这是一篇英国的回归分析与实验设计代码代写

Structure of Examination Paper:

There are Five pages.

There are FOUR questions in total.

There are no appendices.

The mark obtainable for each question is 25.

Instructions to Students :

Answer FOUR questions.

Show clearly all your workings.

In all exercises, dj is the j-th digit of your student number.

Question 1.

1a. Consider the linear regression model Y = Xθ+ε, where θ is an m-vector of unknown parameters, Y is an N-vector of observations, X is a design matrix of size N × m and ε is a vector of normal random errors with mean 0 and covariance matrix σ2IN . Here σ2 is some positive unknown constant and IN is the identity matrix of size N × N. Assume that rank(X) = p ≤ m.

(i) Give a geometrical interpretation of the LSE, the least squares estimator (LSE) of θ.

(ii) Defifine the lasso estimators of θ and discuss a motivation behind these estimators.

(iii) Defifine the ridge estimator of θ and brieflfly discuss its properties.

(iv) Let Z be a vector of size m such that the linear form ZT θ is estimable. Explain the method of constructing confifidence intervals for ZT θ.

(v) Derive the maximum likelihood estimator for σ2.

[10=2+2+2+2+2]

1b. Assume m = 3, N = 5 and the model yj = θ0 + θ1xj + θ2x2j + εj . The table below displays the design points and the corresponding results:

xj 0 u −u v −v

yj 0 1 −1  1 −1

Here u = d4 + 1 and v = d4 + 2, where d4 is the fourth digit of your student number.

(i) Test the hypothesis that the complete regression model is statistically signifificant on the 95% confifidence level. Write down the corresponding ANOVA table and compute R2.

(ii) Test the hypothesis

on the 95% confifidence level.

To answer this question please use 19.0 as an approximate value for the 0.95-quantile of the F-distribution with 2 and 2 degrees of freedom.

[15=7+8]

Question 2.

2a. Consider the linear regression model Y = Xθ+ε, where θ is an m-vector of unknown parameters, Y is an N-vector of observations, X is a design matrix of size N × m and ε is a vector of normal random errors with mean 0 and covariance matrix σ2IN .

In a particular application, m = 4, N = 5 and the model is Y = θ1X1 + θ2X2 + θ3X3 + θ4X4 + ε. The table below displays the design points and the corresponding results

X1 u   -1  1  0 -1

X2 −v   1  0  1 -1

X3 u−v  0  1  1 -2

X4 u+2v  -3  1 -2 1

Y   1  1  1 -1  0