I'm currently taking a course about the insurance mathematics. Unfortunately I wasn't able to attend at the first lecture. I have this simple (I think) homework to do and I'm not sure if I'm doing it correctly. It goes as follows: Assume the de Moivre's law with $\omega=100$. Find ${}_{30} p_{20}$ and ${}_{30|10} q_{40}$. I did it like this:
${}_{t} p_{x}=Pr(T_{x}>t)=\frac{S(x+t)}{S(x)}$, where $S(x)$ is a survival function.
Assuming de Moivre's law we have $S(t)=1-\frac{t}{\omega}$, where $0<t\leq\omega$.
From that we have ${}_{t} p_{x}=\frac{\omega -x -t}{\omega - x}$ so in our case:
$ {}_{30} p_{20} = Pr(T_{20}>30)=\frac{S(20+30)}{S(20)}=\frac{\omega - 20-30}{\omega -20}=\frac{100-50}{100-20}=\frac{5}{8}$.
For the second task we know that:
${}_{t} q_{x}=Pr(T_{x}\leq t)$ and
$ {}_{t|s} q_{x} = {}_{t+s} q_{x} - {}_{s} q_{x}$ so:
$ {}_{30|10} q_{40} = {}_{40} q_{40} - {}_{10} q_{40} = Pr(T_{40}\leq 40) - Pr(T_{10}\leq 40)= $
$= 1 -Pr(T_{40}> 40) - 1 + Pr(T_{10}> 40) =- \frac{S(80)}{S(40)} +\frac{S(50)}{S(40)}=$
$=-\frac{100-80}{100-40} + \frac{100-50}{100-40}=-\frac{1}{3}+\frac{5}{6}=\frac{1}{2}$
Is that correct? I have this feeling that I make some naive and incorrect transitions.
Your answer to the first part is correct.
For the second part, the symbol ${}_{t|s} q_x$ is the probability that the future lifetime of $(x)$ is in $(t, t+s]$ rather than $(s, t+s]$. So the correct formula is $${}_{t|s} q_x = {}_{t+s} q_x - {}_t q_x = \Pr[t < T_{40} \le t+s];$$ that is to say, it is the probability that $(x)$ dies before age $t+s+x$, minus the probability that $(x)$ dies before age $t+x$. We can also write it as $${}_{t|s} q_x = ({}_t p_x )({}_s q_{x+t}) = ({}_t p_x)(1 - {}_s p_{x+t}).$$ This gives me $${}_{30|10} q_{40} = \frac{1}{6}.$$