Click for printer friendely version of this HowTo Linear Models with Multiple Dependent Variables Suppose the observations, or dependent variables, $y_i$s, are vectors with $q$ correlated characteristics instead of single variables, as would be the case of multiple observations made on the same individual. A random sample of $n$ of these vectors could be arranged in a rectangular array to form an $n \times q$ matrix Y, where the first row of Y is the vector of characteristics observed on the first individual, the second row is the vector observed on the second individual and so on. Assuming that the $q$-dimensional observation vector has a multivariate normal distribution, and that the $n$ observations vectors are independent, we can extend the univariate model developed in Sections 3.13.1 through3.13.5 to encompass the $q$ correlated variables. The model now appears as: $${\bf Y} = {\bf X}\boldsymbol{\beta} + \boldsymbol{\epsilon},$$ (3.13.22) which looks exactly like the univariate general linear model in Equation 3.13.3, except in this case, Y is an $n \times q$ matrix and $\boldsymbol{\beta}$ is an $m \times q$ matrix. The matrix X, the design matrix, is the same matrix of known constants that appeared in the univariate model. The hypothesis can be generalized to: $${\bf C}\boldsymbol{\beta}{\bf U} = \boldsymbol{\theta},$$ (3.13.23) where C is an $t \times p$ matrix U is a $q \times u$ matrix, and $\boldsymbol{\theta}$ is an $t \times u$ matrix and C and U are arbitrary matrices designed to yield the appropriate hypothesis. Multiple Regressionno_title Example 3.13.6.1   Example 3.13.6.2 (Multiple Regression)   A series of animals were studied where cardiac output and mean blood pressure were measured while heart rate and respiration were varied. The data from this study can be found in Appendix C.1. We will model with data with the formula: $$(y_{i,1}, y_{i,2}) = \boldsymbol{\beta}_0 + x_{i,1}\boldsymbol{\beta}_1 + x_{i, 2}\boldsymbol{\beta}_2$$ (3.13.24) where $y_{i,1} =$ mean blood pressure of the $i$-th animal, $y_{i,2} =$ cardiac output of the $i$-th animal, $x_{i,1} =$ respiration rate of the i-th animal, $x_{i,2} =$ heart rate of the i-th animal. Thus, $${\bf Y} = \left[ \begin{array}{cc} y_{1,1} & y_{1,2} \vdots & \vdots y_{n,1} & y_{n,2} \end{array} \right], {\bf X} = \left[ \begin{array}{ccc} 1 & x_{1,1} & x_{1,2} \vdots &\vdots & \vdots 1 & x_{n,1} & x_{n,2} \end{array} \right],$$ and $$\boldsymbol{\beta} = \left[ \begin{array}{cc} \beta_{0,1} & \beta_{0,2} \beta_{1,1} & \beta_{1,2} \beta_{n,1} & \beta_{n,2} \end{array} \right].$$ Some questions that we might ask about this data are Does respiration rate affect cardiac output and mean blood pressure? Does heart rate affect cardiac output and mean blood pressure? To answer the first question, we test the hypothesis that the respiration rate regression coefficients are zero: $$\mathrm{H}_0: \beta_{1,1} = \beta_{1,2} = 0.$$ We can convert this hypothesis into matrix form using Equation 3.13.25 by defining C and U such that $${\bf C} = \left[ \begin{array}{ccc} 0 & 1 & 0 \end{array}\r... ... = \left[ \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right],$$ which yields: $$\left[ \begin{array}{ccc} 0 & 1 & 0 \end{array} \right] \left[ \b... ... \end{array} \right] \left[ \begin{array}{cc} 1 & 0 0 & 1 \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \end{array} \right]$$ $$\left[ \begin{array}{cc} \beta_{1,1} & \beta_{1,2} \end{array} \right] \left[ \begin{array}{cc} 1 & 0 0 & 1 \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \end{array} \right]$$ $$\left[ \begin{array}{cc} \beta_{1,1} & \beta_{1,2} \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \end{array} \right]$$ To answer the second question, we test the hypothesis that the heart rate coefficients are zero: $$\mathrm{H}_0: \beta_{2,1} = \beta_{2,2} = 0,$$ for which $${\bf C} = \left[ \begin{array}{ccc} 0 & 0 & 1 \end{array}\r... ... = \left[ \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right].