Jan 16'24
Exercise
A life is subject to the following 3 -year select and ultimate table:
[math][x][/math] | [math]\ell_{[x]}[/math] | [math]\ell_{[x]+1}[/math] | [math]\ell_{[x]+2}[/math] | [math]\ell_{x+3}[/math] | [math]x+3[/math] |
---|---|---|---|---|---|
55 | 10,000 | 9,493 | 8,533 | 7,664 | 58 |
56 | 8,547 | 8,028 | 6,889 | 5,630 | 59 |
57 | 7,011 | 6,443 | 5,395 | 3,904 | 60 |
58 | 5,853 | 4,846 | 3,548 | 2,210 | 61 |
You are also given:
(i) [math]e_{60}=1[/math]
(ii) Deaths are uniformly distributed over each year of age
Calculate [math]\stackrel{\circ}{e}_{[58]+2}[/math].
- 1.5
- 1.6
- 1.7
- 1.8
- 1.9
Jan 16'24
Answer: B
[[math]]
\stackrel{\circ}{e}_{[58]+2}=e_{[58]+2}+0.5
[[/math]]
[[math]]e_{[58]+2}=p_{[58]+2}\left(1+e_{61}\right)=p_{[58]+2}\left[1+\frac{e_{60}}{p_{60}}-1\right][[/math]]
[[math]]=\frac{l_{61}}{l_{[58]+2}} \times \frac{e_{60}}{p_{60}}=\frac{2210}{3548} \times \frac{1}{(2210 / 3904)}=\frac{3904}{3549}=1.100338[[/math]]
[[math]]\stackrel{\circ}{e}_{[58]+2}=1.100338+0.5=1.6[[/math]]