Triangle Notation
- \(a\) = minimum
- \(b\) = maximum
- \(c\) = mode
- \(h\) = density at the mode = \(\frac{2}{b-a}\)
\[f(x) = \left\{ \begin{array}{ll} \frac{h}{c-a}(x-a) & \mbox{if } a \leq x \leq c \\ \frac{h}{c-b}(x-b) & \mbox{if } c < x \leq b \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (1)\]
Integrating the pdf in equation (1) to solve for h
\[\int f(x) dx = 1\]
\[\frac{h}{c-a}\int_{a}^{c} (x-a) dx + \frac{h}{c-b} \int_{c}^{b} (x-b) dx = \frac{h(b-a)}{2}\]
\[h=\frac{2}{b-a}\ \ \ \ (2)\]
Substituting back into equation (1),
\[f(x) = \left\{ \begin{array}{ll} \frac{2}{(b-a)(c-a)}(x-a) & \mbox{if } a \leq x \leq c \\ \frac{2}{(b-a)(c-b)}(x-b) & \mbox{if } c < x \leq b \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (3)\]
Integrating equation (3) to find \(E(x)\),
\[E(X)=\int xf(x)dx = \frac{h}{c-a}\int_a^c (x^2-ax) dx + \frac{h}{c-b}\int_c^b (x^2-bx) dx\]
\[=\frac{a+b+c}{3}\ \ \ \ (3)\]
\[V(X) = E(X^2) - \big(E(X)\big)^2 = \int x^2f(x)dx- \bigg(\frac{a+b+c}{3}\bigg)^2\]
\[=\frac{h}{c-a}\int_{a}^{c} x^2(x-a) dx + \frac{h}{c-b} \int_{c}^{b} x^2(x-b) dx- \bigg(\frac{a+b+c}{3}\bigg)^2\]
\[=\frac{a^2+b^2+c^2-ab-ac-bc}{18}\]
Define:
\[a_l=\log_{\phi}(a),\ \ b_l=\log_{\phi}(b),\ \ c_l=\log_{\phi}(c),\ \ h=\frac{2}{b_l-a_l}, \ \ \phi = \mbox{log base}\]
\[f(z) = \left\{ \begin{array}{ll} \frac{h}{c_l - a_l}(z - a_l) & \mbox{if } a_l \leq z \leq c_l \\ \frac{h}{c_l - b_l}(z - b_l) & \mbox{if } c_l < z \leq b_l \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (4)\]
However,
\[E({\phi}^z) \neq {\phi}^{E(z)}\ \ \ \ (5)\]
Therefore, transforming…
\[Y={\phi}^Z\]
\[Z=\log_{\phi}(Y)\]
\[w(y)=\log_{\phi}(y)\]
\[w'(y)=\frac{dz}{dy} = \frac{1}{y\log({\phi})}\]
\[g(y)=f(w(y))w'(y)\]
\[g(y) = \left\{ \begin{array}{ll} \frac{2}{(c_l-a_l)(b_l-a_l)\log({\phi})}\frac{log_{\phi}(y) - a_l}{y} & \mbox{if } 0 < a \leq y \leq c \\ \frac{2}{(c_l-b_l)(b_l-a_l)\log({\phi})}\frac{log_{\phi}(y) - b_l}{y} & \mbox{if } c < y \leq b \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (5)\]
Define:
\[\beta_1=\frac{2}{(c_l-a_l)(b_l-a_l)}\]
\[\beta_2=\frac{2}{(c_l-b_l)(b_l-a_l)}\]
Finding the CDF,
\[G(y)=\int_{-\infty}^y g(y)dy\]
\[\mbox{for}\ a \leq y \leq c,\ \ G(y) = \frac{\beta_1}{\log({\phi})} \int_a^y \frac{\log(y)}{y\log({\phi})}-\frac{a_l}{y}dy\]
\[=\beta_1 \bigg[\frac{\log_{\phi}^2(y)}{2} - a_l \log_{\phi}(y) - \frac{a_l^2}{2} + a_l^2\bigg]\]
\[\mbox{for}\ c < y \leq b,\ \ G(y) = G(c) + \frac{\beta_2}{\log({\phi})} \int_c^y \frac{\log(y)}{y\log({\phi})} - \frac{b_l}{y}dy\]
\[=G(c) + \beta_2 \bigg[\frac{\log_{\phi}^2(y)}{2} - b_l \log_{\phi}(y) - \frac{c_l^2}{2} + b_l c_l\bigg]\]
Checking that the CDF is 1 at b,
