FM for Actuaries

Stochastic Interest Rates 313
Therefore,
and
Furthermore,
and
E ( ) ( )
ä n⌉ =1+E an−1⌉ , (10.20)
E ( ) )
s n⌉ =1+E
(¨sn−1⌉ . (10.21)
Var ( ) ( )
ä n⌉ =Var an−1⌉ , (10.22)
Var ( ) )
s n⌉ =Var
(¨sn−1⌉ . (10.23)
Example 10.3: Consider a lognormal interest rate model with parameters µ =
0.04 and σ 2 1
=0.016. Find the mean and variance of a(5),
a(5) , ¨s 5⌉, a 5⌉ , s 5⌉ and
ä 5⌉ under this model.
Solution: We directly apply equations (10.6), (10.7), (10.9) and (10.10) to obtain
E[a(5)] = 1.27125,
Var [a(5)] = 0.13460,
[ ] 1
E = 0.85214,
a(5)
[ ] 1
Var = 0.06058.
a(5)
For ¨s 5⌉
, using expressions (10.11) to (10.13), we have
r s = e 0.04+ 1 2 (0.016) − 1=0.04917,
(
vs 2 = e 2(0.04)+0.016) (
e 0.016 − 1 ) =0.01775,
j s = 2r s + rs 2 + v2 s =0.11851,
¨s 5⌉ r s
= 5.78773,
¨s 5⌉ j s
= 7.08477.
Applying equations (10.14) and (10.15), we obtain
E (¨s )
5⌉
= 5.78773,
Var (¨s )
5⌉ = 1.26076.
For a 5⌉ , using expressions (10.16) and (10.17), we have
r a = e 0.04− 1 2 (0.016) − 1=0.03252,
j a = e 2(0.04−0.016) − 1=0.04917,
a 5⌉ r a
= 4.54697,
a 5⌉ j a
= 4.33942.