Isn't it just supposed to represent the instantaneous rate of death at age x? So why do we need to add the t?

Login

+4 votes

Best answer

So firstly we need to consider what mu represents, which is force of mortality at a specific age of time. Furthermore we need to consider surrounding information, which in the case of mortality models is the age of the individual (x) and the time after the age which we are interested in (t).

Thus the force of mortality becomes a function of both age (x) and time (t) and is represented by mu (x+t) and not just mu (x). To simplify this, I will put it in the context of an example, so you can see it's relevance.

Example We have a 40 year old male, and we are interested his probability of survival till age 60. Thus we set x at 40, and and t varies from 0 to 20. The answer would therefore be as follows:

$$e^{\int_{0}^{20} \mu(40+t) \space dt} $$

Alternatively we could have:

$$e^{\int_{40}^{60} \mu(x) \space dx} $$

I do hope his clarifies your issue.

- All categories
- BUS 1003H - Introduction to Financial Risk (31)
- BUS 2016H - Financial Mathematics (38)
- BUS 3018F - Models (40)
- BUS 3024S - Contingencies (22)
- BUS 4028F - Financial Economics (12)
- BUS 4027W - Actuarial Risk Management (1)
- BUS 2029H - Research Project (2)
- Mphil (1)
- Calculus and Pure Mathematics (3)
- Statistics (10)

...

Okay, I understand that. In question 1.1 of the tut pack, it reads as follows:

1.1 Suppose you know only that, for a certain life, μ40 = 0.00922 and μ41 = 0.00966. Use these figures to find a simple approximation to q40.

(Hint: how is p40 related to μ40+t, 0 ≤ t < 1?)

Do we need μ41 in this case? Or do we just use μ40 to calculate p40 and then put q40=1-p40 ? Or is μ40+s a linear function between μ40 and μ41 ?