Detailed Solution to Regression and Distribution Problems

This assignment solution covers regression and distribution problems, including maximum likelihood estimation and Fisher Information Matrix analysis. The solution discusses Model A and Model B for pasture yield prediction, providing calculations and standard errors for maximum likelihood estimato...Read More

Last Name 1
REGRESSION AND DISTRIBUTION PROBLEMS
Student's Name
Professor's Name
Course Name
Date

Last Name 2
Regression and Distribution Problems
Question 1
Part A
The likelihood functions of different parameters can be maximized with respect to parameter ϴ
to get the estimators of these parameters1.
The maximum likelihood function:
^β1=β1=∑ ( Xi− X ) (Y i −Y )
∑ ( Xi −X )
2
^β2=β2=Y −B1 X
^v=∑ Y i
n
v=
∑
i
( Y i− ^Y i ) 2
n
Part B
Fisher Information Matrix
Segment 1
For Model A
I ( β1 , β2 , v)= ∑
−i=1
n
log ( 1
( 2 π σ 2 )
n
2
e( −1
2 σ2 ∑
i=1
n
( y− β1− β2 t )2¿)
)¿
For Model B
1 Montgomery, Douglas C., Elizabeth A. Peck, and G. Geoffrey Vining. Introduction to Linear
Regression Analysis. Hoboken: John Wiley & Sons, 2012.

Last Name 3
I ( β1 , β2 , v)= ∑
−i=1
n
log ( 1
( 2 π σ 2 )
n
2
e( −1
2 σ2 ∑
i=1
n
( y− β1
1 +14 e−β2 t )2
¿)
)¿
Segment 2
For Model A
I ( ^β1 , ^β2 , ^v )−1
=log ¿
For Model B
I ( ^β1 , ^β2 , ^v )−1
=log ¿
Part C
The scatter line and fitted line
Part D
The results are in the Appendix for the estimation computations
Hence for

Last Name 4
Model A
X=0.92655t-0.59207
Therefore, yield at time 0 and 80
X=(0.92655*0)-0.59207
Pasture Yield=-0.59207
X=(0.92655*80)-0.59207
Pasture Yield=73.53193
Model B
X= 72.28575
1+14 e−0.0680184 t
Therefore, yield at time 0 and 80
X= 72.28575
1+ 14 e−0.0680184∗0
Pasture Yield=4.8
X= 72.28575
1+ 14 e−0.0680184∗80
Pasture Yield= 68.14571
Part E
Standard errors for the maximum likelihood estimators
Model A
^β1=0.0450088
^β2=2.20625
Model B
^β1=1.408413

Last Name 5
^β2=0.0018188
Part F
The best model to utilize would have to be model B it fits a majority of the data points and it
explains a significant proportion of the change in the dependent variables (Pasture Yield)
Question 2
Part A
X follows a binomial distribution X Bi(m , p)
Hence the pmf of this can be rewritten as follows
( m
x ) px ( 1− p ) m−x=exp [ ( log p
1− p ) x−mlog (1− p) ] ( m
x )
Giving us the Exponential Family form with n ( p ) =log p
1− p
Part B
X is a complete sufficient statistic because for Eϴ(g(X))=0; for all ϴ, implies therefore that
Pϴ(g(X)=0)= 12
Part C
(i)
If for a sufficient statistic like X, there exists a measurable function where g: ϴ→ X such that
X=g(ϴ)
(ii)
if we let ϴ be equivalent to 0 then Pj(ϴ)=1. As such, X is a complete sufficient statistic.
2 Schervish, Mark J. Theory of Statistics. Berlin: Springer Science & Business Media, 2012.

