Respuesta :
EXPLANATION:
I have attached a table containing point differentials for teams.
Massey’s ratings are based on the principle that for each
game the difference in the ratings should be equal to the point differential.
Mathematically,
ri − rj = yk
[tex]r_{i}\;-\;r{j} = y_k\\\\where \\r_{i}\; = \;rating \;for\; team\; i,\\ r_{j} \;=\;rating\; for \;team\; j\\and y_{k}\; =\; score\; for \;team \;i\; - \;score\; for \;team\; j\; for\;game\; k.[/tex]
For our teams, therefore, we have
[tex]r_{1}\;-r_{2}\;=\;15.2\\\\r_{1}\;-r_{3}\;=\;-12.8\\\\r_{2}\;-r_{3}\;=\;-29.4\\\\r_{2}\;-r_{4}\;=\;9.6\\[/tex]
In Matrix Form;
[tex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;X\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;r\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y\\\\\left[\begin{array}{cccc}1&-1&0&0\\1&0&-1&0\\0&1&-1&0\\0&1&0&-1\end{array}\right] \left[\begin{array}{c}r_{1}&r_{2}&r_{3}&r_{4}\end{array}\right] \;=\;\left[\begin{array}{c}15.2&-12.8&-29.4&9.6\end{array}\right] \\[/tex]
This turns out to be an inconsistent system of equations with no solutions. Therefore, I using method of least squares to find a solution to the normal equations we get by multiplying by [tex]X^{T}[/tex].
Mathematically,
[tex]X^{T} X*r\;=\;X^{T}Y[/tex]
Assuming M = [tex]X^{T} X[/tex] and P = [tex]X^{T} y[/tex], above equation can be written as
[tex]M\;*\;r\;=\;P\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;......\;(1)[/tex]
[tex]M\;=\;X^{T}X\; =\; \left[\begin{array}{cccc}1&1&0&0\\-1&0&1&1\\0&-1&-1&0\\0&0&0&-1\end{array}\right] \left[\begin{array}{cccc}1&-1&0&0\\1&0&-1&0\\0&1&-1&0\\0&1&0&-1\end{array}\right] \\\\\\\\M\;=\;\left[\begin{array}{cccc}2&-1&-1&0\\-1&3&-1&-1\\-1&-1&2&0\\0&-1&0&1\end{array}\right][/tex]
[tex]Similarly,\\\\P\;=\;X^{T}y\; =\; \left[\begin{array}{cccc}1&1&0&0\\-1&0&1&1\\0&-1&-1&0\\0&0&0&-1\end{array}\right] \left[\begin{array}{c}15.2&-12.8&-29.4&9.6\end{array}\right] \\\\\\\\P\;=\;\left[\begin{array}{c}2.4&-35&42.2&-9.6\end{array}\right][/tex]
Unfortunately, M is not invertible. To get around this, Massey replaces the n-th row of the matrix M by a row of 1’s and replacing the final row of P by a zero. The adjusted matrices will be written as M' and P', respectively.
[tex]M'\;=\;\left[\begin{array}{cccc}2&-1&-1&0\\-1&3&-1&-1\\-1&-1&2&0\\1&1&1&1\end{array}\right][/tex]
[tex]and\;\;\;\;\;\;P'\;=\;\left[\begin{array}{c}2.4&-35&42.2&0\end{array}\right][/tex]
Equation (1) will become,
[tex]\;\;\;\;M'\;*\;r\;=\;P'\\\\\therefore\;\;r\;=\;P'M'^{-1}[/tex]
where,
[tex]M'^{-1}\;=\;\left[\begin{array}{cccc}0.666&0.25&0.333&0.25\\0&0.25&0&0.25\\0.333&0.25&0.666&0.25\\-1&-0.75&-1&0.25\end{array}\right][/tex]
[tex]\therefore\;\;r\;=\;\left[\begin{array}{c}2.4&-35&42.2&0\end{array}\right]\left[\begin{array}{cccc}0.666&0.25&0.333&0.25\\0&0.25&0&0.25\\0.333&0.25&0.666&0.25\\-1&-0.75&-1&0.25\end{array}\right]\\\\\\\\.\;\;\;\;\;r\;=\;\left[\begin{array}{c}6.9167&-8.7500&20.1833&-18.3500\end{array}\right][/tex]
Thus, ranking will be as follows;
Rank 1 = DUKE (rating = 20.1833)
Rank 2 = UNC (rating = 6.9167)
Rank 3 = WAKE FOREST (rating = -8.7500)
Rank 4 = NCSU (rating = -18.3500)
