# Triangle Distribution Math

## Triangle Notation • $$a$$ = minimum
• $$b$$ = maximum
• $$c$$ = mode
• $$h$$ = density at the mode = $$\frac{2}{b-a}$$

## Triangle PDF

$f(x) = \left\{ \begin{array}{ll} \frac{h}{c-a}(x-a) & \mbox{if } a \leq x \leq c \\ \frac{h}{c-b}(x-b) & \mbox{if } c < x \leq b \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (1)$

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),

$f(x) = \left\{ \begin{array}{ll} \frac{2}{(b-a)(c-a)}(x-a) & \mbox{if } a \leq x \leq c \\ \frac{2}{(b-a)(c-b)}(x-b) & \mbox{if } c < x \leq b \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (3)$

## Triangle Mean

Integrating equation (3) to find $$E(x)$$,

$E(X)=\int xf(x)dx = \frac{h}{c-a}\int_a^c (x^2-ax) dx + \frac{h}{c-b}\int_c^b (x^2-bx) dx$

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

## Triangle Variance

$V(X) = E(X^2) - \big(E(X)\big)^2 = \int x^2f(x)dx- \bigg(\frac{a+b+c}{3}\bigg)^2$

$=\frac{h}{c-a}\int_{a}^{c} x^2(x-a) dx + \frac{h}{c-b} \int_{c}^{b} x^2(x-b) dx- \bigg(\frac{a+b+c}{3}\bigg)^2$

$=\frac{a^2+b^2+c^2-ab-ac-bc}{18}$

## Logarithmic Triangle distribution

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 $$E(y)$$,

$E(y) = \int y\ g(y)\ dy$

$=\frac{\beta_1}{\log({\phi})} \int_a^c \bigg[\frac{\log(y)}{\log({\phi})} - a_l\bigg]dy + \frac{\beta_2}{\log({\phi})} \int_c^b \bigg[\frac{\log(y)}{\log({\phi})} - b_l\bigg]dy$

$=\frac{c\beta_1}{\log^2({\phi})} \bigg[\log(c) - 1 - \log(a) + \frac{a}{c} \bigg] + \frac{c\beta_2}{\log^2({\phi})} \bigg[\frac{-b}{c} - \log(c) + 1 + \log(b) \bigg]$

## Method of Moments

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

$c = 3E(x) - a - b = 3E(x) - \min(x) - \max(x)$

## Maximum Likelihood Estimation

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 $$c$$.

### Maximizing the Likelihood with respect to $$c$$

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 $$X_{(n+1)} = 1$$.

$\large \max_{0 \le c \le 1} L(x|c) = \max_{r \ \epsilon \ (0,\dots,n)} \ \ \max_{x_{(r)} \le c \le x_{(r+1)}} \ \ L(x|c)$

#### Case 1. $$c$$ is between the first and second to last order statistic $$r \ \epsilon \ (1, \dots, n-1)$$

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.

##### Side note on $$z=c^r(1-c)^{n-r}$$ being unimodal

$\frac{dz}{dc} = rc^{(r-1)}(1-c)^{n-r} + c^r(n-r)(1-c)^{n-r-1}(-1) = c^{(r-1)}(1-c)^{n-r-1}(r - cn)$

$$\frac{dz}{dc} = 0$$ 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 $$(0,1)$$.

#### Case 2. $$c$$ is between 0 and the first order statistic $$r = 0$$

$\large \max L(x|c) = \max_{0 \le c \le x_{(1)}} \prod_{i=1}^{n} \frac{1-x_{(i)}}{1-c} = \prod_{i=1}^{n} \frac{1-x_{(i)}}{1-x_{(1)}}$

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.

#### Case 3. $$c$$ is between the last order statistic $$r = n$$ and 1

$\large \max L(x|c) = \max_{x_{(n)} \le c \le 1} \prod_{i=1}^{n} \frac{x_{(i)}}{c} = \prod_{i=1}^{n} \frac{x_{(i)}}{x_{(n)}}$

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.

#### All Cases

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.

