Week 2 Annotated

ACTL2002/ACTL5101 Probability and Statistics: Week 2 Video Lecture Notes

ACTL2002/ACTL5101 Probability and Statistics

c© Katja Ignatieva

School of Risk and Actuarial StudiesAustralian School of Business

University of New South Wales

[email protected]

Week 2 Video Lecture NotesProbability: Week 1 Week 2 Week 3 Week 4

Estimation: Week 5 Week 6 Review

Hypothesis testing: Week 7 Week 8 Week 9

Linear regression: Week 10 Week 11 Week 12

Video lectures: Week 1 VL Week 3 VL Week 4 VL Week 5 VL

mailto:[email protected]


Bernoulli distribution

Special Discrete Distributions


The Binomial Distribution

The Geometric Distribution

The Negative Binomial Distribution



Bernoulli Distribution

Discrete distribution.

The outcome in a Bernoulli experiment is one of two mutuallyexclusive events, classified as either success (x = 1) or failure(x = 0).

Denote the probability of success (i.e., x = 1) by p,0 < p ≤ 1. Let X denote the Bernoulli random variable sothat:

X =

{1, w.p. p;0, w.p. 1− p.

Notation: X ∼ Bernoulli (p) .

Formulae & Tables (F&T) book page 7.

202/220



p.m.f.: pX (x) = px · (1− p)1−x , for x = 0, 1, and zerootherwise.mean: E [X ] =

∑all x

pX (x) · x

=0 · (1− p) + 1 · p = p.

variance:Var(X ) =E

[X 2]− µ2

X

=∑all x

pX (x) · x2 − p2

=02 · (1− p) + 12 · p − p2 = p · (1− p) .

m.g.f.:MX (t) =E

[eXt]

=∑all x

pX (x) · ext

=(1− p) · e0·t + pe1·t = p · et + (1− p) .203/220




Suppose we have n independent trials such that there is aprobability p of success for each trial, i.e., we have nindependent trials of a Bernoulli r.v..

The random variable X which represents the number ofsuccesses out of n trials has a Binomial distribution and wewrite X ∼ Binomial (n, p).

F&T book, p. 6 (values for cumulative density function givenon pp. 186–188).

204/220



0 10 200

0.1

0.2

0.3

x

prob

abilit

y m

ass

func

tion Binomial(20,0.5) p.m.f.

n = 20, p = 0.5

0 10 200

0.1

0.2

0.3

x

prob

abilit

y m

ass

func


n = 20, p = 0.1

0 100 2000

0.1

0.2

x

prob

abilit

y m

ass

func


n = 200, p = 0.5

0 100 2000

0.1

0.2

x

prob

abilit

y m

ass

func


n = 200, p = 0.1

205/220



0 10 200

0.5

1

xcum

ulat

ive d

ensit

y fu

nctio

n Binomial(20,0.5) c.d.f.

n = 20, p = 0.5

0 10 200

0.5

1

xcum

ulat

ive d

ensit

y fu

nctio


n = 20, p = 0.1

0 100 2000

0.5

1

xcum

ulat

ive d

ensit

y fu

nctio


n = 200, p = 0.5

0 100 2000

0.5

1

xcum

ulat

ive d

ensit

y fu

nctio


n = 200, p = 0.1

206/220



p.m.f.: pX (x) =

(n

x

)· px · (1− p)n−x , for x = 0, 1, . . . , n,

and zero otherwise.

mean: (alternative proof: see slide 211)

E [X ] =E

[n∑

i=1

Yi

]=

n∑i=1

E [Yi ]∗= n · E [Yi ] = n · p.

variance: (alternative proof: see slide 211)

Var(X ) =Var

(n∑

i=1

Yi

)∗=

n∑i=1

Var (Yi ) = n · Var (Yi ) = n · p · (1− p) .

m.g.f.: (alternative proof: see slide 209)

E[eXt]

=E[e∑n

i=1 Yi ·t]∗= E

[(eYi ·t

)n]=(p · et + (1− p)

)n* Using Yi ∼ Ber(p) i.i.d. for i = 1, . . . , n.

207/220




By applying the binomial expansion with n ∈ 1, 2, 3, . . ., andp ∈ [0, 1] we have:

n∑x=0

(n

x

)· px · (1− p)n−x = (p + 1− p)n = 1.

Using the binomial expansion (F&T page 2):

(a + b)n =n∑

k=0

(n

k

)· an−k · bk ,

for any integer n.

Hence, the sum of the probabilities of all events equals one.

208/220




Also, for the moment generating function, we have:

MX (t) =E[eXt]

=n∑

x=0

ext ·(n

x

)· px · (1− p)n−x

=n∑

x=0

(n

x

)·(p · et

)x · (1− p)n−x

∗=(p · et + (1− p)

)n* again, applying the binomial expansion with a = p · et andb = (1− p).

209/220




Recall: MX (t) = (p · et + (1− p))n.

Then we can also use the m.g.f. to find the moments. Since

M ′X (t) = n(p · et + (1− p)

)n−1 · p · et

and

M ′′X (t)∗=n · (n − 1) ·

(p · et + (1− p)

)n−2 ·(p · et

)2

+ n ·(p · et + (1− p)

)n−1 · p · et

=npet(pet + (1− p)

)n−2 ·((n − 1) pet + pet + (1− p)

).

* using product rule: ∂f (x)·g(x)∂x = ∂f (x)

∂x · g(x) + ∂g(x)∂x · f (x).

210/220




We have:

E [X ] = M ′X (0) = n · p

and

E[X 2]

= M ′′X (0) = n · p · ((n − 1) · p + 1) .

The variance is easily shown to be:

Var (X ) =E[X 2]− (E [X ])2

=n · p · ((n − 1) · p + 1)− (n · p)2

=n · p · (1− p) .

