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Q2. The response time in milliseconds was determined for three different types ...
Jun 28, 2024
Solution by Steps
step 1
The given linear model is yij=μ+τi+ϵij y_{ij} = \mu + \tau_i + \epsilon_{ij} for i=1,2,3 i = 1, 2, 3 and j=1,2,3 j = 1, 2, 3 . We need to write this in the form y=Xβ+ϵ \mathbf{y} = X \boldsymbol{\beta} + \boldsymbol{\epsilon}
step 2
The response times are given in the table: Circuits Typeamp;Response Time1amp;9amp;12amp;102amp;20amp;21amp;233amp;6amp;5amp;8 \begin{array}{cccc} \hline \text{Circuits Type} & \text{Response Time} \\ \hline 1 & 9 & 12 & 10 \\ 2 & 20 & 21 & 23 \\ 3 & 6 & 5 & 8 \\ \hline \end{array} We can write the response vector y \mathbf{y} as: y=(91210202123658) \mathbf{y} = \begin{pmatrix} 9 \\ 12 \\ 10 \\ 20 \\ 21 \\ 23 \\ 6 \\ 5 \\ 8 \end{pmatrix}
step 3
The design matrix X X for the model yij=μ+τi+ϵij y_{ij} = \mu + \tau_i + \epsilon_{ij} is: X=(1amp;1amp;0amp;01amp;1amp;0amp;01amp;1amp;0amp;01amp;0amp;1amp;01amp;0amp;1amp;01amp;0amp;1amp;01amp;0amp;0amp;11amp;0amp;0amp;11amp;0amp;0amp;1) X = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix}
step 4
The parameter vector β \boldsymbol{\beta} is: β=(μτ1τ2τ3) \boldsymbol{\beta} = \begin{pmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \end{pmatrix}
step 5
The error vector ϵ \boldsymbol{\epsilon} is: ϵ=(ϵ11ϵ12ϵ13ϵ21ϵ22ϵ23ϵ31ϵ32ϵ33) \boldsymbol{\epsilon} = \begin{pmatrix} \epsilon_{11} \\ \epsilon_{12} \\ \epsilon_{13} \\ \epsilon_{21} \\ \epsilon_{22} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{32} \\ \epsilon_{33} \end{pmatrix}
step 6
Therefore, the model in matrix form is: y=Xβ+ϵ \mathbf{y} = X \boldsymbol{\beta} + \boldsymbol{\epsilon}
Answer
The model in matrix form is y=Xβ+ϵ \mathbf{y} = X \boldsymbol{\beta} + \boldsymbol{\epsilon} .
Key Concept
Linear Model in Matrix Form
Explanation
The linear model yij=μ+τi+ϵij y_{ij} = \mu + \tau_i + \epsilon_{ij} can be expressed in matrix form as y=Xβ+ϵ \mathbf{y} = X \boldsymbol{\beta} + \boldsymbol{\epsilon} , where y \mathbf{y} is the response vector, X X is the design matrix, β \boldsymbol{\beta} is the parameter vector, and ϵ \boldsymbol{\epsilon} is the error vector.
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Solution by Steps
step 1
To find a generalized inverse of XTX \mathbf{X}^T \mathbf{X} , we first compute XTX \mathbf{X}^T \mathbf{X} : XTX=(9amp;3amp;3amp;33amp;3amp;0amp;03amp;0amp;3amp;03amp;0amp;0amp;3) \mathbf{X}^T \mathbf{X} = \begin{pmatrix} 9 & 3 & 3 & 3 \\ 3 & 3 & 0 & 0 \\ 3 & 0 & 3 & 0 \\ 3 & 0 & 0 & 3 \end{pmatrix}
step 2
One possible generalized inverse G \mathbf{G} of XTX \mathbf{X}^T \mathbf{X} is: G=(19amp;0amp;0amp;00amp;13amp;0amp;00amp;0amp;13amp;00amp;0amp;0amp;13) \mathbf{G} = \begin{pmatrix} \frac{1}{9} & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & \frac{1}{3} \end{pmatrix}
Answer
One possible generalized inverse G \mathbf{G} of XTX \mathbf{X}^T \mathbf{X} is: G=(19amp;0amp;0amp;00amp;13amp;0amp;00amp;0amp;13amp;00amp;0amp;0amp;13) \mathbf{G} = \begin{pmatrix} \frac{1}{9} & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & \frac{1}{3} \end{pmatrix}
Key Concept
Generalized Inverse
Explanation
A generalized inverse of a matrix A \mathbf{A} is a matrix G \mathbf{G} such that AGA=A \mathbf{A} \mathbf{G} \mathbf{A} = \mathbf{A} . It is used when A \mathbf{A} is not invertible.
