MSDS 567: Statistical Models and Computing Assignment Solution

This document presents a comprehensive solution to an advance econometrics assignment. The assignment focuses on analyzing data from a rat weight experiment, where rats were subjected to different treatments (control, thiouracil, and thyroxin) and weighed over several weeks. The solution explores different statistical model specifications, including variance and covariance calculations, to assess the impact of the treatments on rat weight gain. The analysis includes tables and figures illustrating the average weight of rats over time for each treatment. Furthermore, the solution delves into the similarities and dissimilarities between different assumptions related to the model and provides a discussion on the most suitable model for the given case study. The assignment also provides a maximum likelihood estimator for the given model. The document provides a detailed analysis of the experiment and the statistical models used.

Running head: ADVANCE ECONOMETRICS
Advance Econometrics
Name of the Student:
Name of the University:
Author’s Note:

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Table of Contents
Introduction:....................................................................................................................................2
Answer 1..........................................................................................................................................2
Answer 1.A......................................................................................................................................2
Answer 1.B......................................................................................................................................3
Answer 1.C......................................................................................................................................4
Similarities...................................................................................................................................4
Dissimilarities..............................................................................................................................4
Answer 1.D......................................................................................................................................4
Answer 1.E......................................................................................................................................6
Answer 2..........................................................................................................................................7

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Introduction:
Answer 1
yij=β0 i+ β1 i tij +eij
Answer 1.A
yij=β0 ,i +b0 i+ β1 i tij +eij
Variance:
var ( yij|aij ) =var (β0 ,i +b0 i + β1 i tij+ eij)
var ( yij|aij ) =var (b0 i +eij)
var ( yij|aij ) =var ( b0 i ) +var ( eij )
var ( yij|aij ) =D+σ2
Covariance
cov ( yij , yij∨aij)=E[ yij−E ( yij ) ]E [ yij∨aij−E( yij∨aij)]
cov ( yij , yij|aij )=0
As, the conditional mean and unconditional mean are not same.
Correlation
corr ( yij , yij|aij ) = cov ( yij , yij|aij )
[ var ( yij ) var ( yij|aij ) ] 1
2

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corr ( yij , yij|aij ) =0
Answer 1.B
yij=β0 ,i +b0 i+(β1 i+b1 i)tij +eij
Matrix form,
y=(β +b)t +e
Variance
var ( y∨a)=var [ ( β +b ) t+ e ]
var ( y∨a)=var [ ( β +b ) t ] +var (e)
var ( y )=tD+σ2
Covariance cov ( y , y |a ) =E { y−E ( y ) } E ¿
cov ( y , y|a )=0
Correlation
corr ( y , y |a ) = cov ( y , y|a )
[ var ( y ) var ( y|a ) ] 1
2
corr ( y , y|a ) =0
Answer 1.C
` In this section, the discussion will be on the similarities and dissimilarities between the
two assumptions. In the first assumption an additional term is introduced to the intercept term in

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the 1st equation and in the second assumption two additional terms are introduced to the intercept
term and slope term in the 1st equation.
Similarities
The covariance and correlation between Yij and Yij|aij are same and equals to zero for the
both of the assumptions as the conditional and unconditional mean of Yij are different.
Dissimilarities
The variances of conditional Yij are different for both the assumptions as the additional
term in slope term has a variance which is associated with the independent variable. It can be
said that the introduction of additional term to the intercept term gives a constant variance and
the additional term to the slope term gives a variance which is dependent on the variable ‘t’.
Answer 1.D
To see the difference in the weights of rats over the week for different treatment the
average value is calculated. The below table shows that the treatment 3 i.e. thyroxin is more
effective in raising the weight of a rat than the other treatments as the average weight of the rat is
greater in the last week for this treatment. It works slow in the beginning phase but gains the
weights more than the others. The treatment 1 i.e. control treatment is good for the beginning
phase as it has higher average in week 1, week 2 and week 3. The average weight of treatment is
always low that means the 2nd treatment is not healthy for the rats.
Table 1: Average weight of rat in each week corresponding to the received treatment
Row
Labels
Average of
Week 0
Average of
Week 1
Average of
Week 2
Average of
Week 3
Average of
Week 4
1 54 78.5 106 130.1 160.6

