ACTL1101 Introduction to Actuarial Studies
Xiao Xu
© University of New South Wales (2023)
School of Risk and Actuarial Studies, UNSW Business School
x.xu@unsw.edu.au
Week 1:
Probability1
1Readings: Sherris 2.2 and 2.31/38
1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
2/38
1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
3/38
Motivation
What is the science we call ‘Probability’?
‘the branch of mathematics that studies the possible outcomes of given events together
with the outcomes’ relative likelihoods and distributions. In common usage, the word
“probability” is used to mean the chance that a particular event (or set of events) will
occur expressed on a linear scale from 0 (impossibility) to 1 (certainty), also expressed
as a percentage between 0 and 100%’.2
2https://mathworld.wolfram.com/Probability.html3/38
Why do we need Probability in Actuarial Science?
The work of actuaries is intrinsically related to uncertainty.
Consider questions such as
- Will life expectancy improve in the next decade?
- At what age are people most likely to retire?
- How many insurance claims will we have this month? How big will they be?
- How much money can we expect to gain (or lose) from our investments in the stock market?
A good understanding of Probability (and Statistics) is necessary to tackle questions of
this type.
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1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
5/38
Sample Space and Events
A random experiment is an experiment whose outcome is not known in advance with
certainty. Example: tossing a dice and recording the result.
The sample space of a random experiment (often denoted Ω) is the set of all possible
outcomes. In the dice tossing example, Ω = {1, 2, 3, 4, 5, 6}.
Any subset A of the sample space is called an event. Example: A = {1, 3, 5} (‘obtaining
an odd number’) is an event in Ω. B = {1} (‘obtaining 1’) is also an event in Ω.
Given a context, we want to determine the probabilities that certain events occur.
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Elementary Rules of Probability
There are a few elementary rules everyone should know.3 Let A and B be events in the same
sample space, then:
0 ≤ Pr[A] ≤ 1
Pr[A] = 1− Pr[Ac]
Pr[A ∪ B] = Pr[A] + Pr[B]− Pr[A ∩ B]
Where Ac mean ‘not A’, A ∪ B means ‘A or B ’, while A ∩ B means ‘A and B ’.
3At least, everyone in this course!6/38
Determining probabilities
If all outcomes of an experiment are equally likely, a useful formula to determine the
probability of an event A is given by
Pr[A] =
# favourable cases to A
# of total cases
, where
- # favourable cases to A: number of outcomes contained by event A
- # of total cases: number of outcomes in the sample space, in total
Example: what is the probability of getting an odd number upon rolling a fair dice?
# favourable cases
# total cases
=
3
6
=
1
2
.
7/38
Determining probabilities: harder problem
Consider a harder problem:
In a random group of 25 people, what is the probability that at least two of them have
the same birthday? [You can consider that there are 365 days in the year4, and that
all people are equally likely to be born on any given day.]
Here it is not immediately obvious what the sample space is, and even less obvious how
many ‘favourable cases’ there are.
For such problems, we need the concepts of permutation and combination.
4Apologies to anyone born February 29th, this is just to
simplify the problem.8/38
1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
9/38
Permutations
Given a list of distinct objects, any distinct arrangement of them is called a
permutation. Example: given the three letters abc, the possible permutations are: abc,
acb, bac, bca, cab, cba.
In general, the number of possible permutations for n different objects is:
n! = n · (n − 1) · (n − 2) · (. . .) · 2 · 1
Proof: See this as allocating n objects within n ‘spots’. Then:
▶ the first spot: n options
▶ the second spot: n − 1 options, as one object is already placed in the first ‘spot’
▶ (. . .)
▶ the last spot: only 1 option
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Permutations: arranging k out of n objects
Given a list of n distinct objects, how many distinct permutations of k of them (k < n)
can be made?
Reasoning as in the previous slide we obtain:
n · (n − 1) · (n − 2) · (. . .) · (n − k + 1) = n!
(n − k)!