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Solution by Steps
step 1
To compute a solution to the normal equations β^=GXTy \hat{\boldsymbol{\beta}} = \mathbf{G} \mathbf{X}^T \mathbf{y} , we use the generalized inverse G \mathbf{G} obtained in part (b): G=(19amp;0amp;0amp;00amp;13amp;0amp;00amp;0amp;13amp;00amp;0amp;0amp;13) \mathbf{G} = \begin{pmatrix} \frac{1}{9} & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & \frac{1}{3} \end{pmatrix}
step 2
Compute XTy \mathbf{X}^T \mathbf{y} : XTy=(9333) \mathbf{X}^T \mathbf{y} = \begin{pmatrix} 9 \\ 3 \\ 3 \\ 3 \end{pmatrix}
step 3
Compute β^ \hat{\boldsymbol{\beta}} : β^=GXTy=(19amp;0amp;0amp;00amp;13amp;0amp;00amp;0amp;13amp;00amp;0amp;0amp;13)(9333)=(1111) \hat{\boldsymbol{\beta}} = \mathbf{G} \mathbf{X}^T \mathbf{y} = \begin{pmatrix} \frac{1}{9} & 0 & 0 & 0 \\ 0 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & \frac{1}{3} \end{pmatrix} \begin{pmatrix} 9 \\ 3 \\ 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}
Answer
The solution to the normal equations is: β^=(1111) \hat{\boldsymbol{\beta}} = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}
Key Concept
Normal Equations Solution
Explanation
The normal equations XTXβ^=XTy \mathbf{X}^T \mathbf{X} \hat{\boldsymbol{\beta}} = \mathbf{X}^T \mathbf{y} can be solved using a generalized inverse G \mathbf{G} to find β^=GXTy \hat{\boldsymbol{\beta}} = \mathbf{G} \mathbf{X}^T \mathbf{y} .
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Solution by Steps
step 1
To estimate τ1−2τ2+τ3 \tau_1 - 2\tau_2 + \tau_3 using β^ \hat{\boldsymbol{\beta}} : β^=(1111) \hat{\boldsymbol{\beta}} = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}
step 2
Extract the estimates τ^1,τ^2,τ^3 \hat{\tau}_1, \hat{\tau}_2, \hat{\tau}_3 : τ^1=1,τ^2=1,τ^3=1 \hat{\tau}_1 = 1, \quad \hat{\tau}_2 = 1, \quad \hat{\tau}_3 = 1
step 3
Compute the estimate: τ^1−2τ^2+τ^3=1−2(1)+1=0 \hat{\tau}_1 - 2\hat{\tau}_2 + \hat{\tau}_3 = 1 - 2(1) + 1 = 0
Answer
The estimate of τ1−2τ2+τ3 \tau_1 - 2\tau_2 + \tau_3 is 0.
Key Concept
Linear Combination of Estimates
Explanation
To estimate a linear combination of parameters, substitute the estimated values into the combination and simplify.
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Solution by Steps
step 1
To determine if the estimate in part (d) is the unique BLUE (Best Linear Unbiased Estimator), we need to check if the conditions for the Gauss-Markov theorem are satisfied
step 2
The Gauss-Markov theorem states that the ordinary least squares (OLS) estimator is the BLUE if the errors ϵij \epsilon_{ij} are uncorrelated, have zero mean, and constant variance σ2 \sigma^2
step 3
Given that ϵij∼N(0,σ2) \epsilon_{ij} \sim N(0, \sigma^2) , the conditions for the Gauss-Markov theorem are satisfied
step 4
Therefore, the estimate τ1−2τ2+τ3 \tau_1 - 2\tau_2 + \tau_3 is the unique BLUE
Answer
The estimate τ1−2τ2+τ3 \tau_1 - 2\tau_2 + \tau_3 is the unique BLUE.
Key Concept
Gauss-Markov Theorem
Explanation
The Gauss-Markov theorem ensures that the OLS estimator is the BLUE if the errors are uncorrelated, have zero mean, and constant variance.
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