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Magazine Chapter 1: Q12E (page 1)
Question:Consider a problem of multiple linear regression in which a patient鈥檚 reaction \({\bf{Y}}\) to a new drug \({\bf{B}}\) is to be related to her reaction \({{\bf{X}}_{\bf{1}}}\) to a standard drug \({\bf{A}}\) and her heart rate . Suppose that for all values \({{\bf{X}}_{\bf{1}}}{\bf{ = }}{{\bf{x}}_{\bf{1}}}\),\({{\bf{X}}_{\bf{2}}}{\bf{ = }}{{\bf{x}}_{\bf{2}}}\), the regression function has the form \({\bf{E(Y) = }}{{\bf{\beta }}_{\bf{0}}}{\bf{ + }}{{\bf{\beta }}_{\bf{1}}}{{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{\beta }}_{\bf{2}}}{{\bf{x}}_{\bf{2}}}\), and the values of \({\bf{10}}\)sets of observations\(\left( {{{\bf{x}}_{{\bf{i1}}}}{\bf{,}}{{\bf{x}}_{{\bf{i2}}}}{\bf{,}}{{\bf{Y}}_{\bf{i}}}} \right)\) are given in Table \({\bf{11}}{\bf{.2}}\), on page \({\bf{696}}\). Under the standard assumptions of multiple linear regression, determine the M.L.E.鈥檚 of \({{\bf{\hat \beta }}_{\bf{0}}}{\bf{,}}{{\bf{\hat \beta }}_{\bf{1}}}{\bf{,}}{{\bf{\hat \beta }}_{\bf{2}}}{\bf{,\;}}{{\bf{\hat \sigma }}^{\bf{2}}}\).
Short Answer
The values of MLE of \({\hat \beta _0},{\hat \beta _1},{\hat \beta _2},{\rm{\;}}{\hat \sigma ^2}\)are \({\hat \beta _0} = - 11.453,\quad {\hat \beta _1} = 0.450,\quad {\hat \beta _2} = 0.173{\rm{\;and\;}}{\hat \sigma ^2} = 0.887\).
Step by step solution
Define linear statistical models
A linear model describes the correlation between the dependent and independent variables as a straight line.
\(y = {a_0} + {a_1}{x_1} + {a_2}{x_2} + \tilde A,\hat A1/4 + {a_n}{x_n}\)
Models using only one predictor are simple linear regression models. Multiple predictors are used in multiple linear regression models. For many response variables, multiple regression analysis models are used
Design the matrix
Based on the given data, we wish to fit a regression function of the form,
\(y = {\beta _0} + {\beta _1}{x_1} + {\beta _2}{x_2}.\)
The design matrix is
\(Z = \left( {\begin{array}{*{20}{c}}1&{{x_{1,1}}}&{{x_{1,2}}}\\1&{{x_{2,1}}}&{{x_{2,2}}}\\ \vdots & \vdots & \vdots \\1&{{x_{10,1}}}&{{x_{10,2}}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{1.9}&{66}\\1&{0.8}&{62}\\1&{1.1}&{64}\\1&{0.1}&{61}\\1&{ - 0.1}&{63}\\1&{4.4}&{70}\\1&{4.6}&{68}\\1&{1.6}&{62}\\1&{5.5}&{68}\\1&{3.4}&{66}\end{array}} \right)\)
and the transpose matrix containing the values of response variable is
\({Y^\prime } = \left( {\begin{array}{*{20}{l}}{0.7}&{ - 1.0}&{ - 0.2}&{ - 1.2}&{ - 0.1}&{3.4}&{0.0}&{0.8}&{3.7}&{2.0}\end{array}} \right)\)
According to the theorem, the MLE is
We can find that
\( \Rightarrow {\hat \beta _0} = - 11.453,{\hat \beta _1} = 0.450,{\hat \beta _2} = 0.173.\)
Hence the desired matrix is \(\left( {\begin{array}{*{20}{c}}{ - 11.453}\\{0.450}\\{0.173}\end{array}} \right)\).
Find the values
The MLE was found to be
Substitute the known values in the formula,
Hence the values are \({\hat \beta _0} = - 11.453,\quad {\hat \beta _1} = 0.450,\quad {\hat \beta _2} = 0.173{\rm{\;and\;}}{\hat \sigma ^2} = 0.887\)
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