Note on “Parameters estimators of irregular right-angled triangular distribution”

Abstract

Simple estimators were given in (Kachiashvili & Topchishvili, 2016) for the lower and upper limits of an irregular right-angled triangular distribution together with convenient formulas for removing their bias. We argue here that the smallest observation is not a maximum likelihood estimator (MLE) of the lower limit and we present a procedure for computing an MLE of this parameter. We show that the MLE is strictly smaller than the smallest observation and we give some bounds that are useful in a numerical solution procedure. We also present simulation results to assess the bias and variance of the MLE.

Keywords

Irregular right-angled triangular distribution maximum likelihood estimator bounds bias variance mean square error

1. Introduction

The triangular distribution may be useful in the context of risk and uncertainty analysis, e.g., project evaluation and review technique (PERT), as a proxy for the beta distribution, whose parameters are not easy to understand and difficult to estimate (Johnson, 1997). Indeed, the three parameters of a regular triangular distribution, namely the minimum, the maximum, and the most likely values, offer more intuitive interpretation to decision makers (Stein & Keblis, 2009; van Dorp & Kotz, 2002). It is recognized that the classical maximum likelihood estimation of triangular distribution parameters is difficult (Joo & Casella, 2001), in particular due to the non-linearity of the first order conditions. This hurdle may be passed through Monte Carlo simulation e.g., (Stein & Keblis, 2009), estimation with order statistics, e.g., (Samuel & Thomas, 2003), non-linear optimization, e.g., (van Dorp & Kotz, 2002), among others.

The irregular right-angled triangular distribution discussed in (Kachiashvili & Topchishvili, 2016) has two parameters $a$ and $b$ , where $a$ is the lower limit and $b$ is the upper limit, with the mode at $b$ . The estimators proposed in (Kachiashvili & Topchishvili, 2016), which are consistent, unbiased and efficient, are based on the order statistics of the experimental sample. Let $x=(x_{1},\ldots,x_{n})$ be the observed values of a sample of size $n$ and let $x_{(1)}$ and $x_{(n)}$ be the smallest and the largest values, respectively. The estimators of (Kachiashvili & Topchishvili, 2016) are based on $\hat{a}=x_{(1)}$ and $\hat{b}=x_{(n)}$ which are claimed to be maximum likelihood estimators (MLE). While it is true that $\hat{b}$ is a MLE, this is not the case of $\hat{a}$ . For example, with the sample $x=(0,1)$ of size $n=2$ the likelihood function of Eq. (3) in (Kachiashvili & Topchishvili, 2016) is not maximized at $a=0$ but rather the MLE is $a^{*}=-1/2$ .

However, the simplicity of the estimators $\hat{a}=x_{(1)}$ and $\hat{b}=x_{(n)}$ together with their easily obtained expectations based on their respective tail distributions $(1-x^{2})^{n}$ and $x^{2n}$ when $a=0$ and $b=1$ , makes their unbiased versions quite attractive in practice. Nonetheless, we feel that a brief analysis of the MLE estimator of the lower limit parameter $a$ constitutes a useful complement to that of (Kachiashvili & Topchishvili, 2016).

2. Maximum likelihood estimator of the lower limit

For convenience we repeat here the density function $p_{T}(y)$ of the triangular distribution and the likelihood function $L(x|a,b)$ for a sample $x$ of size $n$ .

$\displaystyle p_{T}(y)=\frac{2(y-a)}{(b-a)^{2}},\text{ at }a\leqslant y% \leqslant b,$ (1) $\displaystyle L(x|a,b)=\frac{2^{n}}{(b-a)^{2n}}\prod_{i=1}^{n}(x_{i}-a),\text{% subject to }a\leqslant x_{(1)},b\geqslant x_{(n)}.$ (2)

It is clear from Eq. (2) that $L(x|a,b)$ decreases with $b$ so the MLE of $b$ is $\hat{b}=x_{(n)}$ . But $L(x|a,b)$ does not increase everywhere with $a$ so the MLE of $a$ is not $\hat{a}=x_{(1)}$ . Taking the partial derivative of $\ln L(x|a,b)$ with respect to $a$ , the first-order optimality condition is

$\displaystyle\frac{2n}{b-a}-\sum_{i=1}^{n}\frac{1}{x_{i}-a}=0,$ (3)

where $b=x_{(n)}$ . Then a MLE $a^{*}$ can be obtained by solving the nonlinear Eq. (3), using a numerical method when $n>2$ . Clearly, $a^{*}<\hat{a}$ except in trivial cases, otherwise there would be a zero denominator in Eq. (3). This is made more precise in the following proposition.

.

Let $\hat{a}=x_{(1)}$ and $\hat{b}=x_{(n)}$ . Then

$\displaystyle\hat{a}-\frac{n-1}{n}(\hat{b}-\hat{a})\leqslant a^{*}\leqslant% \hat{a}-\frac{1}{n}(\hat{b}-\hat{a}).$ (4)