211/220



Examples of the Binomial Distribution

Question: The probability of one or more accidents in a yearfor high risk (H) insureds is 25%. When the insurer has 10 Hinsureds, what’s the probability that 5 or more insured havean accident?

Solution: X =“number of insureds with at least one accidenta year”. Pr(X > 4) = 1− Pr(X ≤ 4) = 1− 0.9219 = 0.0781.

212/220




Consider a sequence of independent and repeated trials with pdenoting the probability of success.

Suppose X is the number of trials required up to andincluding the first success. Then X has a geometricdistribution and we write X ∼ Geometric (p) where 0 < p ≤ 1denotes the probability of a success for one trial.

F&T book page 9.

Application: the time (in years) to the first earthquake inAustralia with a magnitude larger than 8.

Note: due to i.i.d. trials: the Geometric distribution has thememoryless property, i.e., Pr(X ≤ a|X ≥ b) = Pr(X ≤ a− b)for b < a.

213/220




p.m.f.: pX (x) = p · (1− p)x−1, for x = 1, 2, . . ..

parameter constraints: 0 < p ≤ 1.

mean:E [X ] =

∑all x

x · p · (1− p)x−1 = p ·∞∑x=1

x · (1− p)x−1

=p ·∞∑x=1

∂

∂(1− p)(1− p)x = p · ∂

∂(1− p)

∞∑x=1

(1− p)x

∗=p · ∂

∂(1− p)

(1− p)

1− (1− p)∗∗= p · 1

p2=

1

p.

* Using geometric series:∑∞x=0 a

x = 11−a ⇔

∑∞x=1 a

x = 11−a − 1 = a

1−a .

** Using quotient rule: ∂∂x

(f (x)g(x)

)=

∂f (x)∂x·g(x)− ∂g(x)

∂x·f (x)

(g(x))2 .214/220




variance: Var(X ) = 1−pp2 (similar as mean).

m.g.f.:MX (t) =E

[eX ·t

]=∞∑x=1

ex ·t · p · (1− p)x−1

=p · et ·∞∑z=0

et·z · (1− p)z

=p · et ·∞∑z=0

(et · (1− p)

)z∗=

p · et

1− (1− p) · et.

Note: (1− p) · et < 1⇒ t < log(

11−p

).

* using geometric series:∑∞

z=0 az = 1

1−a .215/220



Example of the Geometric Distribution

0 5 10 150

0.1

0.2

0.3

0.4

x

prob

abilit

y mas

s fun

ction Geo(0.5) p.m.f.

p = 0.5

0 20 400

0.1

0.2

0.3

0.4

x

prob

abilit

y mas

s fun

ction Geo(0.1) p.m.f.

p = 0.1

0 5 10 150

0.5

1

xcum

ulativ

e de

nsity

func

tion Geo(0.5) c.d.f.

p = 0.5

0 20 400

0.5

1

xcum

ulativ

e de

nsity

func

tion Geo(0.1) c.d.f.

p = 0.1

216/220




Consider a sequence of independent and repeated Bernoullitrials again with p denoting the probability of success on asingle trial. Let X represents the random variable that denotethe number of trials required until there are r successes. ThenX has a negative binomial distribution and we writeX ∼ N.B. (r , p).

F&T book page 8.

Special case: when we have r = 1, i.e., X represents therandom variable that denote the number of trials requireduntil there is one success: X ∼ NB (r = 1, p) = Geometric (p).

217/220




p.m.f.: pX (x) =

(x − 1

r − 1

)· pr · (1− p)x−r , for

x = r , r + 1, . . .

parameter constraints: 0 < p ≤ 1 and r = 1, 2, . . .;

mean: E [X ] = E [Y1 + . . .+ Yr ] = r · E [Yi ] = rp ;

variance:Var(X ) = Var(Y1 + . . .+ Yr ) = r · Var(Yi ) = r ·(1−p)

p2 ;

m.g.f.: MX (t) = MY1+...+Yr (t) = M rYi

(t) =(

p·et1−(1−p)·et

)r;

Proof using i.i.d. property and geometric distribution. LetYi ∼ Geometric(p) for i = 1, . . . , r .

218/220



Example of the Negative Binomial Distribution

0 10 200

0.05

0.1

0.15

0.2

x

prob

abilit

y m

ass

func

tion Negative Binomial(3,0.5) p.m.f.

r = 3, p = 0.5

0 50 1000

0.05

0.1

0.15

0.2

x

prob

abilit

y m

ass

func


r = 3, p = 0.1

0 20 400

0.05

0.1

0.15

0.2

x

prob

abilit

y m

ass

func


r = 10, p = 0.5

0 100 2000

0.05

0.1

0.15

0.2

x

prob

abilit

y m

ass

func


r = 10, p = 0.1

219/220



Example of the Negative Binomial Distribution

0 10 200

0.5

1

xcum

ulativ

e de

nsity

func

tion Negative Binomial(3,0.5) c.d.f.

r = 3, p = 0.5

0 50 1000

0.5

1

xcum

ulativ

e de

nsity

func


r = 3, p = 0.1

0 20 400

0.5

1

xcum

ulativ

e de

nsity

func


r = 10, p = 0.5

0 100 2000

0.5

1

xcum

ulativ

e de

nsity

func


r = 10, p = 0.1

220/220

ACTL2002/ACTL5101 Probability and Statistics: Week 2

ACTL2002/ACTL5101 Probability and Statistics

c© Katja Ignatieva

School of Risk and Actuarial StudiesAustralian School of Business

University of New South Wales

[email protected]

Week 2Probability: Week 1 Week 3 Week 4

Estimation: Week 5 Week 6 Review

Hypothesis testing: Week 7 Week 8 Week 9

Linear regression: Week 10 Week 11 Week 12

Video lectures: Week 1 VL Week 2 VL Week 3 VL Week 4 VL Week 5 VL

mailto:[email protected]


Last week

Introduction to probability;

Definition of probability measure, events;

Calculating with probabilities; Multiplication rule,permutation, combination & multinomial;

Distribution function;

Moments: (non)-central moments, mean, variance (standarddeviation), skewness & kurtosis;

Generating functions;

301/354


This week

Special (parametric) univariate distributions;

Discrete distributions: Bernoulli, Binomial, Geometric,Negative Binomial, and Poisson distributions;

Continuous distributions: Exponential, Gamma, Gaussian,Lognormal, Uniform, Beta, Negative Binomial, Weibull, andPareto distributions.

