Revision as of 01:30, 20 January 2024 by Admin (Created page with "For a special fully discrete 25 -year endowment insurance on (44), you are given: (i) The death benefit is <math>(26-k)</math> for death in year <math>k</math>, for <math>k=1,2,3 \ldots 25</math> (ii) The endowment benefit in year 25 is 1 (iii) Net premiums are level (iv) <math>\quad q_{55}=0.15</math> (v) <math>\quad i=0.04</math> (vi) <math>{ }_{11} V</math>, the net premium policy value at the end of year 11 , is 5.00 (vii) <math>{ }_{24} \mathrm{~V}</math>, th...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
ABy Admin
Jan 20'24

Exercise

For a special fully discrete 25 -year endowment insurance on (44), you are given:

(i) The death benefit is [math](26-k)[/math] for death in year [math]k[/math], for [math]k=1,2,3 \ldots 25[/math]

(ii) The endowment benefit in year 25 is 1

(iii) Net premiums are level

(iv) [math]\quad q_{55}=0.15[/math]

(v) [math]\quad i=0.04[/math]

(vi) [math]{ }_{11} V[/math], the net premium policy value at the end of year 11 , is 5.00

(vii) [math]{ }_{24} \mathrm{~V}[/math], the net premium policy value at the end of year 24 , is 0.60

Calculate [math]{ }_{12} V[/math], the net premium policy value at end of year 12 .

  • 3.63
  • 3.74
  • 3.88
  • 3.98
  • 4.09

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 20'24

Answer: E

In the final year: [math]\left({ }_{24} V+P\right)(1+\mathrm{i})=b_{25}\left(q_{68}\right)+1\left(p_{68}\right)[/math]

Since [math]b_{25}=1[/math], this reduces to [math]\left({ }_{24} V+P\right)(1+i)=1 \Rightarrow(0.6+P)(1.04)=1 \Rightarrow P=0.36154[/math]

Looking back to the [math]12^{\text {th }}[/math] year: [math]\left({ }_{11} V+P\right)(1+i)=b_{12}\left(q_{55}\right)+{ }_{12} V\left(p_{55}\right)[/math]

[math]\Rightarrow(5.36154)(1.04)=14(0.15)+{ }_{12} V(0.85) \Rightarrow{ }_{12} V=4.089[/math]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

00