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August 2, 2022 at 5:18 pm #22817
Actuarial Life Contingencies, Proofs of the Relationships Between Life Insurance APVs and Life Annuity APVs
Given that:
apv_(x)_n_death benefit of 1 payable instantly on death = integral (0,n) v^t * t_p_x * mu(x + t) dt
and
apv_(x)_n_life annuity of 1 per year payable continuously = integral (0,n) v^t * t_p_x dt, show that
apv_(x)_n_death benefit of 1 payable instantly on death = 1 – v^n * n_p_x + ln(v) * apv_(x)_n_life annuity of 1 per year payable continuously.
Method
Looking at the integral (0,n) v^t * t_p_x * mu(x + t) dt,
t_p_x = l(x + t)/l(x) and mu(x + t) = – l’(x + t)/l(x + t) and so t_p_x * mu(x + t) = – l’(x + t)/l(x), which is -t_p_x differentiated with respect to t, and so, integrating by parts, this integral becomes
[-t_p_x * v^t] (0,n) – integral (0,n) -ln(v) * v^t * t_p_x dt
= – v^n * n_p_x + 0_p_x * v^0 + ln(v) * integral (0,n) v^t * t_p_x dt
= 1 – v^n * n_p_x + ln(v) * integral (0,n) v^t * t_p_x dt
= 1 – v^n * n_p_x + ln(v) * apv_(x)_n_life annuity of 1 per year payable continuously.
Given that:
apv_(x)_n_death benefit of 1 payable at the end of the 1/m year of death = sum (k=0…m*n-1) v^(k/m + 1/m) * ((k/m)_p_x – (k/m + 1/m)_p_x)
and
apv_(x)_n_life annuity due of 1 per year payable m times per year = sum (k=0…m*n-1) (v^(k/m) * (k/m)_p_x)/m, show that
apv_(x)_n_death benefit of 1 payable at the end of the 1/m year of death = 1 – v^n * n_p_x + m * (v^(1/m) – 1) * apv_(x)_n_life annuity due of 1 per year payable m times per year.
Method
sum (k=0…m*n-1) v^(k/m + 1/m) * ((k/m)_p_x – (k/m + 1/m)_p_x)
= sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m)_p_x – sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m + 1/m)_p_x
= sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m)_p_x – sum (k=1…m*n) v^(k/m) * (k/m)_p_x
= 1 – 1 + sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m)_p_x – sum (k=1…m*n) v^(k/m) * (k/m)_p_x
= 1 – v^(0/m) * (0/m)_p_x + sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m)_p_x – sum (k=1…m*n) v^(k/m) * (k/m)_p_x
= 1 + sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m)_p_x – sum (k=0…m*n) v^(k/m) * (k/m)_p_x
= 1 – v^n * n_p_x + sum (k=0…m*n-1) v^(k/m + 1/m) * (k/m)_p_x – sum (k=0…m*n-1) v^(k/m) * (k/m)_p_x
= 1 – v^n * n_p_x + sum (k=0…m*n-1) v^(1/m) * v^(k/m) * (k/m)_p_x – sum (k=0…m*n-1) v^(k/m) * (k/m)_p_x
= 1 – v^n * n_p_x + v^(1/m) * sum (k=0…m*n-1) v^(k/m) * (k/m)_p_x – sum (k=0…m*n-1) v^(k/m) * (k/m)_p_x
= 1 – v^n * n_p_x + (v^(1/m) – 1) * sum (k=0…m*n-1) v^(k/m) * (k/m)_p_x
= 1 – v^n * n_p_x + m * (v^(1/m) – 1) * sum (k=0…m*n-1) (v^(k/m) * (k/m)_p_x)/m
= 1 – v^n * n_p_x + m * (v^(1/m) – 1) * apv_(x)_n_life annuity due of 1 per year payable m times per year.
Given that:
apv_(x)_n_death benefit of 1 payable at the end of the year of death = sum (k=0…n-1) v^(k + 1) * (k_p_x – (k + 1)_p_x)
and
apv_(x)_n_life annuity due of 1 per year payable once a year = sum (k=0…n-1) v^k * k_p_x, show that
apv_(x)_n_death benefit of 1 payable at the end of the year of death = 1 – v^n * n_p_x + (v – 1) * apv_(x)_n_life annuity due of 1 per year payable once a year.
Method
sum (k=0…n-1) v^(k + 1) * (k_p_x – (k + 1)_p_x)
= sum (k=0…n-1) v^(k + 1) * k_p_x – sum (k=0…n-1) v^(k + 1) * (k + 1)_p_x
= sum (k=0…n-1) v^(k + 1) * k_p_x – sum (k=1…n) v^k * k_p_x
= 1 – 1 + sum (k=0…n-1) v^(k + 1) * k_p_x – sum (k=1…n) v^k * k_p_x
= 1 – v^0 * 0_p_x + sum (k=0…n-1) v^(k + 1) * k_p_x – sum (k=1…n) v^k * k_p_x
= 1 + sum (k=0…n-1) v^(k + 1) * k_p_x – sum (k=0…n) v^k * k_p_x
= 1 – v^n * n_p_x + sum (k=0…n-1) v^(k + 1) * k_p_x – sum (k=0…n-1) v^k * k_p_x
= 1 – v^n * n_p_x + sum (k=0…n-1) v * v^k * k_p_x – sum (k=0…n-1) v^k * k_p_x
= 1 – v^n * n_p_x + v * sum (k=0…n-1) v^k * k_p_x – sum (k=0…n-1) v^k * k_p_x
= 1 – v^n * n_p_x + (v – 1) * sum (k=0…n-1) v^k * k_p_x
= 1 – v^n * n_p_x + (v – 1) * apv_(x)_n_life annuity due of 1 per year payable once a year.
Richard Purvey, proud of having lived at 15/5 Marytree House, Edinburgh, until the age of 12.
July 2022
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