302/354


The Poisson Distribution


Probability Distributions used in Insurance and FinanceThe Poisson Distribution


Exponential & Gamma distributionSpecial case of the Gamma distribution: Exponential distributionGamma Distribution

Bernoulli familyExercises

The Normal (Gaussian) DistributionThe Normal (Gaussian) DistributionThe Standard Normal Distribution TableDeriving the M.G.F. of the NormalLognormal Distribution

Uniform DistributionUniform Distribution

Beta DistributionBeta Distribution

Negative Binomial (another form (continuous) of the distribution)Negative Binomial

Weibull DistributionWeibull Distribution

Pareto DistributionPareto Distribution





F&T book page 7 (values for cumulative probabilitydistribution given on pp. 175–185).

Let X be a random variable which represents the number ofrare events that occur in a time period.

Application: number of car accidents in a week.

We write X ∼ Poisson (λ) when X has a Poisson distribution.

Relationship with Binomial r.v.: Let the time interval go tozero (n→∞), λ = n · p, then either X takes the values zeroor one w.p. 1 (see tutorial).

303/354




0 5 100

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

x

prob

abilit

y mas

s fun

ction

Poisson p.m.f.

λ= 1λ= 2λ= 4

0 5 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cum

ultat

ive d

ensit

y fun

ction

Poisson c.d.f.

λ= 1λ= 2λ= 4

When λ (claim rate) increases E[X ] (expected number of claims)increases ⇒ probability mass shifts to the right.304/354





Properties of Poisson random variables are:

p.m.f.: pX (x) = e−λ·λxx! , for x = 0, 1, . . .;

parameter constraint: λ > 0;

mean:E [X ] =

∑all x

x · pX (x) =∞∑x=0

x · λx · e−λ

x!

∗=λ ·

∞∑x=1

λx−1 · e−λ

(x − 1)!

∗∗=λ ·

∞∑z=0

λz · e−λ

z!∗∗∗= λ;

* 0 · pX (0) = 0; ** z = x − 1 *** using:∑

all x

pX (x) = 1.

305/354




The Poisson Distributionvariance:

Var(X ) =E[X 2]− (E [X ])2 ∗= λ2 + λ− λ2 = λ,

* using:

E[X 2]

=∑all x

x2 · pX (x) =∞∑x=0

(x · (x − 1) + x) · λx · e−λ

x!

=∞∑x=0

x · λx · e−λ

x!+∞∑x=0

(x · (x − 1)) · λx · e−λ

x!

=λ+ λ2 ·∞∑x=2

λx−2 · e−λ

(x − 2)!∗∗= λ+ λ2 ·

∞∑z=0

λz · e−λ

z!

∗∗∗= λ+ λ2;

** z = x − 2 *** using∑all z

pX (z) = 1.306/354




m.g.f.: MX (t) = exp (λ (et − 1)).

Note that:∞∑x=0

e−λ · λx

x!∗= e−λ · eλ = 1,

* using the series expansion for the exponential function (i.e.,exp(λ) =

∑∞x=0

λx

x! ).

The m.g.f. is derived as (note: etx = (et)x):

MX (t) =E[etX]

=∞∑x=0

ext · e−λ · λx

x!=∞∑x=0

e−λ · (λ · et)x

x!

=e−λ · eλ·et ·∞∑x=0

e−λ·et · (λ · et)x

x!︸︷︷︸=∑all y

pY (y) = 1, with Y ∼ POI(λ · et)

= exp(λ ·(et − 1

)).307/354




Exercise

The expected number of claims for personal accidentinsurance is once every ten year.

a. Question: Which distribution would you use to model thenumber of claims for an insured in a year?

b. Question: What is the probability of two or more claimsusing a Poisson distribution?

c. Question: What are the parameters of the Binomialdistribution?

d. Question: What are the differences between the Poissondistribution and the Binomial distribution, especially in thetail?

308/354




Solutiona. Solution: X ∼ POI(λ = 0.1) distribution.b. Solution: Using a. and the p.m.f. we have:

Pr(X ≥ 2) =1− Pr(X < 2) = 1− Pr(X ≤ 1)

=1− (Pr(X = 0) + Pr(X = 1))

=e−0.1 · 0.10

0!+

e−0.1 · 0.11

1!= 1.1e−0.1.

c. Solution: We have n = 1. Option 1: Correct estimate theprobability of no claim:

p = 1− Pr(X = 0) = 1− e−0.1 · 0.10

0!= 1− e−0.1 ≈ 0.0995.

Option 2: Correct mean: p = E [X ] /n = 0.1.d. Solution: The Poisson distribution allows for more claims by

one contract. Therefore, it would have fatter tails.309/354


Exponential & Gamma distribution

Special case of the Gamma distribution: Exponential distribution














Special case: Exponential distribution

See F&T book, p. 11:

Let X ∼ Gamma(α, β) and we set β = λ and α = 1, then wehave a special case of the Gamma distribution. We say that Xis exponentially distributed: X ∼ EXP (λ).

Application: Three busses an hour on average arriverandomly at a bus stop. The time elapsed after the previousbus has arrived doesn’t matter for the time the next bus isarriving. Let X be the r.v. representing the time to the nextbus is arriving.

The density of the r.v. X becomes:

fX (x) = λ · e−λ·x , for x ≥ 0.

