ACMA 425: Lecture 7
Mixed Distribution Assumption for Multiple Decrements

Example 1   For a double decrement table, you are given the double decrement probabilities  q_x⁽¹⁾ ,  q_x⁽²⁾  which satisfy

$$~_tq_x^{(2)}~=~(t)q_x^{(1)}~~,~~~0 \le t \le 1~~~~~~\mbox{(UDD)}$$

$$~_tq_x^{(2)}~=~q_x^{(2)}~~,~~~0 < t \le 1$$

(second decrement occurs all at the beginning of the year).
Find expressions for $~q_x^{\prime (1)}~~,~~q_x^{\prime (2)}~$ in terms of  q_x⁽¹⁾ ,  q_x⁽²⁾ .

Answer:

Example 2   A multiple decrement table has 2 decrements: deaths (d) and withdrawals(w).
Withdrawals occur once a year three-fourths of the way through the year of age. Deaths in the associated single decrement are uniformly distributed over each year of age.
You are given:

1.

$l_x^{(\tau)}=1000$

2.

$~q_x^{\prime (w)}=0.2$

3.

$l_{x+t}^{(\tau)}=720$

Calculate  d_x^(d)

Answer:

General Derivations

:
In a double decrement table, determine the following:

1.

 q_x⁽¹⁾ ,  q_x⁽²⁾  in terms of $~q_x^{\prime (1)}~,~~q_x^{\prime (2)}~$ assuming
$~q_x^{\prime (1)}~$ is UDD and
  q_x⁽²⁾  all occurs at $x+k~,~~~0\le k\le 1$

2.

$~q_x^{\prime (1)}~,~~q_x^{\prime (2)}~$ in terms of  q_x⁽¹⁾ ,  q_x⁽²⁾  assuming
 q_x⁽¹⁾  is UDD and
  q_x⁽²⁾  all occurs at $x+k~,~~~0\le k\le 1$

Solution:

1.

$$\int\limits^{1}_{0}~_tp_x^{(\tau)}~\mu_{x+t}^{(1)}~dt$$ q_x⁽¹⁾  = 

$$\int\limits^{1}_{0}~_tp_x^{\prime (1)}~_tp_x^{\prime (2)}~\mu_{x+t}^{(1)}~dt$$ = 

$$q_x^{\prime (1)}\int\limits^{1}_{0}~_tp_x^{(2)}~dt~~~~\mbox{under UDD on}~q_x^{\prime (1)}$$ = 

$$\mbox{But}~~~_tp_x^{\prime (2)} ~=~\left\{ \begin{array}{ll}1... ...^{\prime (2)}~&~~\mbox{ for}~~~ k \le t \le 1\end{array}\right.$$

$$q_x^{\prime (1)}~[k+(1-k)p_x^{\prime (2)}]$$ q_x⁽¹⁾  = 

$$q_x^{\prime (1)}~[k+(1-k)(1-q_x^{\prime (2)})]$$ = 

$$q_x^{(\tau)}~-~q_x^{(1)}$$ q_x⁽²⁾  = 

$$1~-~p_x^{\prime (1)}~p^{\prime (2)}~-~q_x^{(1)}$$ = 

$$1~-~(1-q_x^{\prime (1)})(1-q_x^{\prime (2)})~-~q_x^{\prime (1)}[k+(1-k)(1-q_x^{\prime (2)})]$$ = 

k=0

:

$$q_x^{\prime (1)}~(1~-~q_x^{\prime (2)})$$ q_x⁽¹⁾  = 

$$1~-~(1-q_x^{\prime (1)})(1-q_x^{\prime (2)})~-~q_x^{\prime (1)}(1-q_x^{\prime (2)})$$ q_x⁽²⁾  = 

$$1~-~(1-q_x^{\prime (2)})~=~ q_x^{\prime (2)}$$ = 

k=1

:

$$q_x^{\prime (1)}$$ q_x⁽¹⁾  = 

$$1~-~(1-q_x^{\prime (1)})(1-q_x^{\prime (2)})~-~q_x^{\prime (1)}$$ q_x⁽²⁾  = 

$$1~-~[1-q_x^{\prime (1)}~-~q_x^{\prime (2)}~+~q_x^{\prime (1)}q_x^{\prime (2)}]~-~q_x^{\prime (1)}$$ = 

$$q_x^{\prime (2)}~-~q_x^{\prime (1)}q_x^{\prime (2)}~=~ q_x^{\prime (2)}(1~-~q_x^{\prime (1)})$$ = 

2.

$$q_x^{\prime (1)}~=1~-~p_x^{\prime (1)}~=~1~-~\exp\left\{-\int\limits^{1}_{0}~\mu_{x+t}^{(1)}~dt\right\}$$

$$\mu_{x+t}^{(1)}~ \frac{\frac{\partial}{\partial t}~_tq_x^{(1)}}{~_tp_x^{(\tau)}~}~=~\frac{q_x^{(1)}}{1~-~(t)q_x^{(1)}~-~_tq_x^{(2)}}$$ = 

$$\left\{\begin{array}{ll}\frac{q_x^{(1)}}{1~-~(t)q_x^{(1)}}~~~&\mb... ...}{1~-~(t)q_x^{(1)}~-~q_x^{(2)}}~~~&\mbox{for}~~~k \le t \le 1\end{array}\right.$$ = 