### Negative Log Likelihood

$nLL = -\log(L) = -\log\left(\prod_i^n f(x_i)\right)$

$= - \sum_i^n \log\left(f(x_i)\right) = - \sum_{i: \ a \le x_i \lt c}^{n_1} \log\left(f(x_i)\right) - \sum_{i: \ c \le x_i \le b}^{n_2} \log\left(f(x_i)\right)$

where $$n = n_1 + n_2$$

#### Case 1 $$a = c \lt b$$

$nLL = - \sum_{i}^{n} \log(2) + \log(b-x_i) - \log(b-a) - \log(b-c)$

$= -n\log(2) + n\log(b-a) + n \log(b-c) - \sum_{i}^{n} \log(b-x_i)$

#### Case 2 $$a \lt c = b$$

$= - \sum_{i}^{n} \log(2) + \log(x_i - a) - \log(b-a) - \log(c-a)$

$= -n\log(2) + n\log(b-a) + n\log(c-a) - \sum_{i}^{n} \log(x_i - a)$

#### Case 3 a c b

$= - \sum_{i: \ a \lt x_i \lt c}^{n_1} \log(2) + \log(x_i - a) - \log(b-a) - \log(c-a) - \sum_{i: \ c \le x_i \lt b}^{n_2} \log(2) + \log(b-x_i) - \log(b-a) - \log(b-c)$

$= -n\log(2) + n\log(b-a) + n_1\log(c-a) + n_2 \log(b-c) - \sum_{i: \ a \lt x_i \lt c}^{n_1} \log(x_i - a) - \sum_{i: \ c \le x_i \lt b}^{n_2} \log(b-x_i)$

### Gradient of the negative Log Likelihood Given $$c$$:

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.

#### Case 1 $$a = c \lt b$$

$\frac{\partial nLL}{\partial a} = - \frac{n}{b-a}$

$\frac{\partial nLL}{\partial b} = \frac{n}{b-a} + \frac{n}{b-c} - \sum_i^{n} \frac{1}{b-x_i}$

#### Case 2 $$a \lt c = b$$

$\frac{\partial nLL}{\partial a} = - \frac{n}{b-a} - \frac{n}{c-a} + \sum_i^{n} \frac{1}{x_i - a}$

$\frac{\partial nLL}{\partial b} = \frac{n}{b-a}$

#### Case 3 a c b

$\frac{\partial nLL}{\partial a} = - \frac{n}{b-a} - \frac{n_1}{c-a} + \sum_i^{n_1} \frac{1}{x_i - a}$

$\frac{\partial nLL}{\partial b} = \frac{n}{b-a} + \frac{n_2}{b-c} - \sum_i^{n_2} \frac{1}{b-x_i}$

### Hessian of the negative Log Likelihood Given $$c$$:

#### Case 1 $$a = c \lt b$$

$\frac{\partial^2nLL}{\partial a^2} = - \frac{n}{(b-a)^2}$

$\frac{\partial^2 nLL}{\partial b^2} = -\frac{n}{(b-a)^2} - \frac{n}{(b-c)^2} + \sum_i^{n} \frac{1}{(b-x_i)^2}$

$\frac{\partial^2 nLL}{\partial a\partial b} = \frac{\partial^2 nLL}{\partial b\partial a} = - \frac{n}{(b-a)^2}$

#### Case 2 $$a \lt c = b$$

$\frac{\partial^2 nLL}{\partial a^2} = - \frac{n}{(b-a)^2} - \frac{n}{(c-a)^2} + \sum_i^{n} \frac{1}{(x_i - a)^2}$

$\frac{\partial^2 nLL}{\partial b^2} = - \frac{n}{(b-a)^2}$

$\frac{\partial^2 nLL}{\partial a\partial b} = \frac{\partial^2 nLL}{\partial b\partial a} = - \frac{n}{(b-a)^2}$

#### Case 3 $$a \lt c \lt b$$

$\frac{\partial^2 nLL}{\partial a^2} = - \frac{n}{(b-a)^2} - \frac{n_1}{(c-a)^2} + \sum_i^{n_1} \frac{1}{(x_i - a)^2}$

$\frac{\partial ^2 nLL}{\partial b^2} = -\frac{n}{(b-a)^2} - \frac{n_2}{(b-c)^2} + \sum_i^{n_2} \frac{1}{(b-x_i)^2}$

$\frac{\partial ^2 nLL}{\partial a\partial b} = \frac{\partial ^2 nLL}{\partial b\partial a} = - \frac{n}{(b-a)^2}$

### MLE Variance Co-variance

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,

$V([\hat{a}, \hat{b}, \hat{c}]) = \begin{bmatrix} H^{-1}[1,1] & H^{-1}[1,2] & 0 \\ H^{-1}[2,1] & H^{-1}[2,2] & 0 \\ 0 & 0 & V(\hat{c}) \\ \end{bmatrix}$

#### $$r^{th}$$ order statistic

$f(x_{(r)}) = \frac{n!}{(r-1)!(n-r)!} f(x) [F(x)]^{(r-1)}[1-F(x)]^{(n-r)}$

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.

$V\left(X_{(r)}\right) = \int_a^b x^2 f(x_{(r)}) dx - \left[\int_a^b x f(x_{(r)}) dx \right]^2$