Regression and Distribution Problems

This text covers topics such as maximum likelihood function, Fisher information matrix, complete sufficient statistic, and UMVU estimator. It also includes solved examples and explanations for regression and distribution problems in statistics.

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

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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.

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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

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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

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^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.

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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 )

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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)

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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.

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Appendix

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_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))))