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SCHOOL OF RISK AND ACTUARIAL STUDIES
TERM 1 2024
Assignment Part 1
ACTL 2131: Probability and Mathematical Statistics
ACTL 5101: Probability and Statistics for Actuaries
INSTRUCTIONS:
1. TIME SUGGESTED—between 1 HOUR 30 MIN and 1 HOUR 45 MIN TO REPLICATE EX-
AMINATION ENVIRONMENT
2. READING TIME—5 MINUTES
3. THIS PAPER HAS 8 PAGES.
4. TOTAL NUMBER OF QUESTIONS—6
5. TOTAL MARKS AVAILABLE— 100
6. MARKS AVAILABLE FOR EACH QUESTION ARE SHOWN IN THE EXAMINATION PA-
PER (AND OVERLEAF). ALL QUESTIONS ARE NOT OF EQUAL VALUE.
7. ANSWER ALL QUESTIONS IN THE SPACE ALLOCATED TO THEM. IF MORE
SPACE IS REQUIRED, USE THE ADDITIONAL PAGES AT THE END.
8. PLEASE ONLY USE THE TEXT “FORMULÆ AND TABLES FOR ACTUARIAL EXAMI-
NATIONS” (ANY EDITION, UNANNOTATED) AND UNSW APPROVED CALCULATOR
TO REPLICATE EXAM CONDITIONS.
Question Total available marks Total marks attained
for the question for the question
1 [6 marks]
2(a) [2 marks]
2(b) [6 marks]
3(a) [4 marks]
3(b) [6 marks]
3(c) [3 marks]
3(d) [8 marks]
4(a) [6 marks]
4(b) [5 marks]
4(c) [6 marks]
4(d) [5 marks]
5 [12 marks]
6(a) [4 marks]
6(b) [3 marks]
6(c) [12 marks]
6(d) [12 marks]
[total: 100 marks]
Page 2 of 8
Question 1 [6 marks]
An auto insurance company insures drivers against accidents. Historical data shows that 5% of the
insured drivers file an accident claim each year. If the company insures 200 drivers, letW be a random
variable representing the number of drivers who file an accident claim in a year. Find the expression
for the probability that more than 10 but less than 20 drivers file an accident claim in a year. [Note:
Although calculating this probability is not required, the probability formula should be simplified to
minimise the number of terms within the summation].
Page 3 of 8
Question 2 [8 marks]
For k > 2, let f(x) := k
xk+1
for x ≥ 1.
(a) [2 marks] Show that f(x) is a probability density function.
(b) [6 marks] Calculate the distribution function, the expected value, and the variance of a random
variable X distributed according to f .
Page 4 of 8
Question 3 [21 marks]
A rod of length 1 is randomly broken (uniform distribution!) into two pieces such that one piece is
larger and one piece is smaller. Let X be the length of the larger piece. The larger piece is then
randomly broken again into two pieces such that one piece is larger and one piece is smaller. Let Y be
the length of the larger piece from the second break.
(a) [4 marks] Provide the distribution and pdf of X.
Hint: X = max(U, 1− U), where U ∼ Uniform(0, 1).
(b) [6 marks] Write down the formula for conditional distribution of Y given X = x and the associ-
ated conditional pdf.
(c) [3 marks] Determine E[Y |X = x].
(d) [8 marks] What is the probability that Y > 12?
Page 5 of 8
Question 4 [22 marks]
Let V be a continuous random variable with the following probability density function (pdf):
f(v) =
{
2v for 0 ≤ v ≤ 1
0 otherwise
(a) [6 marks] Calculate the moment generating function of V , MV (t).
(b) [5 marks] Determine the expectation and variance of V .
(c) [6 marks] Calculate the moment generating function of Z = V 2, MZ(t).
(d) [5 marks] Compute the expectation and variance of Z.
Page 6 of 8
Question 5 [12 marks]
Let X and Z be independent random variables where X ∼ N(0, 1) and P (Z = 1) = P (Z = −1) = 12 .
(a) [6 marks] Show that Y = Z ·X is standard normally distributed.
(b) [3 marks] Determine the covariance between X and Y .
(c) [3 marks] Explain whether X and Y are independent.
Page 7 of 8
Question 6 [31 marks]
Let U and V be continuous random variables with joint probability density function fU,V (u, v) =
4u(1− v) for 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.
(a) [4 marks] Find the marginal probability density functions fU (u) and fV (v).
(b) [3 marks] Find the conditional probability density function fV |U (v|u).
(c) [12 marks] Let W = U + V . Find the probability density function of W , fW (w).
(d) [12 marks] Find the bivariate pdf of their product and ratio, i.e., Z = U · V and R = UV (assuming
V ̸= 0).
Page 8 of 8