception, and I'd be happy to be proven wrong. As it is, in the codes I have written, I have often been surprised to see someone have a use for an option I put in there just for fun (pure fluff in my eyes), never expecting it to be the perfect solution for somone out there.

Answer 2

[Image Analyst]

Are triangular distributions of actual things in nature something that actually occurs? If so, where/when? To me it seems like something that may only be wanted as something like a homework question. Sure triangular/sawtooth waveforms happen in nature but distributions or histograms? I don't know of any. Perhaps you do. If so, tell me where.

[Walter Roberson]

If you have a peg board with pegs in a triangle grid and you drop objects from the top, they pile up in a triangle. But there isn't any variation in that kind of situation.

[Matt Weber]

Again, this is only a humble suggestion. I use triangular as a sensitivity analysis test, or to see what happens if potential influence of a long tail is reduced, and for data scarce situations when it seems silly to fit something more complex. Google Scholar since 2018: 9,570 instances of triangular dist; 1,420 instances of Burr distribution. I can get by with makedist for Triangular rather than fitdist for my purposes, so this is only a suggestion. No biggie.

[Image Analyst]

@Walter Roberson, do you mean like a Pachinko machine? Sites I see say it's a normal distribution

https://www.sciencedirect.com/science/article/pii/S0304397520303121

[Walter Roberson]

Pachinko-like but they pile up, blocking paths. There is a toy that demonstrates, but unfortunately I do not see an example at the moment.

[Paul]

The sum of two, independent, standard uniform, continuous random variables has a triangular pdf.

syms x u v real
f(x) = rectangularPulse(0,1,x);
F(v) = fourier(f(x),x,v);
h(x) = simplify(ifourier(F(v)^2),100)

h(x) = 

fplot(h(x),'Color','r','LineWidth',2)
hold on
histogram(rand(10000,1)+rand(10000,1),'Normalization','pdf','EdgeColor','none')

That must be useful in some applications. For example, suppose we have system composed of a sequence of two consecutive events. If the execution times of each event are standard uniform and independent, the then the total execution time has a triangular density function.

[Frank]

Paul, I just came across this and concur it may not be high, widespread interest, but it does occur in some real world things. Dice were mentioned but more important to me is the case you cited, sum of standard uniforms. This happens in the real world if one is obtaining data from a device that has a resolution limit to which the underlying data is rounded, so that measured values from the device have an uncertainty +- the resolution. If the underlying measurand is varying continuously and fairly randomly over time, i.e. going up and down over some range perhaps cyclicly, then the uncertainties very likely have uniform distribution and sums of the measurements are going to have uncertainties distributed as a triangular distribution. Likely anyone who wants this can code a workaround. Maybe some one of us who is interested enough and wants to invest the time could just post something on the file exchange?

[Image Analyst]

@Paul and @Frank Yes that is very true. The distribution of the sum of two variables is the convolution of the distribution of the two individual variables. (I learned this in my statistical optics grad school class - not sure how many people know this.) So convolving two rect functions will give a triangle. If you sum 3 variables you need to convolve it with a third rect function and the distribution becomes more rounded. As you convolve in more and more functions it becomes very Gaussian looking very quickly. This means that the distribution of a sum of many random variables with uniform distribution is Gaussian. I think the Central Limit Theorem is somehow involved in the proof. Here is code illustrating it

% Demo by Image Analyst
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format short g;
format compact;
fontSize = 22;
markerSize = 15;
pulseWidth = 111;
pulse = [zeros(1, pulseWidth), ones(1, pulseWidth), zeros(1, pulseWidth)]
subplot(6, 1, 1);
plot(pulse, 'b-', 'LineWidth', 2)
title('Original distribution of a single variable', 'FontSize',fontSize)
grid on;
newPulse = pulse;
% Add together more random variables and 
% you'd get the convolution of their individual PDFs.
for k = 1 : 5
    newPulse = conv(newPulse, pulse, 'full');
    subplot(6, 1, k+1);
    plot(newPulse, 'b-', 'LineWidth', 2)
    grid on;
    caption = sprintf('Distribution after summing %d variables', k+1);
    title(caption, 'FontSize',fontSize)
end

The funny thing is, is that the shapes of the individual PDFs usually doesn't matter. It tends towards a Gaussian, as more variables are added, no matter the shape.

[Paul]

The distribution of the sum of two independent variables is the convolution of the distribution of the two individual variables.

For continuous RV's, I'd replace "distribution" with "density function," but maybe that's being pedantic.

If interested, this reference derives a closed form expression for the pdf of the sum of uniform RV's:

S. M. Sadooghi-Alvandi · A. R. Nematollahi, R. Habibi, "On the distribution of the sum of independent uniform random variables"