John Swanson, president of Market Research Inc., has asked you to estimate the coefficients of the modelY = β0 + β1X1 + β2X21 + b3X2where Y is the expected sales of office supplies for a large retail distributor of office supplies, X1 is the total disposable income of residents within 5 miles of the store, and X2 is the total number of persons employed in information-based businesses within 5 miles of the store. Recent work by a national consulting firm has concluded that the coefficients in the model must have the following restriction:b1 + b2 = 2Describe how you would estimate the model coefficients using least squares.

Name: John Swanson, president of Market Research Inc., has asked you to esti
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Description: John Swanson, president of Market Research Inc., has asked you to estimate the coefficients of the modelY = β0 + β1X1 + β2X21 + b3X2where Y is the expected sales of office supplies for a large retail distributor of office supplies, X1 is the total di

Answered: John Swanson, president of Market…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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John Swanson, president of Market Research Inc., has asked you to estimate the coefficients of the model
Y = β₀ + β₁X₁ + β₂X²₁ + b₃X₂
where Y is the expected sales of office supplies for a large retail distributor of office supplies, X₁ is the total disposable income of residents within 5 miles of the store, and X₂ is the total number of persons employed in information-based businesses within 5 miles of the store. Recent work by a national consulting firm has concluded that the coefficients in the model must have the following restriction:

b₁ + b₂ = 2

Describe how you would estimate the model coefficients using least squares.

Expert Solution

Step 1

Given information:

$Y = β_{0} + β_{1} X_{1} + β_{2} {X_{1}}^{2} + β_{3} X_{2}$

The restriction is,

$β_{1} + β_{2} = 2$

Step 2

Using the least square method, the squared error will be,

$E = \sum_{} {(Y_{i} - \hat{Y})}^{2} = \sum_{} {(Y_{i} - (β_{0} + β_{1} X + β_{2} {X_{1}}^{2} + β_{3} X_{2}))}^{2} = \sum_{} {(Y_{i} - (β_{0} + β_{1} X + (2 - β_{1}) {X_{1}}^{2} + β_{3} X_{2}))}^{2}$

Estimation of $β_{0}$ :

On partial differentiating E with respect to $β_{0}$ and putting it equal to zero, the equation becomes,

$\frac{\partial E}{\partial β_{0}} = 0 \frac{\partial [\sum_{} {(Y_{i} - (β_{0} + β_{1} X_{1} + (2 - β_{1}) {X_{1}}^{2} + β_{3} X_{2}))}^{2}]}{\partial β_{0}} = 0 \sum 2 (Y_{i} - (β_{0} + β_{1} X_{1} + (2 - β_{1}) {X_{1}}^{2} + β_{3} X_{2})) (- 1) = 0 - \sum Y_{i} + \sum_{} β_{0} + β_{1} \sum_{} X_{1} + (2 - β_{1}) {\sum_{} X_{1}}^{2} + β_{3} \sum_{} X_{2} = 0 \sum Y_{i} = n β_{0} + β_{1} \sum_{} X_{1} + (2 - β_{1}) {\sum_{} X_{1}}^{2} + β_{3} \sum_{} X_{2} . . . (1)$

Estimation of $β_{1}$ :

On partial differentiating E with respect to $β_{1}$ and putting it equal to zero, the equation becomes,

$\frac{\partial E}{\partial β_{1}} = 0 \frac{\partial [\sum_{} {(Y_{i} - (β_{0} + β_{1} X_{1} + (2 - β_{1}) {X_{1}}^{2} + β_{3} X_{2}))}^{2}]}{\partial β_{1}} = 0 \sum 2 (Y_{i} - (β_{0} + β_{1} X_{1} + (2 - β_{1}) {X_{1}}^{2} + β_{3} X_{2})) (- X_{1} + {X_{1}}^{2}) = 0 \sum Y_{i} (- X_{1} + {X_{1}}^{2}) - \underset{}{β_{0} \sum} (- X_{1} + {X_{1}}^{2}) - β_{1} \sum_{} X_{1} (- X_{1} + {X_{1}}^{2}) - (2 - β_{1}) {\sum_{} X_{1}}^{2} (- X_{1} + {X_{1}}^{2}) - β_{3} \sum_{} X_{2} (- X_{1} + {X_{1}}^{2}) = 0 \sum Y_{i} (- X_{1} + {X_{1}}^{2}) = \underset{}{β_{0} \sum} (- X_{1} + {X_{1}}^{2}) + β_{1} \sum_{} X_{1} (- X_{1} + {X_{1}}^{2}) + (2 - β_{1}) {\sum_{} X_{1}}^{2} (- X_{1} + {X_{1}}^{2}) + β_{3} \sum_{} X_{2} (- X_{1} + {X_{1}}^{2}) . . . (2)$

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