# Question 6.3. Textbook

+1 vote
84 views

I managed to do (a), but I cannot figure out (b) or (c).

My thinking with (b) was to go back to first principles and group my Kx functions together, and then calculate a variance from there. However, since I was working with a term assurance, I had a pure endowment left behind.

With respect to C, I wasn't sure how to even start. It was covered very vaguely in the slides and I haven't crossed it in the book.

answered Aug 15, 2016 by (1,350 points)
selected Aug 15, 2016

b)

Let $$S$$ be the sum assured, from a). Firstly, define $$L_0$$ for each value of $$K_{[41]}$$:

$$L_0 = Sv^{K_{[41]} +1 } - 350 \ddot{a}_\overline{{K_{[41]} +1}|} \text{if} K_{[41]} \leqslant 2$$ $$= -350 \ddot{a}_\overline{{3}|} \text{if} K_{[41]} \geqslant 3$$.

Since we are assuming the equivalence principle holds,$$\mathbb{E}(L_0) = 0$$. So $$Var(L_0) = \mathbb{E}(L_0^2)$$.

Now find $$\mathbb{E}(L_0^2)$$ like you would for any ordinary discrete random variable:

$$Var(L_0) = \mathbb{E}(L_0^2) = \sum_{k=0}^{\infty} l_0^2 \Pr(K_{[41]} = k).$$ Note that this will be a sum of four terms, one for each of $$K_{[41]} = 0,1,2$$ and $$K_{[41]}\geqslant 3$$ respectively. $$l_0$$ are the values which $$L_0$$ takes on.

c)

We require $$\Pr(L_0) > 0$$ i.e of the loss being positive, and as can be seen from the definition of $$L_0$$ above, this occurs when $$K_{[41]} = 0,1,2$$. So the required probability is equivalent to the probability of the life dying in the first three years. If the life survives to the end of the three years, we would have made a profit, not a loss.

So the required probability is $$1 - {}_3 p_{[41]} = 0.0052$$.

Hope this helps :)