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# PLease follow all the instructions i need it on R studio and pdf file knitted by R studio please submit it on time

Modern Computational Statistical Methods Assignment2: Due Week 9, 2022
STAT 8178/7178
Instructions:
This assignment covers from weeks 1 to 8
1. Due on 5th May 2023
2. For all the questions please provide the relevant mathematical derivations, the com-
puter programs (only using R software) and the plots.
3. Please submit on iLearn a single PDF file containing all your work (code, compu-
tations, plots, etc.). Other file formats (e.g. Word, html) will NOT be accepted.
4. Try to use Rmarkdown through Rstudio. But it is not compulsory to use Rmark-
down even if facilitate to reproduce results. Only upload the pdf file.
1 of 4
Modern Computational Statistical Methods Assignment2: Due Week 9, 2022
Question 1: 10 marks
Consider m samples {(x1, y2), . . . , (xm, ym)} with xi ∈ Rd and yi ∈ R.
(a) [2 marks] Show that the following function g(w) is convex:
g(w) =
1
2
m∑
i=1
(yi − wTxi)2,
where w ∈ Rd.
(b) [2 marks] Show that the following function h(w) is convex:
h(w) =
1
2
γ||w||22,
where γ > 0
(c) [2 marks] Using results from (a) and (b), show that f(w) = g(w) + h(w) is convex.
(d) [4 marks] Solve the optimization problem
min
w∈Rd
f(w)
y expressing the minimizer w in terms of the data matrix of x and the vector y.
Question 2: 21 marks
In this question, we consider
east cancer prediction where the label, or outcome variable
diagnosis has been coded as “M” in case of malignant lumps and “B” in case of benign lumps.
A popular dataset in this context is called the Wisconsin Breast Cancer Dataset and is based
on clinical data released in the early 1990’s. The feature vector x is composed of continuous
variables such as radius mean, texture mean, etc., each potentially affecting the probability of
malignancy. We use the 80-20 splitting strategy where we split it randomly between training
data and testing data. We want to build a prediction model to predict malignant lumps based
on main important features.
1. 1 mark Write down the logistic model for this task.
2. 1 mark Load the following file Breast.Rdata using
The above code provides two data frames: train and test. For both data sets, the first
column (named “diagnosis”) is the categorical outcome. How many attributes are available
to predict the outcome? How many samples are included in the two data frames. What is
the distribution of the outcome variable in the two data sets ?
3. 1 mark Run a logistic model (previously defined) using all attributes (Hint: use the glm
function and specify the family argument). You will estimate your model using the train
dataset.
4. 2 marks Provide the confusion matrix for the train set of your classifier using a threshold
of 0.5 and provide the accuracy of the model for the train set.
2 of 4
Modern Computational Statistical Methods Assignment2: Due Week 9, 2022
5. 2 marks Provide the confusion matrix for the test set of your classifier using a threshold
of 0.5 and provide the accuracy of the model for the test set.
6. 2 marks Why the accuracy for the test set is lower than the one for the training set ?
Which accuracy to report for assessing the performance of your classifier?
7. 2 marks We want to get a parsimonious model, meaning that we want to keep the most
elevant features. One way to tackle this challenge is to run a penalized regression model.
One scientist is struggling to choose between a ridge and a lasso regression model. Give
some justification to choose between the two strategies.
8. 2 marks We want to run a penalized logistic regression using a lasso penalty. To do it you
will use the glmnet R package using cv.glmnet and glmnet functions (Hint: do not forget
to use “family=binomial”). Choose the best tuning parameter lambda using a K-fold cross-
validation strategy (Hint: use the argument type.measure=”class” for choosing lambda to
get the smallest miss-classification e
or). Plot the cross-validation e
or according to the
log of lambda.
9. 2 marks Run the penalized logistic regression for the lambda you have chosen at the
previous step. How many attributes are still in the model?
10. 2 marks For this model and a threshold at 0.5, define your classifier (Hint: use the function
with argument type=”response” ). Report the confusion matrix on the test set and the
accuracy of the model.
11. 2 marks Present on the same plot the ROC curves for the logistic model and the penalized
logistic model.
12. 2 marks Report the two AUC (Area Under the ROC curve). What is your preference
etween the two models ?
Question 3: 7 marks
Let X1, ..., Xn be a random sample from a population with the following Bernoulli distribution
P (X = x) = θx(1− θ)1−x, x = 0 or 1, 0 ≤ θ ≤ 0.5.
We know that the maximum likelihood estimator (MLE) for θ is,
θ̂ = min{X̄, 0.5},
and the method of moments estimator (MOM) for θ is given by,
θ̃ = X̄.
1. 1 mark Generate a random sample with 500 observations from the above distribution
function when θ = 0.45.
2. 2 marks Compute bootstrap estimates of θ using the MLE and MOM estimates. Let’s
assume 1000 replications.
3. 2 marks Compare the performance of the θ̂ and θ̃ using your bootstrap samples by com-
puting the bias and standard e
or.
4. 2 marks Find Bootstrap percentile intervals of the θ̂ and θ̃ using your bootstrap samples.
Compare the results. Use the significance level α = 0.05.
3 of 4
Modern Computational Statistical Methods Assignment2: Due Week 9, 2022
Question 4: 7 marks
The remiss.csv data set contains the remission times for 42 leukemia patients in weeks. Some
of the patients were treated with the drug called 6-mercaptopurine (group = 0), and the rest
were part of the control group (group = 1).
1. 1 mark Create a box plot for the remission times for two treatment and control groups.
Compare the remission times of the two groups.
2. 1 mark Use a normal probability plot to check whether the distribution of the remission
times for each group is Normal.
3. 1 mark Write the null hypothesis and alternative hypothesis to test for the equality of
means between the two groups.
4. 4 mark Perform the above hypothesis test using Monte Carlo simulation to get the critical
values. Estimate the p-value. Do you reject H0 or H1? You can assume the distribution
of the remission times for each group is Normal and their population variances are equal.
Use the significance level α = 0.05.
4 of 4
Answered 1 days After Apr 25, 2023

