代做STATS 763 Advanced Regression Methodology FIRST SEMESTER, 2019帮做Python语言

日期：2025-02-05 05:22

STATS 763

STATISTICS

Advanced Regression Methodology

FIRST SEMESTER, 2019

1. (24 marks; 4 each) Define or explain briefly in the context of this class

(a) facetting

(b) ridge regression

(c) collider

(d) apparent error

(e) competing risks

(f) sparse matrix

2. (24 marks) You have data from a series of randomised clinical trials (experiments) of related drugs to prevent heart attacks:  that is, people are recruited into an experiment and then randomly assigned to one of two treatments. Y is a binary variable indicating heart attack, D is a factor variable with m + 1 levels 0, 1, . . . , m indicating which drug the person received (with D = 0 for no treatment), and G is a factor variable with k levels 1, 2, . . . , k indicating which trial the person was in. You fit two models

model1  <- glm(Y~D, family=binomial(log))

model2  <- glm(Y~D+G, family=binomial(log))

(a) (4 marks) What is the interpretation of the coefficient labelled D1 in the output of model 1?

(b) (4 marks) What is the interpretation of the coefficient labelled D1 in the output of model 2?

(c) (8 marks) Explain why the coefficient labelled D1 in model 2 has a causal interpretation, but the coefficient labelled G3 does not have a causal interpretation.

(d) (4 marks) What is the variance function for this generalised linear model?

(e) (4 marks) Does the link function we are using give the natural parameter for the expo- nential family?

3. (24 marks) A popular model for rainfall combines a logistic regression for whether rain occurs or not and a Gamma regression for the amount of rain if any occurs.  Consider these two models for rainfall amount fitted to daily data from Auckland, where Amount . mm .   is the amount of rain in millimetres, rain indicates whether there was rain on that day, and month is the month (1–12)

model0  <-glm(formula = Amount. mm.~1, family = Gamma(log), data = auck, subset=rain==TRUE)

model1<-glm(formula = Amount . mm . ~sin(month * 2 * pi/12) +

cos(month  *  2  * pi/12),  family  =  Gamma(log), data  =  auck,  subset=rain==TRUE)

model2<-glm(formula = Amount. mm.~factor(month),

family  =  Gamma(log),  data  =  auck,  subset=rain==TRUE)

The likelihood ratio p-value testing model 2 against model 1 is 0.056, the likelihood ratio p-value testing model 1 against model 0 is 0.047, the AIC for model 0 is 2478, the AIC for model 1 is 2473, and the AIC for model 2 is 2468.

(a) (4 marks)The intercept in model 0 is 1.86. What is the interpretation of this value? (b) (5 marks) Is model 2 nested in model 1? Explain.

(c) (6 marks) Which model would you expect to have lowest out-of-sample prediction error for the amount of rain on days when there was rain? Explain.

(d) (4 marks) The dispersion parameter in model 0 is estimated as 2.35. What, according to the model, is the variance of the amount of rain on days when there is rain?

(e) (5 marks) An alternative model is a (Normal) linear model for the logarithms of the non-zero rain amounts. How would the interpretation of the coefficients for this model difer from the Gamma model?

4. (24 marks) Researchers studying autism spectrum disorder (ASD) were looking for biochem- icalways to diagnose ASD in infants where current diagnostic methods can’t be used reliably. They measured concentrations of 87 chemicals in the blood of 60 children (not infants) diag- nosed with ASD and 60 control children who did not have any diagnosis of ASD or any other neurological or psychiatric illness. They used the lasso (L1-penalised maximum likelihood) to fit a logistic regression model predicting ASD diagnosis from the 87 predictor variables. Predictions from the model had 95% sensitivity and 95% specificity

(a) (6 marks) Write down the objective function that is being minimised (you can write L(β; Y) for the binomial likelihood at parameter values β).

(b) (6 marks) Give two advantages that are relevant to this analysis of the lasso/L1  penalty approach over stepwise logistic regression

(c) (6 marks) The size of the penalty was chosen by crossvalidation.  Briefly explain what cross-validation is.

(d) (6 marks) In the population as a whole, about 1.5% of children end up diagnosed with ASD. Assuming that the sensitivity and specificity are the same in the population as in the experimental sample, what proportion of those predicted by the model to have ASD would end up with a diagnosis of ASD by existing criteria.