# Combining Force of Mortality for F and M to find $l_x$

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?

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

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

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.)