## Solution

Mohd answered on Apr 26 2023
-
-
-
2023-04-26
li
ary(magrittr)
li
ary(dplyr)
east.Rdata")
1. 1 mark Write down the logistic model for this task.
esponse variable = Diagnosis
family =Binomial
mod<-glm(diagnosis~.,data=train,family=binomial)
2. 1 mark Load the following file Breast.Rdata using
train<-Breast\$train
test<-Breast\$test

plot(train\$diagnosis, main="Diagnosis
Distribution",xlab="Diagnosis",ylab="Count")
3. 1 mark Run a logistic model (previously defined) using all attributes (Hint: use the
glm function and specify the family argument). You will estimate your model using
the train dataset.
trainset<-train

equire(caret)

mod<-glm(diagnosis~.,family = binomial(),trainset)
summary(mod)
##
## Call:
## glm(formula = diagnosis ~ ., family = binomial(), data = trainset)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.37e-04 -2.10e-08 -2.10e-08 2.10e-08 1.44e-04
##
## Coefficients:
## Estimate Std. E
or z value Pr(>|z|)
## (Intercept) -1.947e+03 1.834e+06 -0.001 0.999
## radius_mean -6.292e+01 5.539e+05 0.000 1.000
## texture_mean -4.053e+00 7.211e+03 -0.001 1.000
## perimeter_mean 1.410e+01 8.570e+04 0.000 1.000
## area_mean -4.277e-01 1.794e+03 0.000 1.000
## smoothness_mean 2.603e+03 2.667e+06 0.001 0.999
## compactness_mean -1.881e+03 3.620e+06 -0.001 1.000
## concavity_mean 2.672e+03 1.039e+06 0.003 0.998
## concave.points_mean -1.770e+03 2.392e+06 -0.001 0.999
## symmetry_mean -8.507e+01 1.584e+06 0.000 1.000
## fractal_dimension_mean -7.652e+01 9.290e+06 0.000 1.000
## radius_se -4.560e+02 6.021e+05 -0.001 0.999
## texture_se -1.183e+02 1.263e+05 -0.001 0.999
## perimeter_se 5.475e+01 6.250e+04 0.001 0.999
## area_se 3.338e+00 8.692e+03 0.000 1.000
## smoothness_se 8.261e+03 2.730e+07 0.000 1.000
## compactness_se 9.646e+03 6.792e+06 0.001 0.999
## concavity_se -5.003e+03 4.997e+06 -0.001 0.999
## concave.points_se 1.445e+04 9.061e+06 0.002 0.999
## symmetry_se -4.809e+03 5.275e+06 -0.001 0.999
## fractal_dimension_se -7.965e+04 6.161e+07 -0.001 0.999
## radius_worst 1.402e+02 1.219e+05 0.001 0.999
## texture_worst 1.492e+01 1.078e+04 0.001 0.999
## perimeter_worst -9.594e+00 2.223e+04 0.000 1.000
## area_worst -4.863e-01 9.480e+02 -0.001 1.000
## smoothness_worst -1.703e+02 2.727e+06 0.000 1.000
## compactness_worst -9.309e+02 1.276e+06 -0.001 0.999
## concavity_worst 1.289e+02 7.715e+05 0.000 1.000
## concave.points_worst 4.287e+01 1.767e+06 0.000 1.000
## symmetry_worst 6.023e+02 3.107e+05 0.002 0.998
## fractal_dimension_worst 7.467e+03 5.395e+06 0.001 0.999
##
## (Dispersion parameter...
SOLUTION.PDF