StatRisk solution homework assignment 1
Problem 2.1 Let X = 1 if the stock defaults and X = 0 otherwise. Let W = i if stock i is picked. a) Here
3

P (X = 1|W = i)P (W = i)

P (X = 1) = i=1 3

pi q i

= i=1 = 0.1 × 0.3 + 0.05 × 0.5 + 0.04 × 0.2 = 0.063.
b) Here
P (W = 1|X = 0) =

P (W = 1, X = 0)
P (X = 0|W = 1)P (W = 1)
0.9 × 0.3
=
=
= 0.2900.
P (X = 0)
1 − P (X = 1)
1 − 0.069

Problem 2.2
a)
∞

∞

ak k k=1

a

=

xk−1 dx

k=1 0 a ∞

a

k=1
∞

xk−1 dx

=
0

xk dx

=
0

k=0 a 1 dx 0 1−x
= − log(1 − a).

=

b)
∞

∞

pk = k=1 1 θ (1 − e−θ )k
= − log(1 − (1 − e−θ )) = = 1. kθ θ θ k=1

c)
∞

LX (t) =

e−tk k=1 ∞

=

(1 − e−θ )k kθ 1
(e−t (1 − e−θ ))k θ k=1 k 1
= − log 1 − e−t (1 − e−θ ) θ 1
= − log 1 + (e−θ − 1)e−t . θ 1

Problem 3.2
√
√
√
√
a) FY (y) = Φ( y) − Φ(− y). By symmetry, the density satisfies φ(− y) = φ( y). Therefore, taking the derivative
1
1
√
√
√ φ( y) + √ φ(− y)
2 y
2 y
1
√
√ φ( y) y 1 − 1 −(√y)2 y 2e
2π
y
1
1 y 2 −1 e− 2 .
1
1
2 2 Γ( 2 )

fY (y) =
=
=
=

This is the density of the χ2 (1) distribution.
b) We can assume that X ∼ N (0, 1) since the kurtosis for µ + σX is the same. From the above, Y = X 2 ∼ G( 1 , 1 ) so
2 2
E[X 4 ] = E[Y 2 ] =

1
1 Γ(2 + 2 )
3 1
= 4 · · = 3.
1
2 2
( 1 )2 Γ( 2 )
2

Since Var[X] = 1, we get
E[X 4 ]
= 3.
(Var[X])2

κ=

Problem 3.3 When we write f (x) ∼ h(x) we shall mean that f (x) = ch(x) for some constant
c.
a) Since
1

2

f (x) ∼ e− 2σ2 (x−µ) ,

g(x) ∼ xα−1 e−βx ,

we get log f (x) − log g(x) = c −

1
(x − µ)2 − (α − 1) log x + βx → −∞ as x → ∞.
2σ 2

Therefore, the G(α, β) distribution has the heaviest tail.
c) Here x b

1

f (x) ∼ xb−1 e−( a ) ,

2

g(x) ∼ x−1 e− 2σ2 (log x−µ) .

Then h(x) = log f (x) − log g(x) = c + (b − 1) log x −

x a b

+ log x +

The leading terms are
−

x a b

and
2

1
(log x − µ)2 .
2σ 2

1
(log x − µ)2 .
2σ 2

Consider k(x) =

x b a 1
(log x
2σ 2

− µ)2

=

2σ 2 xb . ab (log x − µ)2

Using L’Hospitals rule we get
2σ 2 bσ 2 bxb−1 xb
= lim b
.
x→∞ ab 2(log x − µ) 1 x→∞ a log x − µ x lim k(x) = lim

x→∞

Using L’Hospitals rule again gives b2 σ 2 b bσ 2 bxb−1 x = ∞.
= lim
1
x→∞ ab x→∞ ab x lim k(x) = lim

x→∞

Therefore, limx→∞ h(x) = −∞ so that the LN (µ, σ 2 ) distribution has the heaviest tail.
d) Here
1

β

2

f (x) ∼ x−1 e− 2σ2 (log x−µ) ,

g(x) ∼ x−(1+α) e− x .

Therefore def h(x) = log f (x)−log g(x) = c−log x−

1 β (log x−µ)2 +(1+α) log x− → −∞ as x → ∞.
2
2σ x Therefore, the IG(α, β) distribution has the heaviest tail.
Problem 3.4
a) We have in general
1−q =

α

θ θ + πq

1

⇒ πq = θ((1 − q)− α − 1).

Therefore,
1

1

1