Decreasing term assurance - find the interest rate

Question

A formula for decreasing term assurance is: $$ V_0 = V_k \times \left(\frac{1-(1+r)^{-n}}{1-(1+r)^{-(n-k)}}\right) $$ where $r$ is the annual rate of interest, $V_0$ is the sum assured at time $t=0$ and $V_k$ is the sum assured at time $t=k$. The formula is normally used to obtain the original sum assured (ie $V_0)$ when the sum assured $V_k$ at time $t=k$ is known and when $r$ is known.

However, if $V_0, V_k, n, k$ are all known, is it possible to determine the annual rate of interest $r$?

As $0< r <1$ initially I thought I could use the negative binomial theorem to rearrange the original expression: $$ (1 + r)^{-n} = 1 - nr + \frac{1}{2}n(n+1)r^2 - \frac{1}{6}n(n+1)(n+2)r^3 + $$ If I let $L = \frac{V_0}{V_k}$ and if I let $m = n-k$, then the original expression is: \begin{eqnarray*} L &=& \frac{1-(1+r)^{-n}}{1-(1+r)^{-m}} \iff L - L(1+r)^{-m} &=& 1-(1+r)^{-n} \end{eqnarray*}

After substituting in expressions for $(1+r)^{-n}$ and $(1+r)^{-m}$ and then some simple tidying up I obtain the expression: $$ L\left(mr - \frac{1}{2}m(m+1)r^2 + \cdots \right) = \left(nr - \frac{1}{2}n(n+1)r^2 + \cdots\right) $$ or equivalently $$ \left(nr - \frac{1}{2}n(n+1)r^2 + \cdots\right) - L\left(mr - \frac{1}{2}m(m+1)r^2 + \cdots \right) = 0 $$ By grouping terms in the same power of $r$: $$ r(n-Lm) - \frac{1}{2}r^2\Big(n(n+1) -Lm(m+1)\Big) + \frac{1}{6}r^3\Big( n(n+1)(n+2) - Lm(m+1)(m+2)\Big) - \cdots = 0 $$ As $r>0$, dividing through by $r$ yields $$ (n-Lm) - \frac{1}{2}r\Big(n(n+1) -Lm(m+1)\Big) + \frac{1}{6}r^2\Big( n(n+1)(n+2) - Lm(m+1)(m+2)\Big) -\cdots = 0 $$

I thought as an approximate answer I could ignore powers higher than $r^2$ and simply consider the above as a quadratic in $r$. However, this doesn't yield a real value fot $r$ as the discriminant is less than $0$.

My questions are:

1. Have I made a mistake in the above, or is the method a dead-end?

2. Is there an alternative mathematical technique that yields better results?

3. Obviously $r$ can be obtained by trial and error when dealing with a few cases. However, this approach is not practical when dealing with millions of cases. Is there perhaps an algorithmic method to obtain $r$, and if so what?

Answer 1

1. I don't see any mistakes. The accuracy of your method depends highly on the coefficient of $r^3$ being small compared to the coefficient of $r^2$. You should write your equation out as a cubic polynomial and examine the coefficients.

2. As John Bentin suggests, if you let $x = \dfrac{1}{1-r}$, then you get $P(x) = x^n - \dfrac{V_0}{V_k}x^{n-k}+\left(\dfrac{V_0}{V_k} - 1 \right) = 0$.

3. The problem is that, I suppose that $r$ us going to be close to $0$. Which means that $x$ is going to be close to $1$. Notice that $P(1)=0$. It's going to be hard to solve $P(1-r)=0$ when $P(1)=0$. The fix might be to compute the polynomial $Q(x) = \dfrac{P(x)}{x-1}$, using synthetic division, and then find the solution to $Q(x) = 1$ and proceed from there. This is not a simple problem.

Answer 2

What you have to solve is a polynomial equation of the form $$x^n+ax^{n-k}+b=0,$$where $x=1/(1+r)$, and $a$ and $b$ are known constants. If $n\geqslant5$, then there is generally no closed-form solution to this equation. Even if $n=3$ or $n=4$, the closed-form solutions are messy. However, such an equation is, for any $n$, efficiently solved to high accuracy by a numerical method such as Newton$-\!$Raphson, and this is what I would recommend.

Added: While the above substitution suffices for a polynomial analysis of the problem, it is more convenient to use the substitution $x=(1+r)^{-n}$ to analyse the convergence of Newton$-\!$Raphson. Thus the problem is to find the zero of $$f(x):=x-\frac{V_0}{V_k}x^{1-k/n}+\frac{V_0}{V_k}-1.$$We have$$f'(x)=1-\left(1-\frac kn\right)\frac{V_0}{V_k}x^{-k/n}.$$Observe that, as $x$ increases from $0$, $f'(x)$ increases continually (from $-\infty$ asymptotically toward $1$), and $f(x)$ will drop from $V_k/V_0-1$, at $x=0$, to hit the first zero, say at $x=\alpha<1$, while $f'(x)$ is still negative and increasing. This ensures that Newton$-\!$Raphson converges, with successive approximations monotonically increasing on the left towards the desired root $\alpha$, provided that our starting value $x_0$ is in the interval $(0\;\pmb,\;\alpha)$. It is easily seen that $$x_0:=\left(1-\frac{V_k}{V_0}\right)^{\!\!n/(n-k)}$$ is a suitable such value in this interval. (Doubtless there are better starting points, but perhaps not so simple to spot and write down.) Note that convergence to the required zero will fail if we start outside this interval.