$$ Since, in general, these two tests will not be independent, we should make them simultaneously. To do this, let $${\bf C} = \left[ \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1 ... ... = \left[ \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right].$$ Using Equation 3.13.25, this yields: $${\bf C} \boldsymbol{\beta} {\bf U} = \left[ \begin{array}{c... ... = \left[ \begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right].$$ We are now ready to use the multivariable version of the general linear models program found in Appendix $\vert\boldsymbol{\vert}$Hotelling T$^2$no_title Example 3.13.6.3   Example 3.13.6.4 (Hotelling T$^2$)   Hotelling T$^2$ tests allow us to run simultaneous paired $t$-tests on $q$ pairs of characteristics. Consider, in this case, that we have measured mean blood flow, mean blood pressure, cerebro-vascular resistance in a series of experimental subjects both before and after the administration of epinephrine. Using a Hotelling T$^2$ we can answer the question: Did the drug change the blood flow, pressure and resistance significantly? The model for this experiment is thus $${\bf Y} = \boldsymbol{\beta}{\bf X}$$ where $y_{i,1} =$ blood flow before $y_{i,2} =$ blood pressure before $y_{i,3} =$ resistance before $y_{i,1} =$ blood flow after $y_{i,2} =$ blood pressure after $y_{i,3} =$ resistance after $$\boldsymbol{\beta} = \left[ \begin{array}{cccccc} \beta_{\... ...begin{array}{c} 1\\ 1\\ 1\\ \vdots\\ 1 \end{array}\right],$$ where the subscript 'b' refers to a measurement taken before the treatment and the subscript 'a' refers to a measurement taken after. To answer our question about whether the drug changed blood flow, pressure and resistance, we ask if the parameters for before and after measurements have changed. Thus, we form the hypothesis: $$\mathrm{H}_0: \beta_{\mathrm{fb}} - \beta_{\mathrm{fa}} = 0$$ $$\beta_{\mathrm{pb}} - \beta_{\mathrm{pa}} = 0$$ $$\beta_{\mathrm{rb}} - \beta_{\mathrm{ra}} = 0$$ This leads us to define C and U as: $${\bf C} = [1] \textrm{ and } {\bf U} = \left[ \begin{array}... ...\\ -1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{array}\right],$$ so that $${\bf C}\boldsymbol{\beta}{\bf U} = [ \begin{array}{ccc} 0 & 0 & 0 \end{array} ],$$ or, in other words, $$\left[ \begin{array}{ccc} (\beta_{\mathrm{bf}}-\beta_{\mathrm{ba}... ..._{\mathrm{pa}}) & (\beta_{\mathrm{rf}}-\beta_{\mathrm{ra}}) \end{array} \right]$$ $$= [ \begin{array}{ccc} 0 & 0 & 0 \end{array} ].$$ The output from our program is... $\vert\boldsymbol{\vert}$ Multivariate ANOVAno_title Example 3.13.6.5   Example 3.13.6.6 (Multivariate ANOVA)   A series of twenty-four animals were studied by dividing them into six groups according to their diet and sex. The cardiac output, heart rate, and initial body weight of the animals were measured. Since body weight was thought to affect the level of response, it is considered a covariate. Cardiac output and heart rate are both dependent variables. Our model is thus, $$\begin{split}(y_{i,1}, y_{i,2}) & = x_{i,1}\beta_1 + x_{i,2}\... ...ta_1 + x_{i,5}\beta_2 + x_{i, 6}\beta_3 + x_{i, c}, \end{split}$$ where $$x_{i,t}\begin{cases}1 & \textrm{if animal received treatment }t, 0 & \textrm{if animal did not receive treatment }t, \end{cases}$$ and $$x_{i,c} = \textrm{initial body weight}.$$ To test if cardiac output and heart rate vary with sex, we construct the contrast matrices: $${\bf C} = \left[ \begin{array}{ccccccc} 1 & 1 & 1 & -1 & -1 & -1 & 0 \end{array}\right]$$ and $${\bf U} = \left[ \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right].$$ Thus, our hypothesis is3.27: $$\left[ \begin{array}{c} (\beta_{1,1} + \beta_{2,1} + \beta_{... ...eta_{4,2} + \beta_{5,2} + \beta_{6,2}) \end{array}\right] = 0.$$ $\vert\boldsymbol{\vert}$ Click for printer friendely version of this HowTo

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