310/354




0 2 4 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

prob

abilit

y den

sity f

uncti

on

Exponential p.d.f.

λ= 1λ= 2λ= 4

0 2 4 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumu

lative

dens

ity fu

nctio

n

Exponential c.d.f.

λ= 1λ= 2λ= 4

When λ (claim rate) increases E[X ] (expected number of claims)increases ⇒ time between claims decreases ⇒ p.d.f. shifts to left.311/354




Special case: Exponential distributionOne of the interesting properties of the exponential is thememoryless property:

Pr (X > a + b |X > a ) =Pr (X > a + b,X > a)

Pr (X > a)

=Pr (X > a + b)

Pr (X > a)

=

∫∞a+b λ · e

−λxdx∫∞a λ · e−λxdx

=e−λ·(a+b)

e−λ·a= e−λ·b

= Pr (X > b) .

Mean: E[X ] = 1λ , Variance: Var(X ) = 1

λ2 ,

and m.g.f: MX (t) =(1− t

λ

)−1.

312/354



Gamma Distribution













Gamma Distribution

0 2 4 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

probab

ility de

nsity fu

nction

Gamma p.d.f.

α= 1, β= 1α= 2, β= 1α= 4, β= 1

0 2 4 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumula

tive de

nsity fu

nction

Gamma c.d.f.

α= 1, β= 1α= 2, β= 1α= 4, β= 1

In insurance claim modelling the Gamma distribution is oftenused. We write X ∼ Gamma (α, β) to denote that we have arandom variable X that has a Gamma distribution withparameters α (shape parameter) and β (scale parameter).

Application: the length of time between α = 5 accidents, theclaim size for a fire insurance claim.

F&T book, page 12.313/354



Gamma Distribution

0 10 200

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

x

prob

abilit

y den

sity f

uncti

on

Gamma p.d.f.

α= 1, β= 0.5α= 2, β= 0.4α= 4, β= 0.4

0 10 200

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumu

lative

dens

ity fu

nctio

n

Gamma c.d.f.

α= 1, β= 0.5α= 2, β= 0.4α= 4, β= 0.4

When β (claim rate) increases E[X ] (expected number of claims)

increases ⇒ time between α claims decreases ⇒ p.d.f. shifts to left.

When α increases ⇒ more claims, more time ⇒ p.d.f. shifts to right.314/354



Gamma Distribution

Gamma Distribution

density: fX (x) =βα

Γ (α)· xα−1 · e−β·x , for x ≥ 0, and zero

otherwise (note that Gamma random variables arenon-negative).

parameter constraints: α > 0, β > 0;

mean: E [X ] = αβ ;

variance: Var(X ) = αβ2 ;

m.g.f.: MX (t) = E[eXt]

=(

ββ−t

)α,

provided t < β (prove: see slide 318);

Proof of mean and variance: Use the moments (see slide 319).315/354



Gamma Distribution

Gamma function

The Gamma function (see F&T page 5) is defined by:

Γ (α) =

∫ ∞0

xα−1 · e−xdx , α > 0.

One can show the following recursive relationships holds:

Γ (α + 1) = α · Γ (α) (use integration by parts)Γ (n) = (n − 1)!, for n = 1, 2, 3, . . . .

Exercise: Determine Γ(3) and Γ(1/2).

Solution: Γ(3) = (3− 1)! = 2 · 1 · 1 = 2 and

Γ(1/2) =∫∞

0 x−1/2 · e−xdx ∗=∫∞

0

√2t · e

−t2/2dt =√2 ·∫∞

0 e−t2/2dt =

√2 ·√π/2 =

√π,

* using x = t2/2, dx = tdt (see slides 328-329).316/354



Gamma Distribution

Gamma density

To show that the Gamma is a proper density, note that:∫ ∞0

βα

Γ (α)· xα−1 · e−β·xdx =

βα

Γ (α)·∫ ∞

0xα−1e−β·xdx

and re-parameterising with z = β · x so that dx = 1βdz , we have:

βα

Γ (α)·∫ ∞

0xα−1 · e−β·xdx =

βα

Γ (α)·∫ ∞

0(z/β)α−1 · e−z · 1

βdz

=βα

Γ (α)· β−α ·

∫ ∞0

zα−1 · e−zdz︸︷︷︸=Γ(α)

=1.

317/354



Gamma Distribution

Gamma m.g.f.

To prove the formula for the m.g.f., we have:1

MX (t) =

∫ ∞0

ex ·t · βα

Γ (α)· xα−1 · e−βxdx

=

∫ ∞0

βα

Γ (α)· xα−1 · e−x ·(β−t)dx

=βα · (β − t)−α ·∫ ∞

0

(β − t)α

Γ (α)· xα−1 · e−x ·(β−t)dx︸︷︷︸

=1, because∫∞

0 fY (y)dy = 1, with Y ∼ Gamma(α, β − t)

and the result follows.

1The gamma variable takes non-negative values, so theintegration limits should be from 0 to ∞.318/354



Gamma Distribution

Gamma moments

There is a useful formula for higher (non-central or raw) momentsof the Gamma distribution.

E [X n] =

∫ ∞0

xn · βα

Γ (α)· xα−1 · e−β·xdx

=

∫ ∞0

βα

Γ (α)· x (n+α−1) · e−β·xdx

=βα

Γ (α)· β−n−α · Γ (n + α) ·

∫ ∞0

βn+α · x (n+α−1) · e−β·x

Γ (n + α)dx︸︷︷︸

=1, because∫∞

0 fY (y)dy = 1, with Y ∼ Gamma(n + α, β)

=1

βn· Γ (n + α)

Γ (α).

