1

For (x), you are given that $$\mu_x= \begin{cases} 0.01, &\text{if}& x<50\\ 0.02, &\text{if}& x>60\end{cases}$$

and $\mu_x$ is continuous ad linear on [50,60]. Calculate $_{10}p_{55}\\$.

Here is part of my solution:

Note that $_{10}p_{55}= e^{-\int_{55}^{65} \mu_y dy}$

Calculating the integral part:

$\int_{55}^{65} \mu_y dy = \int_{55}^{60} \mu_y dy + \int_{60}^{65} \mu_y dy = \int_{55}^{60} \mu_y dy + \int_{60}^{65} 0.02 dy $

I have no idea how to solve the first integral from 55 to 60 as its force of mortality is not given. Maybe by using the assumption that $\mu_x$ is continuous and linear on [50,60], I might be able to solve it?

Jonelle Yu
  • 397
  • 1
    In this case, continuous and linear means that $\mu_x = ax+b$ for some constants $a$ and $b$. (If it weren't continuous it could be represented by several such equations separated by jumps, for example). You know already that $\mu_x = 0.01$ when $x=50$ and $\mu_x = 0.02$ when $x=60$, so this gives you a pair of simultaneous equations to solve. Can you take it from there? – postmortes Feb 13 '19 at 06:42

1 Answers1

2

If $50\le x\le 60$, $$\mu_x= 0.01 + \tfrac{0.02-0.01}{60-50}(x-50)=0.001x-0.04.$$

J.G.
  • 115,835