CT5- Force of mortality

[Bharti Singla]

Hi all
Please anyone help me with this query:
I am really not getting what is mux ? Is it a probability or a rate?
In ch1, page 6, it is written that mu(x+s) ds is the prob. of dying over very small interval of time and mu(x+s) is the annual rate of transfer between alive and dead at age x+s. What does it mean?
If it is the prob. of dying in next instant, then how it can be annual rate? What is rate actually?
And in the tables, page 81 gives values of mux. What these values actually represent? Like- at age 100, mux is given 0.421777.. what does it mean? 42.18% lives are dying in the next instant or in the next year?
Also, I see in the tables(AM 92):
For ages 17 to 24 and 85 to 120 - qx<mux
And for ages 25 to 84 - qx>mux
Why so?
I know I asked many qus., but once I got the definition of mux, I hope all the queries will be resolved.
If anyone can explain, it would be really a great help!

Thanks in advance.

 

[Robert Chadburn]

So mu(x) is the annual rate at which a person is dying at the exact age of x (so if x is an integer, we mean on the exact date at which the person has their x birthday).

This rate is changing continuously as age changes continuously. So, (p81), anyone alive at exact age 100 will be dying at an annual rate of 0.421777. Anyone alive at age 101 will be dying at an annual rate of 0.457202. Someone aged 100.5 would be dying at an annual rate somewhere between the two.

To illustrate the idea of a rate, let's assume the rate (mu) is constant over the year of age (100 to 101). If we do this then we should use the formula on page 14 of Chapter 3 to calculate this "average" mu. So we calculate mu = -ln(p100) which comes to 0.44 (which is somewhere between the actual rates at age 100 and 101, as we'd expect).

Now let's put 1000 people all aged between 100 and 101 in a house. If someone dies (or they become 101) they have to leave the house immediately and are immediately replaced by another person who is then aged between 100 and 101. So, this means there is always 1000 people in this house all aged between 100 and 101 all the time forever. Suppose we count how many people die in the house during a year? We would expect that to be 1000 x 0.44 = 440. The rate of dying would be forever 440 per year as there will always be 1000 people in the house and they will all be subject to the mortality rate of 0.44 pa per person.

Now to compare this with qx. So, now let's start with an empty house and put 1000 people, all aged exactly 100, into the house at the same time. No-one is replaced when they die, the house just gets emptier! Now we just count how many of these people die during the coming year. As before they will all be subject to the same rate of mortality during the year of age (as we have assumed it is constant), but as the number of survivors falls during the year the total deaths we'd expect to see would be fewer than 440. The expected number of deaths in this case would be 1000 x q100, as q100 is the probability of a person aged 100 dying before reaching age 101. So, the expected number of deaths would only be 1000 x 0.3555 = 356.

I'll stop there, and see if that sorts it out for you.
Robert

 

[Bharti Singla]

So mu(x) is the annual rate at which a person is dying at the exact age of x (so if x is an integer, we mean on the exact date at which the person has their x birthday).

This rate is changing continuously as age changes continuously. So, (p81), anyone alive at exact age 100 will be dying at an annual rate of 0.421777. Anyone alive at age 101 will be dying at an annual rate of 0.457202. Someone aged 100.5 would be dying at an annual rate somewhere between the two.

To illustrate the idea of a rate, let's assume the rate (mu) is constant over the year of age (100 to 101). If we do this then we should use the formula on page 14 of Chapter 3 to calculate this "average" mu. So we calculate mu = -ln(p100) which comes to 0.44 (which is somewhere between the actual rates at age 100 and 101, as we'd expect).

Now let's put 1000 people all aged between 100 and 101 in a house. If someone dies (or they become 101) they have to leave the house immediately and are immediately replaced by another person who is then aged between 100 and 101. So, this means there is always 1000 people in this house all aged between 100 and 101 all the time forever. Suppose we count how many people die in the house during a year? We would expect that to be 1000 x 0.44 = 440. The rate of dying would be forever 440 per year as there will always be 1000 people in the house and they will all be subject to the mortality rate of 0.44 pa per person.

Now to compare this with qx. So, now let's start with an empty house and put 1000 people, all aged exactly 100, into the house at the same time. No-one is replaced when they die, the house just gets emptier! Now we just count how many of these people die during the coming year. As before they will all be subject to the same rate of mortality during the year of age (as we have assumed it is constant), but as the number of survivors falls during the year the total deaths we'd expect to see would be fewer than 440. The expected number of deaths in this case would be 1000 x q100, as q100 is the probability of a person aged 100 dying before reaching age 101. So, the expected number of deaths would only be 1000 x 0.3555 = 356.

I'll stop there, and see if that sorts it out for you.
Robert

That's great sir! Thank you so much.