$$\mbox{ie.}~~ q_x^{\prime (1)} ~ 1~-~\exp\left\{-\int\limits^{k}_{0}\frac{ q_x^{(1)}}{1-t q_x^{(1)}}~dt~-~\int\limits^{1}_{k}\frac{ q_x^{(1)}}{1-t q_x^{(1)}- q_x^{(2)}}~dt\right\}$$ = 

$$1~-~\exp\left\{\log(1-t q_x^{(1)})\vert^k_0~+~\log(1-t q_x^{(1)}- q_x^{(2)})\vert^1_k\right\}$$ = 

$$1~-~\exp\left\{\log(1-k q_x^{(1)})~+~\log(\frac{1- q_x^{(1)}- q_x^{(2)}}{1- k q_x^{(1)}- q_x^{(2)}})\right\}$$ = 

$$1~-~\frac{(1-k q_x^{(1)})(1- q_x^{(\tau)})}{1-k q_x^{(1)}- q_x^{(2)}}$$ = 

$$p_x^{(2)}~=~\frac{ p_x^{(\tau)}}{ p_x^{\prime (1)}}~=~\frac{1- q_x^{(\tau)}}{ p_x^{\prime (1)}}$$

$$p_x^{\prime (1)}~=~1- q_x^{\prime (1)}~=~\frac{(1-k q_x^{(1)})(1- q_x^{(\tau)})}{1-k q_x^{(1)}- q_x^{(2)}}$$

$$\mbox{ie.}~~ p_x^{\prime (2)}~ \frac{(1- q_x^{(\tau)})(1-k q_x^{(1)}- q_x^{(2)})}{(1-k q_x^{(1)})(1- q_x^{(\tau)})}$$ = 

$$\frac{1-k q_x^{(1)}- q_x^{(2)}}{1-k q_x^{(1)}}$$ = 

$$q_x^{\prime (2)} ~ 1- p_x^{\prime (2)}~=~1~-~\frac{1-k q_x^{(1)}- q_x^{(2)}}{1-k q_x^{(1)}}$$ = 

$$\frac{ q_x^{(2)}}{1-k q_x^{(1)}}$$ = 

k=0

:

$$q_x^{\prime (1)}~ 1~-~\frac{1- q_x^{(\tau)}}{1- q_x^{(2)}}$$ = 

$$\frac{ q_x^{(\tau)}- q_x^{(2)}}{1- q_x^{(2)}}~=~\frac{ q_x^{(1)}}{1- q_x^{(2)}}$$ = 

$$q_x^{\prime (2)} ~$$ =  q_x⁽²⁾

k=1

:

$$q_x^{\prime (1)}~ 1~-~\frac{(1- q_x^{(1)})(1- q_x^{(\tau)})}{1- q_x^{(1)}- q_x^{(2)}}~=~ q_x^{(1)}$$ = 

$$q_x^{\prime (2)} ~ \frac{ q_x^{(2)}}{1- q_x^{(1)}}$$ = 

Net Single Premiums in a Multiple Decrement Context

:
Assume m decrements. Then the net single premium for an insurance benefit of 1 payable at the moment of exit of an insured (x) is

$$\overline{A}_x~ \int\limits^{\infty}_{0}~v^t~_tp_x^{(\tau)}~\mu_{x+t}^{(\tau)}~dt$$ = 

$$\int\limits^{\infty}_{0}~v^t~_tp_x^{(\tau)}~\left(\sum\limits_{j=1}^{m}\mu_{x+t}^{(j)}\right)~dt$$ = 

$$\sum\limits_{j=1}^{m}\int\limits^{\infty}_{0}~v^t~_tp_x^{(\tau)}~\mu_{x+t}^{(j)}~dt$$ = 

$$\sum\limits_{j=1}^{m}\overline{A}_x^{(j)}$$ = 

Under the assumption of UDD on each multiple decrement,

$$\overline{A}_x^{(j)}~\dot{=}~\frac{i}{\delta} ~\sum\limits_{k=0}^{\infty}~v^{k+1}~_kp_x^{(\tau)}~q_{x+k}^{(j)}$$

Note: $\overline{A}_x^{(j)}~=~$ n.s.p. for an insurance benefit of 1 payable at the moment of exit of an insured (x) by decrement j.

General:

Typically, the insurance benefit varies depending on the cause of decrement.
Let B_x+t^(j) =  insurance benefit at x+t if cause of decrement is j.
Then,

$$\overline{A}_x~=~ \sum\limits_{j=1}^{m}~\int\limits^{\infty}_{0}~B_{x+t}^{(j)}~v^t~_tp_x^{(\tau)}~\mu_{x+t}^{(j)}~dt$$

Under the assumption of UDD on each multiple decrement and the use of the mid-point rule,

$$\int\limits^{\infty}_{0}~B_{x+t}^{(j)}~v^t~_tp_x^{(\tau)}~\mu... ...rac{1}{2}}^{(j)}~v^{k+\frac{1}{2}}~_kp_x^{(\tau)}~q_{x+k}^{(j)}$$

Example 3   A multiple decrement table has two causes of decrement:

1.

death by accident

2.

death other than accident

You are given:

1.

A fully continuous whole life insurance issued to (x) pays 2 if death results by accident and
1 if death results other than by accident.

2.

$\mu_{x+t}^{(1)}~=~\delta~$, the force of interest.

Calculate the net single premium for this insurance.

Answer: