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…
\[Y={\phi}^Z\]
\[Z=log_{\phi}(Y)\]
\[w(y)=log_{\phi}(y)\]
\[w'(y)=\frac{dz}{dy} = \frac{1}{yln({\phi})}\]
\[g(y)=f(w(y))w'(y)\]
\[g(y) = \left\{ \begin{array}{ll} \frac{2}{(c_l-a_l)(b_l-a_l)ln({\phi})}\frac{log_{\phi}(y) - a_l}{y} & \mbox{if } 0 < a \leq y \leq c \\ \frac{2}{(c_l-b_l)(b_l-a_l)ln({\phi})}\frac{lob_{\phi}(y) - b_l}{y} & \mbox{if } c < y \leq b \\ 0 & \mbox{otherwise} \end{array} \right.\ \ \ \ (5)\]
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,
\[G(y)=\int_{-\infty}^y g(y)dy\]
\[\mbox{for}\ a \leq y \leq c,\ \ G(y) = \frac{\beta_1}{ln({\phi})} \int_a^y \frac{ln(y)}{yln({\phi})}-\frac{a_l}{y}dy\]
\[=\beta_1 \bigg[\frac{log_{\phi}^2(y)}{2} - a_llog_{\phi}(y) - \frac{a_l^2}{2} + a_l^2\bigg]\]
\[\mbox{for}\ c < y \leq b,\ \ G(y) = G(c) + \frac{\beta_2}{ln({\phi})} \int_c^y \frac{ln(y)}{yln({\phi})} - \frac{b_l}{y}dy\]
\[=G(c) + \beta_2 \bigg[\frac{log_{\phi}^2(y)}{2} - b_l log_{\phi}(y) - \frac{c_l^2}{2} + b_l c_l\bigg]\]
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 \(E(y)\),
\[E(y) = \int y\ g(y)\ dy\]
\[=\frac{\beta_1}{ln({\phi})} \int_a^c \bigg[\frac{ln(y)}{ln({\phi})} - a_l\bigg]dy + \frac{\beta_2}{ln({\phi})} \int_c^b \bigg[\frac{ln(y)}{ln({\phi})} - b_l\bigg]dy\]
\[=\frac{c\beta_1}{ln^2({\phi})} \bigg[ln(c) - 1 - ln(a) + \frac{a}{c} \bigg] + \frac{c\beta_2}{ln^2({\phi})} \bigg[\frac{-b}{c} - ln(c) + 1 + ln(b)\bigg]\]