Question 265 MCQ · 0.0 marks
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Statistical Computing · End Term · Sep 2024 · FN
Statistical Computing · End Term · Sep 2024 FN
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Question 266 MCQ · 40.0 marks
6. Consider the logistic regression model: Let the covariate vector for the i-th observation
be a; = (1,2i,...,a%p)" for i = 1,...,n and 8 € RP be the corresponding vector of
regression coefficients. Suppose response y; is a realization of Y; with
Y; ~ Bernoulli(p;),
1
eB
where p; = Tate and Y;,’s are independent. Suppose we perform MM algorithm to
en
find the maximum likelihood estimate of 8, where
1(8) == og( +e) + yx 8
i=l 1
L758;
es
3B;
(1 + a ‘)
Use the above expressions to find the minorizing function and write the iterates of the
MM algorithm. [4 marks]
V7U(8) =— > tats
i
7. Consider a dataset x = (21,22, 73,24) from a multinomial distribution with probabilities
The joint distribution of x is given by
II ai!
i=l
Suppose you observe x = (120,68, 20,32). Assume the complete data is (21, 29, 72,73, 74),
where 2 + 2 = 2; with probabilities
a eae a
Sous (oe ios
(a) Write the complete log likelihood of p. {2 marks]
(b) Write the E-step q(p | py.) for the EM algorithm. {5 marks]
(c) Write the M-step for the EM algorithm. [3 marks]
8. The following code estimates the mean of a normal distribution by constructing the
bootstrap confidence intervals. Study it carefully and answer the following sub-questions.
1 # Simulating Dataset
2 Me
3 gig2 <- 4
in <- 100
5s my.samp <- rnorm(n, mean = mu, sd = sqrt(sig2))
6 alp <- .05
7 ends <- c(alp/2, 1 - alp/2)
sB <- 1000 # number of bootstrap samples
io ## Bootstrapping method 1
11 boot_1 <- numeric(length = B)
12 for(b in 1:B)
13 {
14 samp <- sample(1:n, size = n, replace = TRUE)
1s boot.samp <- my.samp[samp]
16 Dboot_ilb] <- mean(boot.samp)
1}
io ## Bootstrapping method 2
20 boot_2 <- numeric(length = B)
a for(b in 1:B)
a {
2 boot.samp <- rnorm(n, mean = mean(my.samp), sd = sqrt(sig2))
21 boot_2[b] <- mean(boot.samp)
as }
a2 # Bootstrap confidence intervals
28 quantile(boot_1, ends)
29 quantile(boot_2, ends)
si mean (my. samp)
s2 var (my.samp)
st # outputs
ss > quantile(boot_1, ends)
36 2.5% 97.5%
sr 0.448572 1.307118
ss > quantile(boot_2, ends)
39 2.5% 97.5%
0 0.4799244 1.2874340
> mean(my.samp)
ss [1] 0.8863481
“> var(my. samp)
ws [1] 5.063528
I have written answers on the answer sheets
Not applicable
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