商业|MSc Actuarial Science Alternative Timed Assessment

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Academic excellence for business
and the professions
Bayes Business School
MSc Actuarial Science
Alternative Timed Assessment
Module Code
SMM047
Module Title
Probability & Mathematical Statistics (For subject CS1 of the Institute and Faculty of Actuaries Examinations)
January 2022 3hrs – 45mins
Division of Marks:
The number of marks allocated is shown at the end of each question.
Where marks have been quoted for parts of questions, these are intended
to be a helpful guide to candidates.
Instructions to students:
Candidates should answer ALL of the questions.
Candidates should begin each question on a new page.
Your work should be in your own words, it should NOT contain material copied straight from
lecture notes, textbooks, or other resources.
If you do rely on external sources within your answer, these should be properly
referenced/cited.
You are expected to show all necessary working to obtain your final solution.
If this is not done, then marks will be deducted even when the correct numerical solution is
obtained.
This paper contains NINE questions and comprises SEVEN pages including the
title page
Internal Examiner: Dr. Russell Gerrard
External Examiner: Professor. Goran Peskir
QUESTION 1
In 2019 15,000 schoolchildren sat a state Mathematics examination. The average mark was 63.7,
standard deviation 14.4. The examiners’ intention when setting the paper was that the mean mark
should be 65, standard deviation 15.
(i) Test whether there is evidence at the 5% level of significance that the examination was more
difficult than the examiners intended. [2 marks]
(ii) The examiners decide to transform the marks in such a way that the transformed sample has
sample mean 65, sample standard deviation 15.
(a) Calculate the parameters of an appropriate linear transformation, Yi = a+bXi
. [2 marks]
(b) The sample coefficient of skewness of the original marks is ?0.03. What is the sample
coefficient of skewness of the transformed data? Give a reason for your answer. [1 mark]
[Total: 5 marks]
QUESTION 2
A discrete random X has probability generating function
GX(t) = 1
2
(1 + t)e
λ(t?1)
(i) Calculate the expectation of X and the probability that X takes the value 0. [4 marks]
(ii) Identify the distributions of two discrete random variables U and V whose probability generating functions are
GU (t) = 1
2
(1 + t), GV (t) = e
λ(t?1)
.
[2 marks]
(iii) Is it possible to deduce that, if U and V are independent, their sum U + V has the same
distribution as X? Give a reason for your answer. [1 mark]
(iv) Write down E[U], E[V ], P[U = 0] and P[V = 0] and use these to verify your answer to (i).
[3 marks]
[Total: 10 marks]
Page 2 of 7
QUESTION 3
A discrete random variable X has probability function
pX(x) = 1
? log(1 ? θ)
·
θ
x
x
, x = 1, 2, . . .
(i) Find an expression for the expectation of X as a function of θ. [3 marks]
(ii) A researcher announces that, based on a set of 50 observations x1, . . . x50 of the variable X,
she has calculated the method of moments estimate of θ to be 0.6. Assuming she is correct,
what is the value of the sample mean x? [2 marks]
(iii) Show that the maximum likelihood estimator for the parameter θ gives the same value as the
method of moments estimator. [4 marks]
[Total: 9 marks]
QUESTION 4
A collection X1, . . . , Xn of observations are taken from a Normal distribution with unknown mean
μ but with variance known to be equal to 1.
Knowing that the quantity being modelled is positive, an analyst applies as the prior distribution
for μ an exponential distribution with rate parameter λ.
(i) Show that the posterior density of μ given the observations satisfies
π(μ; x) ∝
(
exp
?
n
2

