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.

Thank you very much for your help in advance!


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 $$ 1400=\frac{X}{2}\quad\Longrightarrow\quad \boxed{X=2800} $$

  • $\begingroup$ 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). $\endgroup$ – quietkid Jan 4 at 1:43

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