Instead, suppose that beginning at time 1, the insurer faces one of three interest rate scenarios given by
begin{eqnarray*}
textbf{i} = left{
begin{array}{cr}
i_1 = 4 & textrm{with probability 0.25} \
i_2 = 5 & textrm{with probability 0.50} \
i_3 = 6 & textrm{with probability 0.25} \
end{array} right.
end{eqnarray*}
For a fixed interest scenario (i_s), we have that the random future loss variable at policy duration (k) is
begin{eqnarray*}
_k L(i_s) = v_s^{K+1-k} – P_{40} ddot{a}_{overline{K+1-k|}} = left(1+frac{P_{40}}{1-v_s}right) v_s^{K+1-k} – frac{P_{40}}{1-v_s} ,
end{eqnarray*}
where (v_s=1/(1+i_s)). This has conditional (on the interest scenario) mean
begin{eqnarray*}
textrm{E}(_k L(i_s)|i_s) = A_{40+k}@i_s – P_{40} ddot{a}_{40+k}@i_s,
end{eqnarray*} and variance
begin{eqnarray*}
textrm{Var}(_k L(i_s)|i_s) = left(1+frac{P_{40}}{1-v_s}right)^2 left( ^2A_{40+k}@i_s – A_{40+k}^2 @i_sright).
end{eqnarray*}
The following table summarizes the calculations at duration (k=1) based on the Illustrative Life Table (recall (P_{40} = 0.125021) was set at contract initiation and so is constant across scenarios).
Conditional | Conditional | |||||
---|---|---|---|---|---|---|
(i) | prob | (A_{41}) | (ddot{a}_{41}) | Exp Loss | (^2 A_{41}) | Var Loss |
0.04 | 0.25 | 0.282 | 18.658 | 0.049 | 0.105 | 0.042 |
0.05 | 0.50 | 0.216 | 16.461 | 0.010 | 0.072 | 0.034 |
0.06 | 0.25 | 0.169 | 14.686 | -0.015 | 0.052 |
For example, at (i=0.04), the conditional mean loss is (0.282 – 0.012502 (18.658) = 0.049). The conditional variance is
begin{eqnarray*}
left(1+frac{0.012502}{1-1/(1.04)}right)^2 left( 0.105 – (282)^2right) = 0.042.
end{eqnarray*}
Computing quantities over interest scenarios, we may write the expected loss as
begin{eqnarray*}
_1 V = textrm{E}(_1 L(textbf{i})) &=& sum_{j=1}^3 textrm{E}(_1 L(i_s)|i_s) times Pr(textbf{i}=i_s) \
&=& (0.25) (0.049) + (0.50) (0.010) + (0.25) (-0.015) = 0.014 .
end{eqnarray*}
In this case, the expected fund is
begin{eqnarray*}
textrm{E}~FUND_1 &=& N P_{40}(1.05) – N q_{40} – N (1-q_{40}) ~_1 V \
&=& N left{ (0.012502)(1.05) – 0.0027812 – (1-0.0027812) (0.014) right}\
&=& -0.0033 N,
end{eqnarray*}
a sad situation for the insurer.