μ ? x +
λ
n
2

, for μ > 0
0 for μ < 0 [6 marks] (ii) Is it possible to conclude that the exponential distribution is a conjugate distribution for the normal? Give a reason for your answer. [1 mark] (iii) (a) Identify the mode of the posterior distribution. [1 mark] (b) Is it possible to write the mode of the posterior distribution in the form Zx + (1 ? Z)/λ for a credibility factor Z which does not depend on the data? Explain your answer. [2 marks] [Total: 10 marks] QUESTION 5 The duration of the first phase of a disease can be regarded as a random variable, T, with a Gamma distribution, with known shape parameter equal to 3 but with scale parameter which is determined by factors including age (X), gender (Y ) and overall level of health (Z). (i) Show that Gamma(3, λ) is an exponential family distribution. [3 marks] (ii) Identify the canonical link function. [2 marks] (iii) Write down a generalised linear model for the distribution of T as a function of X, Y and Z which is based on the canonical link function. [2 marks] (iv) Indicate the steps you would take when using an analysis of deviance to determine whether gender has a significant impact on the duration of the first phase of the disease. [4 marks] [Total: 11 marks] Page 3 of QUESTION 6 X1 and X2 are independent Pareto random variables with density functions fXi (x) = αi (1 + x) αi+1 , x > 0, i = 1, 2,
where α1 = 2 and α2 = 4. U is defined as U = min{X1, X2}.
(i) For values u > 0, derive an expression for P[U > u]. Hence show that U has a Pareto
distribution. [4 marks]
(ii) Calculate E[Xi
] as a function of αi and deduce the expectation of U. [3 marks]
(iii) (a) Explain why, for any x > u,
P[X2 > x | X1 = U = u] = P[X2 > x]
P[X2 > u]
.
[2 marks]
(b) Calculate the conditional density of X2 given that X1 = U = u. [2 marks]
[Total: 11 marks]
QUESTION 7
Each coin from a package of 250 new 1-Euro coins was weighed individually on a scale accurate to
0.001g.1
In preparation for a goodness-of-fit test for a normal distribution the observations were
grouped into six categories according to weight and are presented below, along with the “Expected”
numbers in each category.
Weight range Observed frequency Expected frequency
Below 7.490g 53 49.5
7.491g–7.505g 39 35.5
7.506g–7.520g 42 42.4
7.521g–7.535g 31 42.0
7.536g–7.550g 36 34.4
Above 7.550g 49 46.2
Total 250 250
(i) Explain carefully how the values in the “Expected frequency” column were obtained from the
raw data. [3 marks]
(ii) Complete the goodness-of-fit test at the 5% level of significance. [4 marks]
(iii) The sample mean of the data was 7.5197g, sample standard deviation 0.0344g. A second
sample of 250 new 1-Euro coins was measured for comparison. The sample mean was 7.5095g,
sample standard deviation 0.0370g. Test at the 5% significance level whether the two samples
come from populations with the same underlying mean and variance. [Note: the 2.5% point
of F249,249 is 1.283.] [6 marks]
[Total: 13 marks]
1Data collected by Z. Shkedy, M. Aerts and H. Callaret of Hasselt University
Page 4 of 7
QUESTION 8
A sample X1, . . . , X41 is taken from a Normal distribution with unknown mean μ and unknown
variance σ
2
. Let S
2 denote the sample variance.
(i) (a) Write down the sampling distribution of S
2
in the form S
2 ~ Gamma(α, β). [3 marks]
(b) Identify the expectation and variance of S
2
. [2 marks]
(c) Show that the random variable U given by
U =

20
S
2
σ
2
? 1

has expectation 0 and variance 1. [1 mark]
(d) Using a Normal approximation to the distribution of U, derive an expression, involving
S
2
, for an approximate 95% confidence interval for σ
2
. [4 marks]
(ii) Derive an expression, involving S
2
, for an exact 95% confidence interval for σ
2
. [3 marks]
(iii) A colleague points out that confidence intervals based on the χ
2 distribution are not robust.
Comment on whether you would expect the approximate confidence interval to be better than
the exact one if it turns out that the observations are not normally distributed. [1 mark]
[Total: 14 marks]
Page 5 of 7
QUESTION 9
The scatter plot labelled Figure 1 shows the annual average atmospheric carbon dioxide reading (in
parts per million at Mauna Loa) on the x-axis and the global average annual temperature (in ?C
relative to a fixed baseline) on the y-axis. The data set consists of 46 pairs of obervations, covering
the years 1975 to 2020.2
You are given the figures n = 46, x = 368.2, y = 0.47, P(xi ? x)
2 = 26452.39, P(yi ? y)
2 = 3.4952
and P(xi ? x)(yi ? y) = 288.3875.
(i) (a) Calculate the correlation coefficient between the temperature and the CO2 levels. [1 mark]
(b) Use a t test to test at the 1% level of significance the null hypothesis that they are
uncorrelated against the alternative that the correlation is positive. [Note: t44,0.001 =
3.29.] [3 marks]
(ii) The R output from fitting this model is included as Figure 2.
Figure 2
Call: lm(formula = tmp ~ co2)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -3.5453035 0.2026118 -17.50 <2e-16 *** co2 0.0109048 0.0005491 19.86 <2e-16 *** Residual standard error: 0.08929 on 44 degrees of freedom Multiple R-squared: 0.8996, Adjusted R-squared: 0.8974 F-statistic: 394.4 on 1 and 44 DF, p-value: < 2.2e-16 2Data from NASA Science and ESRL Global Monitoring Laboratory Page 6 of 7 (a) Estimate the CO2 level which would give rise to a fitted global temperature rise of 1.5?C relative to the baseline. [2 marks] (b) Calculate, for this CO2 level, a 95% confidence interval for the corresponding expected temperature rise. [4 marks] (iii) A second model, including the year as an additional explanatory variable, is fitted using R. The output is included as Figure 3. Figure 3 Call: lm(formula = tmp ~ co2 + yr) Coefficients: Estimate Std. Error t value Pr(>|t|)
(Intercept) 10.751893 18.815343 0.571 0.5707
co2 0.015296 0.005805 2.635 0.0117 *
yr -0.007967 0.010484 -0.760 0.4515
Residual standard error: 0.08972 on 43 degrees of freedom
Multiple R-squared: 0.901, Adjusted R-squared: 0.8964
F-statistic: 195.6 on 2 and 43 DF, p-value: < 2.2e-16 Based on the second model: (a) Write down the fitted model for the temperature rise as a function of the CO2 level and the year. [1 mark] (b) What combinations of CO2 level and year lead to a forecast temperature rise of 1.5?C? [2 marks] (c) Calculate a value C such that, if the CO2 level remains below C for all years from 2021 onwards, the forecast temperature rise will remain no higher than 1.5?C. [2 marks] (iv) Does the second model fit better than the first? Give two reasons for your answer. [2 marks] [Total: 17 marks] Page 7 of 7

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