319/354



Gamma Distribution

ExerciseThe Gamma(α,β) distribution models the time required for αevents to occur, given that the events occur randomly in aPoisson process with a mean time between events of β. Weknow that major flooding occurs in Queensland on averageevery six years. You have to valuate a reinsurance contractthat pays:

- Zero, if there are less than two major floods in the next tenyear;

- $100 million, if there are two major floods in the next ten year;- $150 million, if there are three or four major floods in the next

ten year;- $200 million, if there are more than four major floods in the

next ten year.

a. Question: What is the expected value of the contract?

b. Question: The price of the contract is $65 million, would yourecommend the insurer to buy the contract?320/354



Gamma Distribution

Pr(Gamma(2, 6) ≤ 10) = 0.4963, Pr(Gamma(3, 6) ≤ 10) = 0.2340,Pr(Gamma(4, 6) ≤ 10) = 0.0883, and Pr(Gamma(5, 6) ≤ 10) = 0.0275.

a. Solution: Let X ∼ Gamma(2, 6) the r.v. for the time untilthe second claim is filed; Y ∼ Gamma(3, 6) the r.v. for thetime until the third claim is filed, and Z ∼ Gamma(5, 6) ther.v. for the time until the fifth claim is filed.

Two or more claims filed in ten years: Pr(X ≤ 10) = FX (10).Probability of exactly two claims:Pr(X ≤ 10)− Pr(Y ≤ 10) = FX (10)− FY (10).

Probability of exactly three or four claims:Pr(Y ≤ 10)− Pr(Z ≤ 10) = FY (10)− FZ (10).

Probability of more than four claims: Pr(Z ≤ 10).

(FX (10)− FY (10)) · 100 + (FY (10)− FZ (10)) · 150 + FZ (10) · 200

=100 · FX (10) + 50 · FY (10) + 50 · FZ (10)

=49.63 + 11.70 + 1.38 = $62.71 million.

b. Solution: Depends on your portfolio. Only this product: yes?

321/354


Bernoulli family

Exercises












Bernoulli family

Exercises

Exercises

A life insurance company offers an annuity. In order to reducelongevity risk, the payout after age 90 depends on the numberof survival in the pool. Each pool has 20 insured.

The probability of surviving to 90, conditional on reaching 65 is:Female Male Both One Only

Prob surviving to 90 41.2% 29.8% 12.3% 58.7%

You want to buy a joint and survivor annuity, which pay out$100 if both spouses are alive and $70 if only one is alive.

- The payout is reduced with 20% if for more than 17 contractsat least one of the spouses is alive.

- The payout is increased with 20% if for less than 10 contractsat least one of the spouses is alive.

322/354


Bernoulli family

Exercises

Exercises

a. What is the probability that the payout increases when youreach 90?

b. What is the probability that the payout decreases when youreach 90?

You want to reduce your risk of a reduction in your payout.

c. How many insurance contract can you buy in order to have aprobability of 5% that none of the contracts will reduce yourpayments? (Hint: use Geometric distribution)

d. How many insurance contract do you need to buy in order tohave a probability of 95% that at least 3 the contracts willincrease your payments?

323/354


Bernoulli family

Exercises

Solutiona. Pr(X ≤ 9) = 0.0132.

b. Let X ∼ Bin(20, 0.71) be the number of contracts with atleast one of the spouses is alive at age 90.1− Pr(X ≤ 17) = 1− 0.9567 = 0.0433.

c. Let Y ∼ Geo(0.0433) be number of contracts which do notget a reduction in the payment.

We have: Pr(Y ≤ 1) = 0.0433, Pr(Y ≤ 2) = 0.0847,Pr(Y ≤ 3) = 0.1244, and Pr(Y ≤ 4) = 0.1623.

Thus, you can buy one contracts.

d. Let Z ∼ NBin(3, 0.0132) be number of contracts needed toget a at least 3 contracts with an increase in the payment.

Pr(Z ≤ 62) = 0.0489, Pr(Z ≤ 63) = 0.0509,Pr(Z ≤ 474) = 0.9496, and Pr(Z ≤ 475) = 0.9501.

Thus, you need to buy 475 contracts.324/354


The Normal (Gaussian) Distribution
















Continuous distribution;

The normal distribution plays a very central role inmathematical statistics.

Notation: X ∼ N(µ, σ2

)denotes a normally distributed

random variable with mean µ (location parameter) andvariance σ2 (scale parameter).

Example: the normal distribution approximates the Binomialdistribution when n · p is large enough (see week 5).

325/354




The standard Normal (Gaussian) Distribution

Standard Normal: When µ = 0 and σ2 = 1, then we have astandard normal random variable and we use Z to denotesuch, that is, Z ∼ N (0, 1).

We can always standardise a normal random variableX ∼ N

(µ, σ2

)as follows:

Z =X − µσ

∼ N (0, 1) , and X = σZ + µ.

E [Z ] = E[X−µσ

]= E[X ]−µ

σ = 0.

Var(Z ) = Var(X−µσ

)= Var(X )

σ2 = 1.

F&T book, page 11 (values for cumulative probabilitydistribution given on pages 160-162).

326/354





density: fX (x) =1√

2π · σexp

(−1

2·(x − µσ

)2)

, for

−∞ < x <∞.

parameter constraints: −∞ < µ <∞ and σ > 0.

mean: E [X ] = µ.

variance: Var(X) = σ2.

m.g.f.: MX (t) = E[eXt]

= exp(µ · t + 1

2 · σ2 · t2

)(prove, see slides 335–337).

Prove of mean and variance: use the m.g.f.!

327/354




Prove of legitimate density (OPTIONAL)

Prove that the normal distribution is a legitimate density by

showing:

∫ ∞−∞

1√2π · σ

· e−12·( x−µ

σ )2

dx = 1.

Transform z =x − µσ

and consider the case of the standard

normal where µ = 0 and σ2 = 1.

Consider then the integral:

I =

∫ ∞−∞

1√2π· e−

12·z2dz .

