So I have the definition of the force of mortality as $\mu_x=-\frac{1}{l_x}\frac{d l_x}{dx}$ and I am given two different forces of mortality, one for females as $\mu_x^f=0.05(1.2)^x$ and the other for males as $\mu_x^m=0.07(1.2)^x$.

I am then asked to calculate an expression for $l_x$. So I am assuming that I am to assume that there is an equal distribution of females and males.

Do I now want to proceed calculating $l_x$ as follows?

$$l_x=l_x^f+l_x^m=l_0^f\exp\left (\int_0^x \mu_t^f dt\right )+l_o^m\exp\left ( \int_0^x\mu_t^mdt\right )$$

  • 1
    $\begingroup$ Presumably the original "$\mu_t$" is supposed to be $\mu_x$ and this is supposed to be considered a function of $x$? $\endgroup$
    – whuber
    Apr 20, 2015 at 15:53

1 Answer 1


Now, we divide Numerator & Denominator of RHS of $\mu_x=-\frac{1}{l_x}\frac{d}{dx}(l_x)$ by $l_0$. Because we will need lower limit of integration to solve.

$$\mu_x=-\frac{1}{\frac{l_x}{l_0}}\frac{d}{dx}\Big(\frac{l_x}{l_0}\Big)$$ then,$$\mu_x=-\frac{d}{dx}\ln \frac{l_x}{l_0}$$ after integration and anti-log,$$\frac{l_x}{l_0}=\exp\Bigg(-\int\limits_0^x\mu_tdt\Bigg)$$ Note:-as upper limit has x, I changed integral w.r.t. to t.

as $l_x=l_0\times\frac{l_x}{l_0}$

$l_x$:total population living at age x,$l_x^f$:population of females living at age x, $l_x^m$:population of males living at age x. Hence $l_x=l_x^f+l_x^m$

And mortality rate of males & females considered seperately. $$\therefore~~~~~~~~~~~~~~~~~~~~~~~l_x=l_0^f\exp\left (-\int_0^x \mu_t^f dt\right )+l_o^m\exp\left (- \int_0^x\mu_t^mdt\right )$$

Note:- You don't need to assume that there is an equal distribution of females and males to get expression.(but if you want final answer and don't know $l_0^f~and~l_0^m$, then you can assume.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.