July 2015 |
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September 2015 |
Assume the score is a linear function of some observable 'x': \begin{equation} y = m \cdot x + c \label{eq:01} \end{equation} Given two data points: (x1, y1), (x2, y2). Therefore, $$y_1 = m \cdot x_1 + c $$ \begin{equation} \Rightarrow c = y_1 - m \cdot x_1 \label{eq:02} \end{equation} Also, $$ y_2 - y_1 = m \cdot (x_2 - x_1) $$ \begin{equation} \Rightarrow m = \frac{y_2 - y_1}{x_2 - x_1} \label{eq:03} \end{equation} Substituting \eqref{eq:02} and \eqref{eq:03} into\eqref{eq:01}, we get the following general formula
\begin{equation} y = \bigg (\frac{y_2 - y_1}{x_2 - x_1} \bigg ) \cdot \big (x - x_1 \big ) + y_1 \label{eq:04} \end{equation}
Assume the score is a logarithmic function of observable 'x': \begin{equation} y = m \log(x + b) + c \label{eq:11} \end{equation} Given two data points: (x1, y1), (x2, y2). Therefore, $$y_1 = m \log(x_1 + b) + c $$ \begin{equation} \Rightarrow c = y_1 - m \log(x_1 + b) \label{eq:12} \end{equation} Also, $$ y_2 - y_1 = m \log \bigg (\frac{x_2+b}{x_1+b} \bigg ) $$ \begin{equation} \Rightarrow m = \frac{y_2 - y_1}{\log \big ( \frac{x_2+b}{x_1+b}\big)} \label{eq:13} \end{equation} Substituting \eqref{eq:12} and \eqref{eq:13} into \eqref{eq:11}, we get the following general formula \begin{equation} y = \frac{y_2 - y_1}{\log \big ( \frac{x_2+b}{x_1+b}\big)} \cdot \log \bigg (\frac{x+b}{x_1+b} \bigg ) + y_1 \label{eq:4} \end{equation} Let's choose the constant term b such that \begin{equation} \frac{x_2+b}{x_1+b} = 11 \label{eq:5} \end{equation} $$\Rightarrow b = \frac{x_2 - 11 x_1}{10} $$ Then
\begin{equation} y = \frac{y_2 - y_1}{\log (11)} \cdot \log \bigg (\frac{10x}{x_2-x_1} + \frac{x_2-11x_1}{x_2-x_1}\bigg ) + y_1 \label{eq:6} \end{equation}
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