We note that I > 0 and that:

I 2 =

(∫ ∞−∞

1√2π· e−

12·z2dz

)2

=1

2π·∫ ∞−∞

∫ ∞−∞

e−12·(y2+z2)dydz .

328/354




This double integral can be evaluated by changing to polarcoordinates by setting

y = r · cos(θ) and z = r · sin(θ)

so that

y2 + z2 = r2 ·(sin2(θ) + cos2(θ)

)︸︷︷︸=1

= r2

and

dydz = det

([ ∂y∂r

∂y∂θ

∂z∂r

∂z∂θ

])drdθ = det

([cos(θ) −r sin(θ)sin(θ) r cos(θ)

])drdθ = rdrdθ.

(Check your calculus book for verification of this polar coordinatetransformation). We have:

I 2 =1

2π

∫ 2π

0

∫ ∞0

e−12r2rdrdθ =

1

2π

∫ 2π

0

[−e

−r2

2

]∞0︸︷︷︸

=0−(−1)=1

dθ =1

2π· 2π = 1.

329/354




−5 0 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

x

prob

abilit

y den

sity f

uncti

on

Normal p.d.f.

µ= −2, σ= 1µ= 0, σ= 1µ= 2, σ= 1

−5 0 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumu

lative

dens

ity fu

nctio

n

Normal c.d.f.

µ= −2, σ= 1µ= 0, σ= 1µ= 2, σ= 1

When µ (average claim size/number of claims) increases ⇒ p.d.f. shifts

to right. When σ2 (variance of claim size/number of claims) increases ⇒p.d.f. shifts away from µ.330/354




−5 0 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

x

proba

bility d

ensity

funct

ion

Normal p.d.f.

µ= 0, σ= 0.5µ= 0, σ= 1µ= 0, σ= 2

−5 0 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumula

tive de

nsity f

unctio

n

Normal c.d.f.

µ= 0, σ= 0.5µ= 0, σ= 1µ= 0, σ= 2

We can easily verify the following (check from F&T page 160-161):

Pr (−1 ≤ Z ≤ 1) =0.6826

Pr (−2 ≤ Z ≤ 2) =0.9544

Pr (−3 ≤ Z ≤ 3) =0.9974331/354



The Standard Normal Distribution Table














The Standard Normal distribution table: F&T book pages160-162. Notation:

Φ (z) = Pr (Z ≤ z)

is used to denote the c.d.f. of a standard normal distribution.

For example, if Z ∼ N (0, 1), then:

Pr (Z ≤ 1.56) = Φ(1.56) = 0.9406.

0

x

f(x)

zα

z1−α

← Pr(Z>z1−α

)=α

Pr(Z<z1−α

)=1−α →

0

x

f(x)

zα

z1−α

Pr(Z<zα)=α→

← Pr(Z>zα)=1−α

symmetry property: Pr(Z>z1−α

)=Pr(Z<zα)

332/354




Example: the Standard Normal Distribution Table

The tables usually gives values only for positive zp but usingthe symmetry property, we can easily derive for negativevalues.

Note that:

Pr (Z ≤ −z) = 1− Pr (Z ≤ z)

so that, for example:

Pr (Z ≤ −1.56) = 1− 0.9406 = 0.0594.

We can get probability values like:

Pr (−1.12 ≤ Z ≤ 1.56) =Φ (1.56)− Φ (−1.12)

=0.9406− (1− 0.8686)

=0.8092.333/354





For non-standard normal distribution, we can alwaysstandardise using the property that if X ∼ N

(µ, σ2

), then

Z =X − µσ

∼ N (0, 1) .

Say, X ∼ N (100, 25), then the probability that X will bebetween 92 and 112 is:

Pr (92 ≤ X ≤ 112) = Pr

(92− 100

5≤ Z ≤ 112− 100

5

)=Φ (2.4)− Φ (−1.6)

=0.9918− 0.0548 = 0.9370.

334/354



Deriving the M.G.F. of the Normal















Derive the moment generating function of the normal distribution:

MX (t) =E[eXt]

=

∫ ∞−∞

ex ·t · 1√2π · σ

· exp

(−1

2·(x − µσ

)2)

︸︷︷︸=fX (x)

dx

=

∫ ∞−∞

1√2π · σ

· exp

(x · t − 1

2·(x − µσ

)2)dx .

Next slide: rewrite the blue part.

335/354




Deriving the M.G.F. of the NormalConsider

x · t − 1

2·(x − µσ

)2

=x · t − 1

2σ2·(x2 − 2µx + µ2

)=− 1

2σ2·

x2︸︷︷︸a2

−2(µ+ σ2 · t

)· x︸︷︷︸

−2ba

− 1

2σ2· µ2

∗=− 1

2σ2·(x −

(µ+ σ2 · t

))2 − 1

2σ2·(µ2 −

(µ+ σ2 · t

)2)

=− 1

2σ2·(x −

(µ+ σ2 · t

))2+

(µ · t +

1

2· σ2 · t2

).

* using (a− b)2 = a2 + b2 − 2ab, with a = x and b = µ+ σ2 · t.336/354




Deriving the M.G.F. of the NormalFrom slide 335 we have:

MX (t) =

∫ ∞−∞

1√2π · σ

· exp

(x · t − 1

2·(x − µσ

)2)dx .

Using the previous slide, replace the blue part by the red part:

MX (t) =

∫ ∞−∞

1√2πσ

exp

(− 1

2σ2

(x −

(µ+ σ2 · t

))2+

(µt +

1

2σ2t2

))dx

=e(µ·t+ 12·σ2·t2) ·

∫ ∞−∞

1√2π · σ

· e− 1

2·(

x−(µ+σ2·t)σ

)2

dx︸︷︷︸=1, because

∫∞0 fY (y)dy = 1, with Y ∼ N(µ+ σ2 · t, σ2)

=e(µ·t+ 12·σ2·t2).