\[G(b) = \frac{c_l^2 - 2a_l c_l + a_l^2}{(c_l-a_l)(b_l-a_l)} + \frac{-b_l^2-c_l^2+2b_lc_l}{(c_l-b_l)(b_l-a_l)}\]
\[= \frac{c_l-a_l}{b_l-a_l} + \frac{-(c_l-b_l)}{b_l-a_l} = 1\]
Now calculating \(E(y)\),
\[E(y) = \int y\ g(y)\ dy\]
\[=\frac{\beta_1}{\log({\phi})} \int_a^c \bigg[\frac{\log(y)}{\log({\phi})} - a_l\bigg]dy + \frac{\beta_2}{\log({\phi})} \int_c^b \bigg[\frac{\log(y)}{\log({\phi})} - b_l\bigg]dy\]
\[=\frac{c\beta_1}{\log^2({\phi})} \bigg[\log(c) - 1 - \log(a) + \frac{a}{c} \bigg] + \frac{c\beta_2}{\log^2({\phi})} \bigg[\frac{-b}{c} - \log(c) + 1 + \log(b) \bigg]\]
\[E(x) = \frac{a + b + c}{3}\]
\[c = 3E(x) - a - b = 3E(x) - \min(x) - \max(x)\]
The procedure for maximum likelihood estimation involves maximizing the likelihood with respect to \(c\) for a fixed \(a\) and \(b\), followed by minimizing the negative log likelihood with respect to \(a\) and \(b\) for a fixed \(c\).
This discussion follows the results from Samuel Kotz and Johan Rene van Dorp. Beyond Beta
For the purposes of this section, with a fixed \(a\) and \(b\), the sample can be easily rescaled to \(a=0\) and \(b=1\). This section will proceed on \([0,1]\) with the mode at \(0 \le c \le 1\)
\[w(x) = \left\{ \begin{array}{ll} \frac{2x}{c} & \mbox{if } 0 \le x \lt c \\ \frac{2(1-x)}{1-c} & \mbox{if } c \le x \leq 1 \\ 0 & \mbox{otherwise} \end{array} \right.\]
\[L(x|c) = \prod_{i}^{n} w(x|c)\]
Assume that the sample is ordered into order statistics \(X_{(1)} \lt \dots \lt X_{(n)}\). Also, note that \(X_{(r)} \le c \lt X_{(r+1)}\). In other words, the mode falls between the \(r^{th}\) and \(r+1\) order statistics.
\[L(x|c) = \prod_{i=1}^{r} \frac{2x_{(i)}}{c} \prod_{i=r+1}^{n} \frac{2(1-x_{(i)})}{1-c} = \frac{2^n \prod_{i=1}^{r} x_{(i)} \prod_{i=r+1}^{n} (1-x_{(i)})}{c^r(1-c)^{n-r}}\]
To maximize the likelihood, we can first maximize with respect to \(r\) and then locate \(c\) between the \(r^{th}\) and \(r+1\) order statistics. For notation purposes, also define \(X_{(0)} = 0\) and \(X_{(n+1)} = 1\).
\[\large \max_{0 \le c \le 1} L(x|c) = \max_{r \ \epsilon \ (0,\dots,n)} \ \ \max_{x_{(r)} \le c \le x_{(r+1)}} \ \ L(x|c)\]
Noticing that maximizing the likelihood is equivalent to minimizing the denominator:
\[\large \max L(x|c) = \max_{r \ \epsilon \ (1,\dots,n-1)} \ \ \min_{x_{(r)} \le c \le x_{(r+1)}} \ \ c^r(1-c)^{n-r}\]
Since \(c^r(1-c)^{n-r}\) is unimodal with respect to \(c\), it should be sufficient to test the end points of an interval to find the minimum on the interval
\[\large = \max_{r \ \epsilon \ (1,\dots,n-1)} \ \ \min_{c \ \epsilon \ (x_{(r)},\ \ x_{(r+1)})} \ \ c^r(1-c)^{n-r}\]
Therefore, for this case, it is sufficient to test the likelihood using \(c\) at each of the sampled points and find the largest.
