FM for Actuaries

Stochastic Interest Rates 309
and
s 5⌉ =1+(1.01) + ···+(1.01)(1.015)(1.02)(1.02) = 5.1474,
¨s 5⌉
=(1.01) + (1.01)(1.015) + ···+(1.01)(1.015)(1.02)(1.02)(1.03) = 5.2459.
We repeat the calculations for the other two scenarios and summarize the results in
the following table.
Scenario Probability a(5)
1
a(5)
a 5⌉
ä 5⌉
s 5⌉
¨s 5⌉
1 0.1 1.0986 0.9103 4.6855 4.7753 5.1474 5.2459
2 0.6 1.1762 0.8502 4.5630 4.7128 5.3670 5.5433
3 0.3 1.2400 0.8064 4.4466 4.6402 5.5141 5.7541
Therefore,
[ ] 1
E =(0.1 × 0.9103) + (0.6 × 0.8502) + (0.3 × 0.8064) = 0.8431,
a(5)
and
[ ] 1
Var
a(5)
= 0.1 × (0.9103 − 0.8431) 2 +0.6 × (0.8502 − 0.8431)
+0.3 × (0.8064 − 0.8431) 2
= 0.00089.
The mean, the variance and the standard deviation of other variables can be computed
similarly. The results are given in the following table.
a(5)
1
a(5)
a 5⌉
ä 5⌉
s 5⌉
¨s 5⌉
Mean 1.1876 0.8431 4.5403 4.6973 5.3892 5.5768
Variance 0.00170 0.00089 0.00505 0.00173 0.01082 0.02105
Std Dev 0.0412 0.0298 0.0711 0.0416 0.1040 0.1451