337/354



Lognormal Distribution















F&T book, page 14.

Consider X ∼ Lognormal(µ, σ2

).

Density is:

fX (x) =1

x · σ ·√

2πexp

(−1

2·(

log(x)− µσ

)2), x > 0.

Parameter constraints: −∞ < µ <∞, 0 < σ <∞.

E[X ] = exp

(µ+

1

2σ2

)Var(X ) =e(2µ+σ2) ·

(eσ

2 − 1).

Note: parameter µ is not E[X ], also σ2 is not Var(X )!338/354




Lognormal DistributionIf Y ∼ N

(µ, σ2

)and X = eY then X is lognormal; or

log(X ) ∼ N(µ, σ2), i.e., “log is normally distributed”.

X = exp (Y ) ∼ Lognormal(µ, σ2

)is said to have a lognormal

distribution with parameters µ and σ2.

0 5 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

x

probab

ility de

nsity f

unction

LogNormal p.d.f.

µ= 0, σ= 1µ= 0, σ= 0.5µ= −1, σ= 0.5

0 5 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumula

tive de

nsity f

unction

LogNormal c.d.f.

µ= 0, σ= 1µ= 0, σ= 0.5µ= −1, σ= 0.5

339/354




Application: Models of stock prices (or returns in general)are often based on the lognormal distribution.

Application: This distribution is also often used to modelclaim sizes.

Property: Product of independent lognormal randomvariables are lognormal (can you explain why?).

Property: To calculate probabilities for a lognormal randomvariable, restate them as probabilities about the associatednormal random variable.

Pr(X ≤ a) = Pr(log(X ) ≤ log(a))

= Pr

(log(X )− µ

σ≤ log(a)− µ

σ

)= Pr

(Z ≤ log(a)− µ

σ

).

340/354




Exercise Lognormal DistributionLosses from large fires can often be modelled using alognormal distribution.

Suppose that the average loss due to fire for buildings of aparticular type is $25 million and the standard deviation of theloss is $10 million.

Question: Determine the probability that a large fire resultsin losses exceeding $40 million.

Solution: Let X be the loss, we haveE[X ] = 25m,Var(X ) = (10m)2.

σ2 = log(

1 + Var(X )E[X ]2

)= log(1 + 100

625 ) = log(1.16),

µ = log(E[X ])− 12σ

2 = log(25)− log(1.16)2 .

Pr(X > 40) = 1− Pr (Y ≤ log (40)) =

1−Pr

(Z ≤ log(40)−log(25)+ log(1.16)

2√log(1.16)

)= 1−Φ(1.4126) = 0.0789.

341/354


Uniform Distribution















F&T book, page 13.

Consider an experiment where all outcomes in a range [a,b]are equally likely to happen, and outcomes outside this rangehave probability zero.

The variable X is called Uniform distributed and we writeX ∼ UNIF(a, b).

density: fX (x) = 1(b−a) , for a ≤ x ≤ b, and zero otherwise.

0 1 20

0.5

1

1.5

2

x

probabil

ity dens

ity func

tion

Uniform p.d.f.

a= 0, b= 0.5a= 0, b= 2a= 1, b= 2

0 1 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumulat

ive den

sity func

tion

Uniform c.d.f.

a= 0, b= 0.5a= 0, b= 2a= 1, b= 2

342/354




Uniform Distributioncumulative distribution:

FX (x) =

∫ x

−∞fX (x)dx =

∫ x−∞ 0dx = 0, if x < a;∫ xa

1(b−a)dx = x−a

b−a , if a ≤ x ≤ b;∫ ba

1(b−a)dx = 1, if x > b.

parameter constraints: a < b.

mean:E [X ] =

∫ ∞−∞

x · fX (x)dx

=

∫ b

ax · 1

(b − a)dx =

[1

2

x2

b − a

]ba

=a + b

2.

variance:Var(X ) =E

[X 2]− (E [X ])2

=

∫ b

ax2 1

(b − a)dx −

(a + b

2

)2

=

[1

3

x3

b − a

]ba

−(a + b

2

)2

=(b − a)2

12.

343/354


Beta Distribution

Beta Distribution












Beta Distribution

Beta Distribution

We write X ∼ Beta (a, b) to denote a Beta random variable Xwith parameters a (shape parameter) and b (shapeparameter).

Generally used to model proportions because the range of x isbetween 0 and 1

Application: percentage of loss on default of a company onits debt.

F&T book, page 13.

See Excel file for relation with Binomial (e.g. application onslide 347).

Beta function:B(α, β) = Γ(α)·Γ(β)

Γ(α+β)

∗=∫ 1

0 xα−1 · (1− x)β−1dx∗∗= (α−1)!·(β−1)!

(α+β−1)! .

* using density of Beta function (next slide),** if α and β are integers.

344/354


Beta Distribution

Beta Distribution

Beta Distribution

density: fX (x) =Γ (a + b)

Γ (a) Γ (b)︸︷︷︸1/B(a,b)

xa−1 (1− x)b−1 , for 0 ≤ x ≤ 1.

parameter constraints: a > 0, b > 0.

mean:E [X ] =

∫ ∞−∞

x · fX (x)dx =

∫ 1

0x ·(xa−1 · (1− x)b−1

B (a, b)

)dx

=1

B (a, b)·∫ 1

0xa · (1− x)b−1dx =

B (a + 1, b)

B (a, b)

=Γ (a + b)

Γ (a) · Γ (b)· Γ (a + 1) Γ (b)

Γ (a + b + 1)

=Γ (a + b)

Γ (a) · Γ (b)· aΓ (a) · Γ (b)

(a + b)Γ (a + b)=

a

a + b

variance: Var(X ) = a·b(a+b)2(a+b+1)

.345/354


Beta Distribution

Beta Distribution

0 0.5 10

0.5

1

1.5

2

2.5

x

prob

abilit

y den

sity f

uncti

on

Beta p.d.f.

a= 2, b= 4a= 0.2, b= 1a= 4, b= 2a= 0.5, b= 0.5

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumu

lative

dens

ity fu

nctio

n

Beta c.d.f.

a= 2, b= 4a= 0.2, b= 1a= 4, b= 2a= 0.5, b= 0.5

When a (number of successes) increases or b (number of failures)

decreases ⇒ proportion of successes increases ⇒ p.d.f. shifts to right.346/354


Beta Distribution

Beta Distribution

ApplicationAn insurance company offers nuclear incident insurance forn = 10 years.

