Risk Analysis in Finance and Insurance, 2nd Edition

Insurance and Reinsurance Risks 219

Also

γ

H

t

=



l

x

− N

x

t



B

−1

0

ν

N

p

x+t−1

(N − t +1)

S

∗

t−1



exp{σ

2

}−1



×

$

S

∗

t−1

B

0

(1 + r)

N



Φ

!

ln



S

t−1

/K



+ σ

2

+(N − t +1)



δ + σ

2

/2



σ

√

N − t +1

"

−Φ

!

ln



S

t−1

/K



+(N −t +1)



δ + σ

2

/2



σ

√

N − t +1

"

+K



Φ

!

ln



S

t−1

/K



+(N − t +1)



δ −σ

2

/2



σ

√

N − t +1

"

−Φ

!

ln



S

t−1

/K



+ σ

2

+(N −t +1)



δ −σ

2

/2



σ

√

N − t +1

")

,

and

β

H

t

= V

π

t

− γ

H

t

S

∗

t

,t=1, 2,...,N.

Worked Example 7.5 A one-step model (7.2) with l

x

=2, N =1and

B

0

= 100,S

0

= 100,K= 100,r=0.01,μ=5,σ=0.5,p

x

(1) = 0.999996.

Solution The contingent claim is

H =

max{S

1

,K}I

{ω: T

1

>1}

B

1

+

max{S

1

,K}I

{ω: T

2

>1}

B

1

.

We have

δ ≈ 0.00995,ν=

100

101

,

and

V

π

0

≈ 2.383,γ

H

1

=1.245.

Note that, since Φ(∞)=1andΦ(−∞) = 0, we obtain

V

π

1

=(2− N

x

1

)

1

B

1

max{S

1

,K