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Q2. The response time in milliseconds was determined for three different types o...

Jun 28, 2024

Q2. The response time in milliseconds was determined for three different types of circuits that could be used in an automatic valve shutoff mechanism. The results are shown in the following table. Circuits Type Response Time 1 9 12 10 2 20 21 23 3 6 5 8 Consider the linear model yij = µ + τi + ϵij , for i = 1, 2, 3 and j = 1, 2, 3 where where µ is the overall mean, τi is the circuit type content effects and ϵij ∼ N(0, σ2 ) is the random error. (a) Write model above in the form y = Xβ +ϵ. Do not impose any restriction on the parameters. This assignment consists of 4 questions on 4 printed pages 3 MEME16203Linear Models (b) Obtain one of the generalized inverse of XTX, G. (c) Use the generalized inverse you obtained in part(b) to compute a solution to the normal equations, βˆ =      µˆ τˆ1 τˆ2 τˆ3      . (d) Using your solution βˆ to the normal equation from part (c), estimates τ1 − 2τ2 + τ3. (e) Is your estimate in part(d) the unique BLUE? Explain. (25 marks)

Solution by Steps

step 1

Define the linear model in matrix form. The model given is

y_{ij} = \mu + \tau_i + \epsilon_{ij}

, where

i = 1, 2, 3

and

j = 1, 2, 3

. We can write this in the form

y = X\beta + \epsilon

step 2

Construct the design matrix

X

. For three circuit types and three observations each, the design matrix

X

is:

X = \begin{pmatrix} 1 &amp; 1 &amp; 0 &amp; 0 \\ 1 &amp; 1 &amp; 0 &amp; 0 \\ 1 &amp; 1 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 0 &amp; 1 \\ 1 &amp; 0 &amp; 0 &amp; 1 \\ 1 &amp; 0 &amp; 0 &amp; 1 \\ \end{pmatrix}

step 3

Define the parameter vector

\beta

and the response vector

y

. The parameter vector

\beta

is:

\beta = \begin{pmatrix} \mu \\ \tau_1 \\ \tau_2 \\ \tau_3 \\ \end{pmatrix}

The response vector

y

is:

y = \begin{pmatrix} 9 \\ 12 \\ 10 \\ 20 \\ 21 \\ 23 \\ 6 \\ 5 \\ 8 \\ \end{pmatrix}

step 4

Write the model in the form

y = X\beta + \epsilon

. Combining the design matrix

X

, parameter vector

\beta

, and response vector

y

, we get:

y = X\beta + \epsilon

Answer

The linear model in matrix form is

y = X\beta + \epsilon

Key Concept

Linear Model in Matrix Form

Explanation

The linear model

y_{ij} = \mu + \tau_i + \epsilon_{ij}

can be written in matrix form as

y = X\beta + \epsilon

, where

X

is the design matrix,

\beta

is the parameter vector, and

y

is the response vector.

Solution by Steps

step 1

Compute

X^T X

. The transpose of

X

is:

X^T = \begin{pmatrix} 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 1 &amp; 1 &amp; 1 \\ \end{pmatrix}

Multiplying

X^T

X

X^T X = \begin{pmatrix} 9 &amp; 3 &amp; 3 &amp; 3 \\ 3 &amp; 3 &amp; 0 &amp; 0 \\ 3 &amp; 0 &amp; 3 &amp; 0 \\ 3 &amp; 0 &amp; 0 &amp; 3 \\ \end{pmatrix}

step 2

Find a generalized inverse

G

X^T X

. One possible generalized inverse is:

G = \begin{pmatrix} \frac{1}{9} &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; \frac{1}{3} &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; \frac{1}{3} &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; \frac{1}{3} \\ \end{pmatrix}

Answer

One generalized inverse of

X^T X

is:

G = \begin{pmatrix} \frac{1}{9} &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; \frac{1}{3} &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; \frac{1}{3} &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; \frac{1}{3} \\ \end{pmatrix}

Key Concept

Generalized Inverse

Explanation

A generalized inverse of a matrix

A

is a matrix

G

such that

AGA = A

. It is used when

A

is not invertible.

Solution by Steps

step 1

Compute

\hat{\beta} = G (X^T y)

. First, calculate

X^T y

X^T y = \begin{pmatrix} 114 \\ 31 \\ 64 \\ 19 \\ \end{pmatrix}

Then, multiply by

G

\hat{\beta} = G (X^T y) = \begin{pmatrix} \frac{1}{9} &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; \frac{1}{3} &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; \frac{1}{3} &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; \frac{1}{3} \\ \end{pmatrix} \begin{pmatrix} 114 \\ 31 \\ 64 \\ 19 \\ \end{pmatrix} = \begin{pmatrix} 12.67 \\ 10.33 \\ 21.33 \\ 6.33 \\ \end{pmatrix}

Answer

The solution to the normal equations is:

\hat{\beta} = \begin{pmatrix} 12.67 \\ 10.33 \\ 21.33 \\ 6.33 \\ \end{pmatrix}

Key Concept

Normal Equations Solution

Explanation

The solution to the normal equations

\hat{\beta} = (X^T X)^{-1} X^T y

can be computed using a generalized inverse when

X^T X

is not invertible.

Solution by Steps

step 1

Use the solution

\hat{\beta}

to estimate

\tau_1 - 2\tau_2 + \tau_3

. From the solution

\hat{\beta}

, we have:

\hat{\tau}_1 = 10.33, \quad \hat{\tau}_2 = 21.33, \quad \hat{\tau}_3 = 6.33

step 2

Calculate

\hat{\tau}_1 - 2\hat{\tau}_2 + \hat{\tau}_3

\hat{\tau}_1 - 2\hat{\tau}_2 + \hat{\tau}_3 = 10.33 - 2(21.33) + 6.33 = 10.33 - 42.66 + 6.33 = -26

Answer

The estimate of

\tau_1 - 2\tau_2 + \tau_3

-26

Key Concept

Linear Combination of Parameters

Explanation

The estimate of a linear combination of parameters can be obtained by substituting the estimated values of the parameters into the linear combination.

Solution by Steps

step 1

Determine if the estimate is the unique BLUE. The Best Linear Unbiased Estimator (BLUE) is unique if the design matrix

X

has full column rank

step 2

Check the rank of

X

. The design matrix

X

has 4 columns and 9 rows. The rank of

X

is 4, which is the number of columns

step 3

Conclude that the estimate is the unique BLUE. Since

X

has full column rank, the estimate is the unique BLUE

Answer

The estimate is the unique BLUE because the design matrix

X

has full column rank.

Key Concept

Unique BLUE

Explanation

The Best Linear Unbiased Estimator (BLUE) is unique if the design matrix has full column rank, ensuring that the normal equations have a unique solution.