Last Name 6
Question 3
Part A
Give a statistic T and we have Varθ ( T ) < ∞ for all θ, We also have S= X 1+ X 2+ X 3
0<l ( θ ) < ∞. With this in mind we can state that for all θ
Varθ (T )≥ |φ ' (θ)|
2
l(θ)
Where φ ( θ )= θ
θ +1
So that φ' ( θ ) =
d ( θ
θ+1 )
dθ
φ' ( θ ) =Cov ⌈ ∂ log p(S ; θ)
dθ , T ⌉ for S=X 1+ X 2+ X 3
And we know the
cov (P , Q)≜ E ⌈ ( P−E [ P ] ) ( Q−E ⌈ Q ⌉ ) ⌉
We also know that
l ( θ ) =−E p(x , θ) { ∂2
∂ θ2 log p( S ;θ) } for S= X1+X2+X3
As such
Varθ ( T ) ≥
Cov ⌈ ∂ log p ( S ;θ )
dθ , T ⌉
−E p (x , θ ) { ∂2
∂ θ2 log p ( S ;θ ) } for S=X 1+ X 2+ X 3
By substituting in the values of S=X1+X2+X3 into the inequality above and simplifying we get:
varθ (T )≥ 1
( 0+1 ) 4 { 1
θ ⌈ 1
θ+1 +1+ 1
1−e−1 ⌉ − e−θ
( 1−e−θ )
2 − 1
( θ+1 ) 2 }
−1
Therefore, proving our assumption about T being an unbiased estimator of φ ( θ )

Last Name 7
Part B
We need to prove that for when θ assumes a value of zero S will assume a value of 1
Where S=X1+X2+X3
Therefore at least one of the three has to equate to one (while others assume values of 0)
Focusing on X1
We get that S is a complete sufficient statistic because f1(g(t))=0 for all ϴ we see that
S(g(t)=0)=1 (NB: X2 and X3 will be equivalent to 0)
We can test the preposition made above as follows
For g ( θ )=0 hence j=0 and in all cases where j { j=0,1 for a bernoulli distribution
j=0,1, … k −1 for a Poisson distribtion }
Let X1= f 1 ( 0|θ ) = 1
θ+1 for θ=0 we get X 1=1
Let X2= f 2 ( 0|θ )= e−θ θ j
j ! for θ=0 we get X 2=0
Let X3= f 3 ( 0|θ )= e−θ θj
j !(1−e−θ ) we get X 3=0 becausedoes not have g (θ )=0 i . e . j=1,2 , …
Hence
S=X1+X2+Xe
S=1+0+0
S=1
Since S=1, we conclude that S is a complete sufficient statistic of θ
Part C
Since S is a complete sufficient statistic
Therefore the expected value of the following expression will tend to φ (θ)

Last Name 8
Exp⌈ S ( 2S−1−1 )
2S−1+S(2S−1−1) ⌉
This will create a 1:1 relationship with φ (θ)
Thus implying the expression about is a good UMVU estimator of φ (θ)
Bibliography
Montgomery, Douglas C., Elizabeth A. Peck, and G. Geoffrey Vining. Introduction to Linear
Regression Analysis. Hoboken: John Wiley & Sons, 2012.
Schervish, Mark J. Theory of Statistics. Berlin: Springer Science & Business Media, 2012.

Last Name 9
Appendix

Last Name 10
_cons -.5920715 2.20625 -0.27 0.796 -5.809024 4.62488
t .92655 .0450088 20.59 0.000 .820121 1.032979
X Coef. Std. Err. t P>|t| [95% Conf. Interval]
Total 4648.06336 8 581.007919 Root MSE = 3.2848
Adj R-squared = 0.9814
Residual 75.5288929 7 10.7898418 R-squared = 0.9838
Model 4572.53446 1 4572.53446 Prob > F = 0.0000
F(1, 7) = 423.78
Source SS df MS Number of obs = 9
. regress X t
/b2 .0680184 .0018188 37.40 0.000 .0637177 .0723191
/b1 72.28575 1.408413 51.32 0.000 68.95538 75.61612
X Coef. Std. Err. t P>|t| [95% Conf. Interval]
Total 18223.42 9 2024.82447 Res. dev. = 24.62788
Root MSE = 1.077813
Residual 8.1317662 7 1.16168088 Adj R-squared = 0.9994
Model 18215.288 2 9107.64422 R-squared = 0.9996
Number of obs = 9
Source SS df MS
Iteration 12: residual SS = 8.131766
Iteration 11: residual SS = 8.131766
Iteration 10: residual SS = 8.131767
Iteration 9: residual SS = 8.133913
Iteration 8: residual SS = 11.73754
Iteration 7: residual SS = 104.6428
Iteration 6: residual SS = 141.8845
Iteration 5: residual SS = 173.5564
Iteration 4: residual SS = 188.2788
Iteration 3: residual SS = 220.8791
Iteration 2: residual SS = 1470.832
Iteration 1: residual SS = 2153.957
Iteration 0: residual SS = 4648.063
(obs = 9)
. nl (X = {b1}/(1 +(14*exp(-1*{b2}*t))))