CS2 Assignment X2 (2019 Vers Tb)

[sadikxwhfnc]

Hi,

I am attempting the CS2 Assignment X2 of the 2019 Textbook and I do not understand the solution to Qn 2.2.

The question is as follows:
In a certain population, the force of mortality at age x is given by:

70 < x <= 75 : 0.02
75 < x <= 80 : 0.04
80 < x <= 85 : 0.07

Find the probability that a life now aged exactly 73 will die between exact age 79 and exact age 82.

The solution is as follows:

Required probability = 6P73 * 3Q79

6P73 = 2P73 * 4P75 = exp(-2 * 0.02) * exp(-4 * 0.04) = exp (-0.2)

I have two questions with regard to the solution:

1. Why is 2P73 * 4P75 = exp(-2 * 0.02) * exp(-4 * 0.04) used instead of 3P73 * 3P76 = exp(-3 * 0.02) * exp (-3 * 0.04)??
This is because I thought that the force of mortality for age 73, 74 , and 75 is 0.02??

2. What is wrong with the method I used to solve as follow:
I find the probability of death at age = 79 , 80 , 81 , and 82.
P(Age = 79) = exp(-3*0.02) * exp(-3*0.04) * (1 - exp(-0.04)) = 0.032751
P(Age = 80) = exp(-3*0.02) * exp(-4*0.04) * (1 - exp(-0.04))
P(Age = 81) = exp(-3*0.02) * exp(-5*0.04) * (1 - exp(-0.07))
P(Age = 82) = exp(-3*0.02) * exp(-5*0.04) * * exp(-1 * 0.07) * (1 - exp(-0.07))

Many thanks for your help in advance.

 

[Andrew Martin]

Hello

1. Why is 2P73 * 4P75 = exp(-2 * 0.02) * exp(-4 * 0.04) used instead of 3P73 * 3P76 = exp(-3 * 0.02) * exp (-3 * 0.04)??
This is because I thought that the force of mortality for age 73, 74 , and 75 is 0.02??

I believe you're only thinking of integer ages here. According to the given information, the force of mortality for someone aged 75.0000001 is 0.04 for example.

2. What is wrong with the method I used to solve as follow:
I find the probability of death at age = 79 , 80 , 81 , and 82.
P(Age = 79) = exp(-3*0.02) * exp(-3*0.04) * (1 - exp(-0.04)) = 0.032751
P(Age = 80) = exp(-3*0.02) * exp(-4*0.04) * (1 - exp(-0.04))
P(Age = 81) = exp(-3*0.02) * exp(-5*0.04) * (1 - exp(-0.07))
P(Age = 82) = exp(-3*0.02) * exp(-5*0.04) * * exp(-1 * 0.07) * (1 - exp(-0.07))

In general this approach is fine. You need to be careful with the calculation above from your first question. However we can indeed write the probability by considering different ages of death as follows:

6p73 * 3q79 = 6p73 q79 + 7p73 q80 + 8p73 q81
= exp(-2 * 0.02) * exp(-4 * 0.04) * (1 - exp(-0.04))
+ exp(-2 * 0.02) * exp(-5 * 0.04) * (1 - exp(-0.07))
+ exp(-2 * 0.02) * exp(-5 * 0.04) * exp(-0.07) * (1 - exp(-0.07))
= 0.13487

Again be careful with the integer vs. continuous ages here. The final probability in your calculation relates to dying between 82 and 83, which we don't need as we're told they die between 79 and 82.

Hope this helps

Andy