Example: Out of 10 people, how many different teams of 4 people can you make, provided
that all roles in the team are distinct?
10/38
Permutations with duplicates
Assume we have n objects, of which r are identical. How many different permutations of
those n objects can we create?
n!
r !
Assume we have n objects in total, of which n1 are alike between them, n2 are alike
between them, . . ., nk are alike between them. In that case, how many different
permutations of those objects can we create?
n!
n1!n2! · · · nk !
11/38
Combinations
Given a list of distinct objects, a selection of them for which the order of the selected
items does not matter, is called a combination.
Given n distinct objects, the number of ways to select k (k ≤ n) of them (where the order
of selection does not matter) is (
n
k
)
:=
n!
k!(n − k)!
Example: If there are initially 49 participants at The Voice, and Delta Goodrem needs to
create a team of 10, how many different teams can she possibly create?
12/38
What is the difference between permutations and combinations?
Combinations do not care about ordering!
The UNSW gym lockers require a 4 digit code to unlock. Let’s say I use the code 1234.
How many 4-number permutations of 1234 are there?
How many 4-number combinations of 1234 are there?
What about duplicates?
How many 4-number permutations of 1233 are there?
How many 4-number combinations of 1233 are there?
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Example: Back to our ‘Birthday Problem’
Question: In a random group of 25 people, what is the probability that at least two of
them have the same birthday? [You can consider that there are 365 days in the year and
that all people are equally likely to be born on any given day.]
Solution: It is easier to find the probability of
Ac := ‘no two people have the same birthday’.
Indeed, the number of cases favourable to event Ac is the number of permutations of 25
days in a choice of 365 days, i.e.
365!
(365− 25)!
The number of possibilities in total is: 36525
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Example: Back to our ‘Birthday Problem’
Hence,
Pr[Ac] =
# favourable cases to Ac
# total cases
=
365!
(365− 25)!36525 ≈ 0.4313.
So finally,
Pr[A] = 1− Pr[Ac] ≈ 0.5687.
15/38
1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
16/38
Conditional Probabilities
For two events A and B in the same sample space Ω, we denote Pr[A|B] the conditional
probability that A occurs given that B has occurred.
If Pr[B] > 0, we have that
Pr[A|B] = Pr[A ∩ B]
Pr[B]
.
16/38
Conditional Probabilities: Example with Venn Diagram
If B ⊆ A, Pr[A] = 0.25,Pr[B] = 0.1,Pr[C ] = 0.4,Pr[A ∩ C ] = 0.1,
Pr[A|B] = 1
Pr[B|A] = Pr[B ∩ A]/Pr[A] = 0.1/0.25 = 0.4
Pr[A|C ] = Pr[A ∩ C ]/Pr[C ] = 0.1/0.4 = 0.25
17/38
Independence
Two events A and B are said to be independent events if
Pr[A ∩ B] = Pr[A] Pr[B]
If A and B are independent, then
Pr[A|B] = Pr[A ∩ B]
Pr[B]
= Pr[A]
Likewise, Pr[B|A] = Pr[B]
In words, A and B are independent if knowledge that B has occurred does not change the
probability of A to occur (and vice-versa).
18/38
Bayes Theorem
This important theorem5 states that for two events A and B ,
Pr[A|B] = Pr[B|A] Pr[A]
Pr[B]
.
Where it is often useful to compute Pr[B] as:
Pr[B] = Pr[B|A] Pr[A] + Pr[B|Ac] Pr[Ac].
A common medical example: A rare disease affects 1 in every 10,000 people. For someone
infected, a diagnostic test is positive 95% of the time. For someone without the disease, the
test is positive 1% of the time. You test positive. Knowing this, what is the chance that you
actually have the disease?
5Click here to read about this theorems’ history.19/38
Bayes medical example continued
Let A :=‘I have the disease’, and B :=‘The test is positive’.
We need
Pr[B] = Pr[B|A] Pr[A] + Pr[B|Ac] Pr[Ac]
= 0.95 · 0.0001+ 0.01 · 0.9999 = 0.010094
From which we get
Pr[A|B] = Pr[B|A]P[A]
Pr[B]
=
0.95× 0.0001
0.010094
≈ 0.0094
20/38
1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
21/38
Random Variables
A random variable X is a function that assigns values to the possible outcomes of a
random experiment.
Example: in the experiment of tossing two dice, a random variable X could represent the sum
of both dice. Another random variable Y could represent the minimum of the two dice. Etc.
Discrete random variables: the number of possible values is finite or countably infinite.
Continuous random variables: the possibles values appear on a continuous spectrum
(hence are uncountably infinite).
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Random Variables: Examples in Actuarial Studies
Discrete random variables
Number of motor vehicle accidents occurring during a particular month
Number of lives who are lost to cancer during a particular time period
Continuous random variables
Time until first accident on a car insurance policy
Claim payment for a fire insurance policy
22/38
Random Variables: Example
Assume the probability that any of 5 independent people has an insurance claim in a given
year is 5%. If X represents the total number of claims, find the probabilities:
Pr[X = 0] = (1− 0.05)5 ≈ 77.38%
Pr[X = 1] = 0.05 · (1− 0.05)4 · 5 ≈ 20.36%6
In general, out of n independent ‘events’, each having a probability p of ‘happening’, the
probability to observe k (0 ≤ k ≤ n) events is(
n
k
)
pk(1− p)n−k
6We need 1 ‘claim’ and 4 ‘no claim’, but there are 5 distinct
choices for the person with no claim.23/38
Probability Mass Function (PMF)
For a discrete random variable X , the PMF p(x),7 is the probability that X takes a
particular value x :
p(x) = Pr(X = x),
with
p(x) ≥ 0,
and ∑
all x
p(x) = 1.
7Sometimes written pX (x).24/38
Probability Density Function (PDF)
For a continuous random variable X , the PDF f (x),8 is defined such that
Pr (a ≤ X ≤ b) =
∫ b
a
f (x)dx ,
with
f (x) ≥ 0 for −∞ < x <∞,
and ∫ ∞
−∞
f (t)dt = 1.
8Sometimes written fX (x).25/38
Cumulative Distribution Function (CDF)
For any random variable X , the CDF, denoted F (x),9 is defined as
F (x) = Pr (X ≤ x) .
It gives the probability a random variable X is less than or equal to a specified value x .
The CDF always satisfies:
F (−∞) = 0,
and
F (+∞) = 1.
9Sometimes written FX (x).26/38
CDF: Discrete vs Continuous Case
For a discrete random variable X , the CDF of X can be computed as
F (x) = Pr (X ≤ x)
=
∑
k≤x
p(k).
For a continuous random variable X , the CDF of X can be computed as
F (x) = Pr (X ≤ x)
=
∫ x
−∞
f (t)dt,
and we also have
f (x) =
d
dx
F (x).
27/38
Expectation
The expectation (or expected value) of a random variable X , denoted E[X ] (and
sometimes µ) is the weighted average of all possible values X can take, each value being
weighted by the probability that X assumes it.
For discrete random variables:
E [X ] = µ =
∑
all x
xp(x)
For continuous random variables:
E [X ] = µ =
∫ ∞
−∞
xf (x)dx
28/38
Expectation of a function of X
It is often useful to compute the expectation of a function of a random variable, g(X ):
Discrete random variable
E [g (X )] =
∑
all x
g (x) p(x)
Continuous random variable
E [g (X )] =
∫ ∞
−∞
g (x) f (x)dx
For instance, moments (about the origin) of a random variable are given by
E [X r ] r = 1, 2, 3, . . .
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Law of total expectation
For two random variables X and Y , we have the so called Law of total expectation:
E[X ] = E[E[X |Y ]].
A consequence of this result is that, for events A1, . . . ,An forming a partition of a sample
space Ω, we have
E[X ] =
n∑
i=1
E[X |Ai ] Pr[Ai ].
(events Ai form a partition if they are all mutually exclusive with ∪ni=1Ai = Ω.)
30/38
Law of total expectation: Example
Question: Let Y be the number of drinks Josephine has the night before an exam, and
let X be her result in said exam (out of a 100), with
Pr[Y = 0] = 0.5, Pr[Y = i ] = 0.1 for i ∈ {1, 2, 3, 4, 5}
E[X |Y ] = 90− 2Y 2.
What is the expectation of her result in this exam?
Answer:
E[X ] = E[E[X |Y ]] = E[90− 2Y 2]
= 90 · 0.5+ (88+ 82+ 72+ 58+ 40) · 0.1
= 79.
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Variance
The variance, denoted V[X ] (or Var[X ], and sometimes σ2) is another key property of a
random variable X . It is defined as
V [X ] = σ2 = E
[
(X − µ)2
]
= E[X 2]− E[X ]2
The variance is a measure of the ‘dispersion’, ‘spread’, or ‘volatility’ of a random variable.
We call standard deviation the square root of the variance: σ
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Properties of the Expectation and Variance
For X a random variable and a, b two constants:
E[aX + b] = aE[X ] + b
V[aX + b] = a2V[X ]
For many random variables X1,X2, . . . ,Xn, we have:
E[X1 + . . .+ Xn] = E[X1] + . . .+ E[Xn]
If the random variables X1,X2, . . . ,Xn are independent, then:
E[X1 × . . .× Xn] = E[X1]× . . .× E[Xn]
V[X1 + . . .+ Xn] = V[X1] + . . .+ V[Xn]
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1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
34/38
Common Probability Distributions in Actuarial Studies
Discrete distributions
▶ Poisson
▶ Binomial
▶ Geometric and Negative Binomial
Continuous distributions
▶ Exponential and Gamma
▶ Normal and Log-Normal
▶ Weibull
Review the document common_distributions.pdf on Moodle.
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1 Motivation
2 Sample Space & Events
3 Permutations & Combinations
4 Conditional Probabilities and Independence
5 Random Variables
6 Common Probability Distributions in Actuarial Studies
7 Typical Exam Questions
35/38
Questions taken from Quiz 2020
In a tutorial of 20 students, there are 12 domestic and 8 international students. The tutor
creates, totally randomly, 4 groups of 5 students each. Those groups (1,2,3,4) are considered
distinct. We call an ‘arrangement’ a specific allocation of all students to those four distinct
groups, where the ordering of students within a group does not matter. Careful: because the
groups are distinct, five students forming Group 1 is not equivalent to those same five students
forming Group 2.
1 (1pt) In total, how many different arrangements are they?
2 (2pt) How many arrangements yield that one of the groups is international-only?
3 (1pt) What is the probability that one of the
groups is international-only?
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Solution I
1 The total number of arrangements is:(
20
5
)
·
(
15
5
)
·
(
10
5
)
·
(
5
5
)
=
20! · 15! · 10!
15! · 5! · 10! · 5! · 5! · 5!
=
20!
(5!)4
≈ 1.1733 · 1010
This is because, for the first group, you can choose 5 people out of 20 (and order does not
matter). Once you’ve done this, there are only 15 people left to choose from (for the
second group), and etc.
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Solution II
2 There are 4 choices for the group in which all students are international. Once you have
chosen which of the 4 groups is international-only, you must pick 5 out of 8 students to
form that group. Hence, the number of arrangements is:
4 ·
(
8
5
)
·
(
15
5
)
·
(
10
5
)
·
(
5
5
)
=
4 · 8! · 15! · 10!
3! · 5! · 10! · 5! · 5! · 5!
=
4 · 8! · 15!
3! · (5!)4 =
4 · 8 · 7 · 15!
(5!)3
≈ 1.6951 · 108
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Solution III
3 The answer is simply the ratio of part 2 divided by part 1, i.e.
# cases favourable
# total cases
=
224 · 5! · 15!
20!
≈ 0.01445
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