In these ten years there have been x = 3 years with a claim.

The insurer does not price idiosyncratic risk, but does pricesystematic risk (i.e., uncertainty in p).

The insurer has only limited information of the true value p.Assume claims are $1 billion each.

a. Question: What is the price of the contract when the price isthe mean half the standard deviation?

b. Question: What is the price of the contract when the price isthe 75% quantile?

a. Solution: (See Excel file) $0.398705 billion.

b. Solution: (See Excel file) $0.420471 billion.347/354


Negative Binomial (another form (continuous) of the distribution)

Negative Binomial













Negative Binomial

Negative binomial (another form of the distribution)We look at a model of heterogeneous risks.

Pr (X = x) =

(k + x − 1

x

)· pk · (1− p)x for x = 0, 1, 2, . . ..

Consider an insurance portfolio of risks, e.g., the spectrum ofdrivers, from good drivers to bad drivers.

For each, assume the number of claims, conditional on λ, isPoisson(λ).

Assume λ has a Gamma distribution, i.e.,

Pr (u < λ < u + du) =β

Γ (α)· e−β·u · (β · u)α−1 du.

We have already assumed (X |λ) is Poisson distributed withparameter λ. Or, that (X |λ = u) ∼ Poisson(u). Thus:

Pr (X = x |u < λ < u + du) = e−u · ux

x!.

348/354



Negative Binomial

Probability a policyholder chosen at random has x claims is:

Pr (X = x)LTP=

∫ ∞0

e−u · ux

x!· β

Γ (α)· e−β·u · (β · u)α−1 du

∗=

(α + x − 1

x

)·(

β

1 + β

)α·(

1

1 + β

)x

i.e., NB(p =β

1 + β, k = α), using:∫∞

0 uα+x−1e−u(β+1)du∗∗=∫∞

0 e−z · zx+α−1 ·(

11+β

)x+α−1· 1

1+βdz = Γ(α+x)(β+1)α+x .

(** change of variables: z = u(β + 1), du = dzβ+1 )

* using

(α + x − 1

x

)=

(x + α− 1)!

x! · (α− 1)!=

1

x!· Γ(α + x)

Γ(α).

349/354


Weibull Distribution
















F&T book, page 15.

Application: Survival function: used for modelling lifetimes(time to failure).

Let X ∼Weibull(c , γ) is Weibullly distributed variable.

probability density: fX (x) = c · γ · xγ−1 · exp (−c · xγ), forx > 0, and zero otherwise.

cumulative distribution function:FX (x) = 1− exp (−c · xγ).

moments: E [X r ] = Γ(

1 + rγ

)· 1c r/γ

.

Parameters: γ < 1 (γ = 1/γ > 1): decreasing(constant/increasing) failure rate over time.

350/354


Pareto Distribution

Pareto Distribution












Pareto Distribution

Pareto Distribution

Pareto DistributionHeavy tailed distribution (an extreme value distribution).

Often used for reinsurance purposes. The Pareto distributiontapers away to zero much more slowly than LogNormal.Hence, it is more appropriate for estimating reinsurancepremium in respect of very large claims.

F&T book page 14–15. Used to model r.v. with very largevalues with very low probabilities (e.g. incomes).

Cumulative distribution function (α > 0, λ > 0):

FX (x) = 1−(

λ

λ+ x

)α, x > 0.

Probability density function:

fX (x) =α · λα

(λ+ x)α+1=

α

λ · (1 + x/λ)α+1.

351/354


Pareto Distribution

Pareto Distribution

0 2 40

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

x

prob

abilit

y den

sity f

uncti

on

Pareto p.d.f.

α= 3, λ= 1α= 3, λ= 2α= 3, λ= 4α= 1, λ= 4

0 2 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

cumu

lative

dens

ity fu

nctio

n

Pareto c.d.f.

α= 3, λ= 1α= 3, λ= 2α= 3, λ= 4α= 1, λ= 4

When λ increases ⇒ p.d.f. shifts to right.

When α increases ⇒ p.d.f. shifts to left.352/354


Pareto Distribution

Pareto Distribution

Pareto Distribution

Moments do not always exist:

E [X r ] =Γ (α− r) · Γ (1 + r)

Γ (α)·λr , r = 1, 2, and 3, provided r < α.

For example:

Var (X ) =α · λ2

(α− 1)2 · (α− 2)

and only exists if α > 2.

Prove: see exercise 8.

353/354


Pareto Distribution

Pareto Distribution

Software packages (Not in exam)Distributions are not uniquely specified.

Common differences:- Geometric (r = 1) & Negative Binomial: change of variables:z = x − r . Support: z ≥ 0.

- Exponential distribution: κ = 1/λ. Parameter constraint:κ > 0.

- Gamma distribution: κ = 1/β. Parameter constraint: κ > 0.

- Normal & Lognormal distribution defined as: N(µ, σ) andLN(µ, σ) instead of N(µ, σ2) and LN(µ, σ2).

- Pareto distribution: z = x + β. Support: z > β (see exercise8).

Be careful how to use pre-programmed c.d.f. and p.d.f.functions in software packages!

In this course we will follow notation from F&T.354/354

Week 2 Annotated

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