# Parameter estimates for the triangular distribution

A question was posted here (now deleted) in relation to estimating the parameters of the triangular distribution, which has density

$$f(x;a,b,c)=\begin{cases} \quad 0 & \text{for } x < a, \\ \frac{2(x-a)}{(b-a)(c-a)} & \text{for } a \le x \le c, \\ \frac{2(b-x)}{(b-a)(b-c)} & \text{for } c < x \le b, \\\quad 0 & \text{for } b < x. \end{cases}$$

But the question is worth asking, so I am asking it myself.

What are good ways to estimate the parameters for this distribution?

Discussion of MLE is good but other estimators can make for fruitful answers.

Note 1: Many documents related to PERT seem to use $X_{(1)}$ and $X_{(n)}$ to estimate $a$ and $b$ and then (given that) use method of moments for $c$. If you advocate this approach in particular, some discussion of efficiency would be most helpful, but at the least some reason for the choice (or one similar to it) would be important.

Note 2:

[Perhaps this should be the start of an answer but I will place it here as guidance on answers relating to ML for the present.]

Note that for MLE setting derivatives of log-likelihood to zero won't work.

For example for known $a$ and $b$ (which wlog we can take as 0,1 by simple rescaling), see the discussion on MLE for $c$ here: MLE for triangle distribution? .

Additionally, in general the ML estimates for the endpoints $a$ and $b$ are not the extreme order statistics. See, for example here (1)

(1) Kotz, Samuel, and Johan Rene van Dorp (2004),
The Triangular Distribution, (Chapter 1)
Beyond Beta—Other Continuous families of Distributions with Bounded Support and Applications,
World Scientiﬁc, NJ
(sample chapter)

Using the extreme-order statistics as estimators for the boundaries $a,b$ and then using

$$E(X) = \frac {a+b+c}{3}$$

to estimate $c$ by method of moments is so ...maddeningly easy,

$$\hat a = X_{(1)},\;\; \hat b = X_{(n)},\;\;\hat c = 3\bar X - \hat a - \hat b$$ it made me think how I could start by estimating $c$ first, just for the twist of it. Here it is but not yet with any properties of the estimator. I will make this community wiki in case any one is interested in working it further.

1) Obtain the empirical quartiles $\hat q_1, \hat q_3$ and form the Interquartile Range $\text{IQR} = \hat q_3 - \hat q_1$

2) Use the Friedman-Diaconis rule to bin the data.

$$\text{Bin size}=2\, { \text{IQR} \over{ {n^{1/3}} }}$$

3) Form the empirical histogram and estimate $\hat c$ as the mid-point of the bin with the highest empirical frequency.

4) Then solve for $a,b$ the system of equations

$$q_1 = a + \frac {\sqrt{(c-a)(b-a)}}{2}$$ $$q_3 = b + \frac {\sqrt{(b-c)(b-a)}}{2}$$

using the estimated $\hat q_1, \hat q_3, \hat c$ (the inverse CDF expressions I took from the book chapter the OP links to, page 8).

• I doubt that the usual suggested bandwidth guidance is even applicable here. The triangle distribution has been referred to as the "lack of knowledge" distribution. So, with very small samples, say 8 points, the rule you suggested yields a bandwidth equal to IQR (!?), which implies just use the mid-point of the IQR. Jun 13, 2021 at 14:40
• @AJKOER Thanks for the comment. I don't think I know of many methods that are applicable in all circumstances. So indeed, for very small samples the approach I am describing may not be very useful. Jun 14, 2021 at 19:58

Assume that one has a sufficient sample size $$n$$ where the sample mode and mean suggest an estimate of $$c$$.

Proceeding conditionally, based on this information, there are thus $$h$$ values of the triangular random deviate observed that fall below the assumed $$c$$ value, and also correspondingly $$k$$ above.

Then, the derivative of the conditional log-likelihood function, set equal to zero, facilely renders the answer:

$$\hat c = \frac {h*a + k*b}{n}$$

which equates to the known value of $$c$$ in the case of the Symmetric Triangular Distribution:

$$c = \frac {a + b}{2}$$