\[\frac{dz}{dc} = rc^{(r-1)}(1-c)^{n-r} + c^r(n-r)(1-c)^{n-r-1}(-1) = c^{(r-1)}(1-c)^{n-r-1}(r - cn)\]
\(\frac{dz}{dc} = 0\) at \(c=0,\ 1,\ \frac{r}{n}\). At \(0 < c < \frac{r}{n}\), \(z\) is positive, and at \(\frac{r}{n} < c < 1\), \(z\) is negative. Therefore, \(z\) is unimodal on \((0,1)\).
\[\large \max L(x|c) = \max_{0 \le c \le x_{(1)}} \prod_{i=1}^{n} \frac{1-x_{(i)}}{1-c} = \prod_{i=1}^{n} \frac{1-x_{(i)}}{1-x_{(1)}}\]
Choosing the largest endpoint in the interval, creates the smallest denominator, and the largest likelihood.
Therefore, for this case, it is sufficient to test the likelihood using \(c\) at the first sampled point.
\[\large \max L(x|c) = \max_{x_{(n)} \le c \le 1} \prod_{i=1}^{n} \frac{x_{(i)}}{c} = \prod_{i=1}^{n} \frac{x_{(i)}}{x_{(n)}}\]
Choosing the smallest option in the denominator creates the largest likelihood. Again, it is sufficient to test the likelihood using \(c\) at the largest sample point.
For all cases, it is sufficient to compute the sample likelihood using \(c\) equal to each of the samples, and choosing the largest likelihood from the \(n\) options to find the corresponding \(c\). This calculation is performed with a fixed \(a\) and \(b\), so the test must be performed iteratively as \(a\) and \(b\) are separately optimized.
\[nLL = -\log(L) = -\log\left(\prod_i^n f(x_i)\right)\]
\[ = - \sum_i^n \log\left(f(x_i)\right) = - \sum_{i: \ a \le x_i \lt c}^{n_1} \log\left(f(x_i)\right) - \sum_{i: \ c \le x_i \le b}^{n_2} \log\left(f(x_i)\right)\]
where \(n = n_1 + n_2\)
\[ nLL = - \sum_{i}^{n} \log(2) + \log(b-x_i) - \log(b-a) - \log(b-c)\]
\[ = -n\log(2) + n\log(b-a) + n \log(b-c) - \sum_{i}^{n} \log(b-x_i)\]
\[ = - \sum_{i}^{n} \log(2) + \log(x_i - a) - \log(b-a) - \log(c-a) \]
\[ = -n\log(2) + n\log(b-a) + n\log(c-a) - \sum_{i}^{n} \log(x_i - a)\]
\[ = - \sum_{i: \ a \lt x_i \lt c}^{n_1} \log(2) + \log(x_i - a) - \log(b-a) - \log(c-a) - \sum_{i: \ c \le x_i \lt b}^{n_2} \log(2) + \log(b-x_i) - \log(b-a) - \log(b-c)\]
\[ = -n\log(2) + n\log(b-a) + n_1\log(c-a) + n_2 \log(b-c) - \sum_{i: \ a \lt x_i \lt c}^{n_1} \log(x_i - a) - \sum_{i: \ c \le x_i \lt b}^{n_2} \log(b-x_i)\]
The negative log likelihood is not differentiable with respect to \(c\) because the limits of the sum (\(n_1\) and \(n_2\)) are functions of \(c\). There the gradient and hessian are derived as if \(c\) is fixed.
