Solutions for Spatial Linear Models for Environmental Data, 1st Edition by Zimmerman (All Chapters included)

Spatial Linear Models for
Environmental Data, 1st Edition by
Dale L. Zimmerman



Complete Chapter Solutions Manual
are included (Ch 1 to 12)




** Immediate Download
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** All Chapters included

,Table of Contents are given below


1. Introduction

2. An Introduction to Covariance Structures for Spatial Linear Models

3. Exploratory Spatial Data Analysis

4. Provisional Estimation of the Mean Structure by Ordinary Least Squares

5. Generalized Least Squares Estimation of the Mean Structure

6. Parametric Covariance Structures for Geostatistical Models

7. Parametric Covariance Structures for Spatial-Weights Linear Models

8. Likelihood-Based Inference

9. Spatial Prediction

10. Spatial Sampling Design

11. Analysis and Design of Spatial Experiments

12. Extensions

, Solutions to Exercises in Spatial Linear Models for
Environmental Data
1 Introduction
There are no exercises for this chapter.


2 An Introduction to Covariance Structures for Spatial Linear
Models

SOLUTION TO EXERCISE 2.1.

Cov(Y1 , Y2 ) = E[(Y1 − µ1 )(Y2 − µ2 )]
= E[Y1 Y2 − Y1 µ2 − µ1 Y2 + µ1 µ2 ]
= E(Y1 Y2 ) − µ2 E(Y1 ) − µ1 E(Y2 ) + µm u2
= E(Y1 Y2 ) − µ1 µ2 ,

thus establishing (2.7). To obtain (2.8), first observe that E(a1 Y1 +b1 ) = a1 µ1 +b1 and similarly E(a2 Y2 +b2 ) =
a2 µ2 + b2 . Then using the definition of the covariance between two random variables and (2.6),

Cov(a1 Y1 + b1 , a2 Y2 + b2 ) = E{[a1 Y1 + b1 − (a1 µ1 + b1 )][a2 Y2 + b2 − (a2 µ2 + b2 )]}
= E[a1 (Y1 − µ1 )a2 (Y2 − µ2 )]
= a1 a2 σ12 .

Also observe that E(a1 Y1 + a2 Y2 ) = a1 µ1 + a2 µ2 . Then using (2.4),

Var(a1 Y1 + a2 Y2 ) = E[(a1 Y1 + a2 Y2 )2 ] − (a1 µ1 + a2 µ2 )2
= E(a21 Y12 + a22 Y22 + a1 a2 Y1 Y2 ) − (a21 µ21 + a22 µ22 + a1 a2 µ1 µ2 )
= a21 [E(Y12 ) − µ21 ] + a22 [E(Y22 ) − µ22 ] + a1 a2 [E(Y1 Y2 ) − µ1 µ2 ]
= a21 σ12 + a22 σ22 + a1 a2 σ12 ,

which establishes (2.9).


SOLUTION TO EXERCISE 2.2.
Using (2.9), and assuming that σ12 and σ22 exist,
1
γ12 = Var(Y1 − Y2 )
2
1
= [Var(Y1 ) + Var(Y2 ) − 2Cov(Y1 , Y2 )]
2
1 2
= (σ + σ22 − 2σ12 ).
2 1
If σ12 = σ22 = σ 2 , then γ12 = 12 (σ 2 + σ 2 − 2σ12 ) = σ 2 − σ12 = σ 2 (1 − ρ12 ).


SOLUTION TO EXERCISE 2.3.



2

, By (2.15),
 
4 X 4
1 X
Var(Y ) = Var  Y (i, j)
16 i=1 j=1
4 4 4 4
1 XXXX
= Cov(Y (i, j), Y (k, l))
162 i=1 j=1
k=1 l=1
4 4 4 4
1 X X X X |i−k|+|j−l|
= 0.5 .
162 i=1 j=1
k=1 l=1

The values of |i − k| + |j = l| in the exponent range from 0 to 6, and occur with the following frequencies:

Value of i − k| + |j = l| Possible values of |i − k| and |j − l| Frequency
0 (0,0) 4 × 4 = 16
1 (1,0) and (0,1) (6 × 4) + (4 × 6) = 48
2 (2,0) and (1,1) and (0,2) (4 × 4) + (6 × 6) + (4 × 4) = 68
3 (3,0) and (2,1) and (1,2) and (0,3) (2 × 4) + (4 × 6) + (6 × 4) + (4 × 2) = 64
4 (1,3) and (2,2) and (3,1) (6 × 2) + (4 × 4) + (2 × 6) = 40
5 (3,2) and (2,3) (2 × 4) + (4 × 2) = 16
6 (3,3) 2×2=4
Therefore,
1
Var(Y ) = [(16 × 0.50 ) + (48 × 0.51 ) + (68 × 0.52 ) + (64 × 0.53 ) + (40 × 0.54 )
162
+ (16 × 0.55 ) + (4 × 0.56 )]
.
= 0.266.
1
Under the standard assumption of uncorrelated observations, Var(Y ) = 16 = 0.067. The standard 95%
confidence interval for the mean, derived under the uncorrelatedness assumption, is
1.96
Y ± √ = Y ± 0.49.
16
The nominal coverage probability of this interval is 0.95, but the true coverage probability is
 
0.49 Y −µ 0.49
P (Y − 0.49 < µ < Y + 0.49) = P √ <√ <√
0.266 0.266 0.266
= Φ(0.950) − Φ(−0.950)
= 0.658.



SOLUTION TO EXERCISE 2.4.
First observe that Y − Y (s0 ) may be written as aT y where a = ( n1 , n1 , . . . , n1 , −1)T is of length n + 1 and
y = (Y (s1 ), Y (s2 ), . . . , Y (sn ), Y (s0 ))T . Observe that the elements of a sum to 0. Then by (2.22),
 
n n+1 n−1 n n
X X 1 X X 1 X
Var(aT y) = −2 ai aj γij = −2  2 γij − γi0 
i=1 j=i+1
n i=1 j=i+1
n i=1



3