I haven't been able to get any traction on this question. I'm relatively new to life contingencies and I have a feeling I'm just not seeing a relationship between the survival function and the force of mortality.
A gently nudge in the right direction or a complete solution would be appreciated.
Question from SOA 2000 Fall #25 Exam, for reference.
The key for part (a) is quite simple if you study the hint. What condition (ii) tells us is that the force of mortality for (26) is the same for either group, because if $t > 1$, you get $\mu_{25+t}^N = \mu_{25+t}^M$. Put another way, given that an individual in each group survives to age 26, their forces of mortality are exactly the same at every time point moving forward. Therefore, the curtate expected future lifetime for (26) in either case is also the same: $$e_{26}^N = e_{26}^M.$$ Then we easily conclude using the hint that $$\frac{e_{25}^N}{e_{25}^M} = \frac{p_{25}^N (1 + e_{26}^N)}{p_{25}^M (1 + e_{26}^M)} = \frac{p_{25}^N}{p_{25}^M}.$$
For part (b), we are given condition (iii), so it is natural to use the result from part (a). To this end, recall $$\mu_x(t) = -\frac{d}{dt}\left[\log {}_t p_x\right] \iff {}_t p_x = \exp \left( -\int_{u=0}^t \mu_x(u) \, du \right). $$ It follows that $$p_{25}^N = \exp\left(-\int_{u=0}^1 \mu_{25+u}^M + \frac{1-u}{10} \, du \right) = p_{25}^M \exp\left(-\int_{u=0}^1 \frac{1-u}{10} \, du \right) = p_{25}^M e^{-1/20}.$$ The rest is straightforward: $$e_{25}^N = e_{25}^M \frac{p_{25}^N}{p_{25}^M} = 10 e^{-1/20} \approx 9.51229.$$ Note this is consistent with condition (ii), which stipulates that the force of mortality for (25) in group $N$ in the first year is strictly greater than the corresponding force of mortality of (25) in group $M$; thus the expected future lifetime must be less in group $N$ compared to $M$.