Credibility Theory
Lecture Solutions – Week 1

1.1
Let

A be the event that a student is an actuarial student.
C be the event that a student is a Credibility Theory student.

P(C)

= 0.02

⇒

P( C ) = 0.98

P(A | C) = 0.98
P(A | C ) = 0.07

P( A | C ) P(C )
P ( A | C ) P(C ) + P( A | C ) P(C )

P (C | A) =
=

0.98 × 0.02
0.98 × 0.02 + 0.07 × 0.98

=

0.0196
0.0882

= 0.2222

= 22%

1.2
Let

A1 = “The science department is a heavy polluter”.
A2 = “The science department is a light polluter”.
A3 = “The science department is not a polluter”.
B = “The waterways are visibly polluted”.

P(A1) = 0.1;

P(A2) = 0.2;

P(A3) = 0.7.

P(B |A1) = 0.9;

P(B |A2) = 0.6;

P(B |A3) = 0.3.

P(B)

= P(B |A1)P(A1) + P(B |A2)P(A2) + P(B |A3)P(A3)
= 0.9*0.1 + 0.6*0.2 + 0.3*0.7
= 0.42

P ( A1 | B) =

P( B | A1 ) P( A1 ) 0.9 × 0.1
=
= 0.21
P( B)
0.42

P ( A2 | B) =

P( B | A2 ) P( A2 ) 0.6 × 0.2
=
= 0.29
P( B)
0.42

P( A3 | B) =

P( B | A3 ) P( A3 ) 0.3 × 0.7
=
= 0.5
P( B)
0.42

1.3
The Bayesian model is: p ~ U(0,1) and (X|p) ~ Bin(n,p).
Thus, f(p) = 1, 0 ≤ p ≤ 1

and

⎛n⎞ f(x|p) = ⎜ ⎟ p x (1 − p) n − x .
⎝x⎠

⎛n⎞ x n −x
⎜ ⎟ p (1 − p) x f(p|x) = ⎝ ⎠
1⎛ n ⎞ x n −x
∫0 ⎜ x ⎟ p (1 − p) dp
⎝ ⎠
=

p x (1 − p) n − x
1

∫p
0

Recall:

1

∫p
0

α−1

x

(1 − p) n − x dp

(1 − p)β−1 dp =

Γ(α)Γ(β)
Γ(α + β)

Therefore, f(p|x) =

Γ(n + 2) p x (1 − p) n − x ,
Γ(x + 1)Γ(n − x + 1)

0≤p≤1

(p|x) ~ Beta(x+1, n – x + 1)

1.4
Let

X be an indicator variable such that X = 1 if a card drawn from the urn (with replacement) is observed to be white and X = 0, otherwise.
W be the number of white cards in the urn.

Given p, X is a Bernoulli random variable with parameter p.

Since we have not been given any information regarding the distribution of the number of white cards, we assume that all numbers of white cards from 1 to (n – 1) are equally likely. Thus, since p = w/n, p is uniformly distributed on {1/n,…, (n-1)/n}.
The Bayesian model is then:
(X | p) ~ Bern(p)

⇒

f(x | p) = px(1 – p)1 – x, x = {0, 1}

f(p) = 1/(n – 1),

p = {1/n, …, (n – 1)/n}

Using Bayes’ theorem: f(p| x) =

f ( x | p) f ( p)
∑ f ( x | p) f ( p)

⎛ 1 ⎞ p x (1 − p )1− x ⎜
⎟
⎝ n −1 ⎠
=
⎛ 1 ⎞
∑ p x (1 − p)1− x ⎜ n − 1 ⎟
⎝
⎠
∀p
⎛ 1 ⎞ p⎜ ⎟
⎝ n −1 ⎠
=
⎛ 1 ⎞
∑ p ⎜ n −1 ⎟
⎝
⎠
∀p

=

[Since we observed x = 1]

r

p
(n − 1) / 2

[Note:

∑i = i =1

r (r + 1) and 2

1

n −1

∑ p = n ∑w =
∀p

w=1

Therefore, the probability that the proportion p of white cards is p0 is:

f ( p = p0 | x = 1) =

2 p0
,
n −1

p0 = {1/n, …, (n – 1)/n}.

1.5 f (λ ) = θ e −θλ , λ > 0

f(y)

f ( y | λ ) = λ e−λ y

∞

= ∫ f ( y | λ ) f (λ ) d λ
0

∞

= ∫ θ e−θλ λ e− λ y d λ
0

, y > 0.

n −1
]
2

∞

= θ ∫ λ e− λ ( y +θ ) d λ
0

∞

− λ ( y +θ )
∞e
⎡ λ e − λ ( y +θ ) ⎤ dλ =θ ⎢
+θ ∫
⎥
0 ( y +θ )
⎣ −( y + θ ) ⎦ 0
∞

⎡ e − λ ( y +θ ) ⎤
=θ ⎢
2⎥
⎣ −( y + θ ) ⎦ 0

=

f (λ | y )

=

θ
( y + θ )2

,

y > 0.

f ( y | λ ) f (λ ) f ( y)

θ e−θλ λ e − λ y
=
θ ( y + θ ) −2
= ( y + θ ) 2 λ e − λ ( y +θ )
=

( y + θ )2 λ 2−1e− λ ( y +θ )
,
Γ(2)

λ>0

⇒ (λ | y) ~ Gamma(2, (y + θ)).

1.6 f(λ|y) ∝ f(λ)f(y|λ)
= θ e −θλ λ e − λ y
= θλ e − λ ( y +θ )
∝ λ e − λ ( y +θ ) ,

λ > 0.

You could finish solving this problem by integrating this expression with respect to λ and then using the answer as a normalizing factor.

∫

∞

0

∞

λe

− λ ( y +θ )

dλ

− λ ( y +θ )
∞e
⎡ λ e − λ ( y +θ ) ⎤
= ⎢ dλ +∫
⎥
0 ( y +θ )
⎣ −( y + θ ) ⎦ 0
∞

⎡ e − λ ( y +θ ) ⎤
= ⎢
2⎥
⎣ −( y + θ ) ⎦ 0

= (y + θ)-2

Therefore, f(λ|y) = ( y + θ ) 2 λ e − λ ( y +θ )
=

( y + θ )2 λ 2−1e− λ ( y +θ )
,
Γ(2)

λ > 0.

⇒ (λ | y) ~ Gamma(2, (y + θ)).

Alternatively: f(λ|y) ∝ λ e − λ ( y +θ )
= λ 2 −1e − λ ( y +θ ) ,

λ > 0.

This is the kernel of the Gamma(2, (y + θ)) distribution. Therefore,
(λ | y) ~ Gamma(2, (y + θ))

[posterior distribution]

( y + θ )2 λ 2−1e− λ ( y +θ )
, λ>0 f(λ|y) =
Γ(2)

[posterior density function].