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PRIFYSGOL CYMRU / UNIVERSITY OF WALES
ABERYSTWYTH
INSTITUTE OF MATHEMATICAL AND PHYSICAL SCIENCES
SEMESTER 1 EXAMINATIONS, JANUARY/FEBRUARY 2008
MA36510 – Linear Statistical Models
Time allowed – 2 hours
All questions may be attempted
Marks gained from questions in Section B will be given greater consideration in assessing a first class performance.
Calculators are permitted, provided they are silent, self-powered, without communications facilities, and incapable of holding text or other material that could be used to give a candidate an unfair advantage. They must be made available on request for inspection by invigilators, who are authorised to remove any suspect calculators.
Statistical Tables will be provided
Information Unless otherwise stated you may quote without proof: The n-dimensional multivariate Normal MVN(μ,Σ) distribution with probability density function f(y) = (2π)–n/2|Σ|–1/2exp(–Q/2) where Q = (y–μ)TΣ–1(y–μ) = yTΣ–1y – 2yTΣ–1μ + μTΣ–1μ. In turn, Q has a chi-squared distribution on n degrees of freedom.
The Normal equations for ordinary least squares estimation in the linear model E[Y]=Xβ are given by XTXβ = XTy.
Section A
1 Y 1 , Y 2 and Y 3 have respective means of +1, 0, +2 and each has variance 4. Y 1 is uncorrelated with Y 2 , the covariance between Y 1 and Y 3 is +2 and the correlation between Y 2 and Y 3 is –¼. (i) Write down the dispersion matrix of Y = (Y 1 Y 2 Y 3 )T. U and V are defined as U=Y 1 +Y 2 ―Y 3 and V = Y 2 +2Y 3. (ii) Deduce the mean vector and the dispersion matrix of (U V)T. (iii) Calculate the expected value of Q = 6X 12 + 5X 22 + Y 32 – 8Y 1 Y 3 + 4Y 2 Y 3
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2 The bivariate random vector (X Y)T^ has the probability density function
f x y ( , ) = 21 π^ exp {− 12 ( x 2 + 5 y 2 + 4 xy − 2 y + 2 )}
(a) Show that the dispersion matrix of
X
Y
is
and calculate the
mean vector. (b) Calculate the probability that X+Y takes a negative value. (c) What is the median of the distribution of Q = X^2 +5Y^2 +4XY-2Y+2?
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3 X,^ Y^ and^ Z^ are uncorrelated standard Normal random variables. Find the value of a such that a suitable multiple of Q = X^2 + aY 2 + Z 2 – 4XY + 2XZ – aYZ has a chi-squared distribution and find the upper quartile of Q in this case. [8]
4 Six observations each of mean θ+φ were taken, and a further four observations each had expectation 2θ–φ. The first six observations totalled 90 whereas the other four added to 60. The ten observations are uncorrelated, each having variance σ^2 =10. Find the ordinary least squares estimates of θ and φ. What does the Gauss-Markov Theorem say about your estimator of θ? Verify that (Y 1 +Y 7 )/3 is also an unbiased estimator of θ. What is the efficiency of this estimator? [10]
9 When fitting a two parameter model to 18 observations with
( T ) 1
1 2
T T
β β
− = ^ − ^ = =
− ^
X X X y y y
a 95% confidence for the ratio θ= β 1 /β 2 of the two parameters was calculated as 0.638<θ<2.305. Without carrying out detailed calculations, explain clearly how this was done. [8]
10 The strength of wooden beams is thought to vary with the specific gravity and the moisture content of the wood. Ten beams were tested, these three variables measured on each one.. A linear model was then fitted to the data, giving the results shown at the end of this paper. (a) Give an unbiased estimate of the variance of the data. (b) The estimated standard error of the coefficient β 1 is has been blanked out; calculate this value. (c) Would you consider a model with any of the terms omitted? Why? (d) The column headed hii gives the leverages. What is the sum of these values? Why? [You should NOT need to add these numbers together!] (e) Three kinds of residuals are given; two of these for the sixth observation are missing. Calculate them. Explain also how the Studentized residual for this observation was calculated. (f) Consider carefully all the results and plots given and comment on the analysis. You should pay particular attention to any unusual observations with regard to undue influence or any that may be regarded as outliers.
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Formulae:
Standardized: ( )
ˆ ˆ ˆ 1 i^ i ii
e = σ ε− h Studentized: * [ ]
ˆ ˆ (^1) i^ i i hii
ε =^ ε σ −
Wood beam data
Model: E[strength] = β 0 + β 1 sg + β 2 moist
- 1 11.14 0.499 11.1 0.418 11.584 -0.444 -2.114 -3. beam strength sg moist h ii fit ord stand stud - 2 12.74 0.558 8.9 0.242 12.671 0.069 0.286 0. - 3 13.13 0.604 8.8 0.417 13.089 0.041 0.196 0. - 4 11.51 0.441 8.9 0.604 11.677 -0.167 -0.966 -0. - 5 12.38 0.550 8.8 0.252 12.630 -0.250 -1.050 -1. - 6 12.6 0.528 9.9 0.148 12.150 0.450 1.769 2. - 7 11.13 0.418 10.7 0.262 11.003 0.127 0.538 0. - 8 11.7 0.480 10.5 0.154 11.583 0.117 0.463 0. - 9 11.02 0.406 10.5 0.316 10.954 0.066 0.290 0. - 10 11.41 0.467 10.7 0.187 11.419 -0.009 -0.036 -0. - 10 4.951 98.8 47.40782 -38.27536 -2. - XTX = 4.951 2.488995 48.5849 (XTX)-1 = -38.27536 41.99840 1. - 98.8 48.5849 984 -2.87021 1.76943 0. - 118.76 Parameter Estimate ESE yTy = 1415. - XTy = 59.20694 β 0 10.301524 1. - 1168.445 β 1 8.4947108 1.7850238 RES(β) = 0. - β 2 -0.2663214 0. - +0.418 ―0.002 +0.080 ―0.274 ―0.046 +0.181 +0.129 +0.222 +0.050 +0.
- ―0.002 +0.242 +0.292 +0.136 +0.243 +0.128 ―0.041 +0.033 ―0.035 +0.
- +0.080 +0.292 +0.417 ―0.019 +0.273 +0.187 ―0.126 +0.044 ―0.153 +0.
- H = ―0.274 +0.136 ―0.019 +0.604 +0.197 ―0.038 +0.168 ―0.021 +0.275 ―0.
- ―0.046 +0.243 +0.273 +0.197 +0.252 +0.111 ―0.030 +0.019 ―0.010 ―0.
- +0.181 +0.128 +0.187 ―0.038 +0.111 +0.148 +0.042 +0.117 +0.012 +0.
- +0.129 ―0.041 ―0.126 +0.168 ―0.030 +0.042 +0.262 +0.145 +0.277 +0.
- +0.222 +0.033 +0.044 ―0.021 +0.019 +0.117 +0.145 +0.154 +0.120 +0.
- +0.050 ―0.035 ―0.153 +0.275 ―0.010 +0.012 +0.277 +0.120 +0.316 +0.
- +0.242 +0.004 +0.004 ―0.028 ―0.010 +0.111 +0.174 +0.168 +0.148 +0.