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2 54.7 76.3 95.8 108.2 124
3 55.57 75.86 104.86 132.71 162.86
The below figure is derived from the table 1 which presents the weight in vertical axis
and the time i.e. the weeks are presented in the horizontal axis. The diagram is represented to
show that the treatment 2 is unhealthy and the treatment 1 is more effective and the 3rd treatment
wins the race in gaining weight.
Average of Week 0 Average of Week 1 Average of Week 2 Average of Week 3 Average of Week 4
0
20
40
60
80
100
120
140
160
180
Effect of Treatments Over the Week
Series1 Series2 Series3
Figure 1: Average weight of rat for different treatments over the weeks
The table 2 shows the variance co-variance matrix of weights of the rats over weeks and
table 3 shows the variance co-variance matrix of average weight of rats for treatments. The
variance from both tables have a huge difference. Covariance of week 0 and week 1 is 4401.04
form the table 2 and 6314.05 from table 3.

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Table 2: Variance Co-variance Matrix of rat weights over weeks
Week 0 Week 1 Week 2 Week 3 Week 4
Week 0 3123.69 4401.04 5815.00 6989.85 8402.35
Week 1 4401.04 6221.88 8217.00 9869.77 11862.69
Week 2 5815.00 8217.00 10899.08 13138.65 15821.27
Week 3 6989.85 9869.77 13138.65 15908.85 19191.38
Week 4 8402.35 11862.69 15821.27 19191.38 23205.15
Table 3: Variance co-variance matrix of average weight of rats in terms of treatment over weeks
Week 0 Week 1 Week 2 Week 3 Week 4
Week 0 4498.14 6314.05 8395.66 10159.53 12252.70
Week 1 6314.05 8869.12 11792.35 14267.92 17211.09
Week 2 8395.66 11792.35 15704.33 19036.10 22989.77
Week 3 10159.53 14267.92 19036.10 23123.17 27962.16
Week 4 12252.70 17211.09 22989.77 27962.16 33845.40
Answer 1.E
The figure 1 shows that the average weight is increasing over the time depending on the
treatment that is provided to the rat. Different treatments have different pace of growth in rat
weight. That means a factor is associated with the time, which depends on the treatment
(Kripfganz 2016). That means the coefficient of time consist that factor and it can be represented
as below
β1 i=β1 ,1+ b1 i where ai=1
The intercept term presents the initial weight which also consist some variance which is
shown in the table 2 and table 3. For proper estimation an additional term will be helpful which
can be presented as below:
β0 i=β0 ,1 +b0 i where where ai=1

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The above discussion indicates that the assumption 2 that is presented in Q1.b, is the
proper model for the given case study.
Answer 2
The given model is
yij=β +bi +eij
Assumptions:
The rearranged model is as follows:
Y A +XB=U
In the above, model Y matrix contains yij, A is the unit matrix, X contains bi, B matrix contains
the β and a unit matrix and U is the random error matrix. The rows of U are independently and
normally distributed. Vector mean is zero and the nonsingular variance matrix is V.
Let’s assume δ denotes the unknown elements of matrix ( A
B ), y denotes the columns of Y, δ is the
coefficient matrix of Z, Z contains the explanatory variable and G is Kronnecker product of V-1
and unit matrix I (Silverman 2018).
The simplest form for the maximum likelihood estimator is as follows:
^Z ' ^G Z ^δ= ^Z ' ^G y

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Reference
Silverman, Bernard W. Density estimation for statistics and data analysis. Routledge, 2018.
Kripfganz, Sebastian. "Quasi–maximum likelihood estimation of linear dynamic short-T panel-
data models." The Stata Journal 16, no. 4 (2016): 1013-1038.