# during the first 4 years, interest is credited using a simple

I'm having trouble with the following problem from actuarial exam FM: During the first 4 years, interest is credited using a simple interest rate of $$5\%$$ a year. After 4 years, interest is credited at a force of interest: $$\delta_t = \frac{0.2}{1+0.2t}, t \geq 4$$ The following are numerically equal: (i) the current value at time $$t = 4$$ of payments of 1000 at time $$t =2$$ and 400 at time $$t = 7$$; and (ii) the present value at time $$t = 0$$ of a payment of $$X$$ at time $$t = 10$$.

I have two questions about the solution

1. The solution says the current value of (i) $$= 1000[1 + 2(.05)] + 400\frac{a(4)}{a(7)}$$. I was wondering why the first term isn't $$1000\frac{a(4)}{a(2)}$$.
2. I thought that the value of (ii) would be $$X\cdot \frac{1}{1+.05(4)} \cdot \frac{1.8}{1+.2(6)}$$, where the last fraction is the inverted $$a(t)$$ you get from the force of interest. But the solutions say something different. I'm wondering why my representation is not correct.

For the point (i)

1. The current value $$V'$$ at time $$t=4$$ of payments of $$1000$$ at time $$t=2$$ is the future value of $$1000$$ at the simple interest rate $$i=5\%$$ for $$2$$ years using the formula $$a(n)=a(0)(1+in)$$ $$V'=1000\,(1+2\times 5\%)=1000\times 1.1$$ and $$1.1=(1+2\times 5\%)=\frac{a(4)}{a(2)}$$.
2. The current value $$V''$$ at time $$t=4$$ of payments of $$400$$ at time $$t=7$$ is the present value of $$400$$ at the force of interest rate $$\delta_t$$ for $$3$$ years using the formula $$a(t)=a(t_0)\mathrm{e}^{\int_{t_0}^t\delta_\tau\mathrm d \tau}$$. Observing that $$\mathrm{e}^{\int_{t_0}^t\delta_\tau\mathrm d \tau}=\mathrm{e}^{\int_{t_0}^t\frac{0.2}{1+0.2\tau}\mathrm d \tau}=\mathrm{e}^{\left(\log(\tau+5)\big|_{t_0}^t\right)}=\frac{t+5}{t_0+5}$$, we have $$\frac{a(t)}{a(t_0)}=\frac{t+5}{t_0+5}$$ $$V''=400\times\frac{a(4)}{a(7)}=400\times \frac{4+5}{7+5}=400\times \frac{9}{12}$$
3. The current value at $$t=4$$ then is $$V=V'+V''=1400$$

For the point (ii)

The present value $$W$$ at time $$t=0$$ of a payment of $$X$$ at time $$t=10$$ is the discounted value $$W'=X\cdot\frac{a(4)}{a(10)}$$ at the interest force $$\delta_t$$ at time $$t=4$$, which is then discounted at the simple interest $$i$$ $$W=\frac{W'}{1+4i}=X\cdot\frac{1}{1+4i}\cdot\frac{a(4)}{a(10)}$$ that is $$W=X\cdot \frac{1}{1+4\times 0.05}\cdot \frac{4+5}{10+5}= X\cdot \frac{1}{1.2}\cdot \frac{9}{15}=X\cdot \frac{0.6}{1.2}=\frac{X}{2}$$

Find $$X$$

We know that $$V=W$$, so we have $$1400=\frac{X}{2}\quad\Longrightarrow\quad \boxed{X=2800}$$

• Thank you so much! For the first part, in the manual I don't think it's mentioned future value yet/I'll have to review that. So in general for simple interest the method of shifting is equivalent to just calculating the future value? :O The second part makes sense now! I forgot I have to discount by a(4)/a(10) vs just starting from a(0). – quietkid Jan 4 at 1:43