\[\frac{\partial nLL}{\partial a} = - \frac{n}{b-a}\]
\[\frac{\partial nLL}{\partial b} = \frac{n}{b-a} + \frac{n}{b-c} - \sum_i^{n} \frac{1}{b-x_i}\]
\[\frac{\partial nLL}{\partial a} = - \frac{n}{b-a} - \frac{n}{c-a} + \sum_i^{n} \frac{1}{x_i - a}\]
\[\frac{\partial nLL}{\partial b} = \frac{n}{b-a}\]
\[\frac{\partial nLL}{\partial a} = - \frac{n}{b-a} - \frac{n_1}{c-a} + \sum_i^{n_1} \frac{1}{x_i - a}\]
\[\frac{\partial nLL}{\partial b} = \frac{n}{b-a} + \frac{n_2}{b-c} - \sum_i^{n_2} \frac{1}{b-x_i}\]
\[\frac{\partial^2nLL}{\partial a^2} = - \frac{n}{(b-a)^2}\]
\[\frac{\partial^2 nLL}{\partial b^2} = -\frac{n}{(b-a)^2} - \frac{n}{(b-c)^2} + \sum_i^{n} \frac{1}{(b-x_i)^2}\]
\[\frac{\partial^2 nLL}{\partial a\partial b} = \frac{\partial^2 nLL}{\partial b\partial a} = - \frac{n}{(b-a)^2}\]
\[\frac{\partial^2 nLL}{\partial a^2} = - \frac{n}{(b-a)^2} - \frac{n}{(c-a)^2} + \sum_i^{n} \frac{1}{(x_i - a)^2}\]
\[\frac{\partial^2 nLL}{\partial b^2} = - \frac{n}{(b-a)^2}\]
\[\frac{\partial^2 nLL}{\partial a\partial b} = \frac{\partial^2 nLL}{\partial b\partial a} = - \frac{n}{(b-a)^2}\]
\[\frac{\partial^2 nLL}{\partial a^2} = - \frac{n}{(b-a)^2} - \frac{n_1}{(c-a)^2} + \sum_i^{n_1} \frac{1}{(x_i - a)^2}\]
\[\frac{\partial ^2 nLL}{\partial b^2} = -\frac{n}{(b-a)^2} - \frac{n_2}{(b-c)^2} + \sum_i^{n_2} \frac{1}{(b-x_i)^2}\]
\[\frac{\partial ^2 nLL}{\partial a\partial b} = \frac{\partial ^2 nLL}{\partial b\partial a} = - \frac{n}{(b-a)^2}\]
For the optimization of \((a,b)\) given \(c\), we can use the inverse of the hessian of the negative log likelihood for an estimate of the covariance matrix of \(\hat{a}\) and \(\hat{b}\). For the variance in \(\hat{c}\), we use the variance of the \(r^{th}\) order statistic which corresponds to \(c\). The covariance of \((a,b)\) and \(c\) is not computed because the negative log likelihood is not differentiable with respect to \(c\).
Let \(H\) denote the Hessian matrix, and let \(H^{-1}[1,1]\) be the \(V(\hat{a})\), \(H^{-1}[2,2]\) be the \(V(\hat{b})\), and \(H^{-1}[1,2] = H^{-1}[2,1]\) be the \(Cov(\hat{a}, \hat{b})\). Then,
\[ V([\hat{a}, \hat{b}, \hat{c}]) = \begin{bmatrix} H^{-1}[1,1] & H^{-1}[1,2] & 0 \\ H^{-1}[2,1] & H^{-1}[2,2] & 0 \\ 0 & 0 & V(\hat{c}) \\ \end{bmatrix} \]
\[f(x_{(r)}) = \frac{n!}{(r-1)!(n-r)!} f(x) [F(x)]^{(r-1)}[1-F(x)]^{(n-r)}\]
A closed form solution to \(V(X_{(r)})\) is not easily obtainable for
the triangle, so numerical integration is used with \(f(x)\) as dtriangle and \(F(x)\) as ptriangle.
\[V\left(X_{(r)}\right) = \int_a^b x^2 f(x_{(r)}) dx - \left[\int_a^b x f(x_{(r)}) dx \right]^2\]