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I am working on the following problem:

A borrower has a mortgage that calls for level annual payments of 1 at the end of each year for 20 years. At the time of the seventh regular payment an additional payment is made equal to the amount of principal that according to the original amortization schedule would have been repaid by the eighth regular payment. If payments of 1 continue to be made at the end of the eighth and succeeding years until the mortgage is fully repaid, show that the amount saved in the interest payments over the full term of the mortgage is $1 - v^{13}$.

If one pays $v^{13}$ at time $t=7$, then it will accumulate to $v^{13}(1 + i)^{13} = 1$ at time $t = 20$. Hence by paying $v^{13}$ at $t = 7$, one would not have to pay the final payment of 1. The interest portion of the final payment would have been $i\cdot{}a_1 = 1 - v$, where $a_n$ is an n-year annuity immediate.

It seems to me that one is actually saving $1 - v$ in interest. Could I get some clarification on why it is actually $1 - v^{13}$.

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Start decomposing this problem.

Level Annuity Case

Consider some level annual payments of $1$ over $20$ years at interest rate $i$. For each payment $X_t = 1$ for $t = 1, 2, \dots, 20$, we define $I_t$ to be the interest portion of $X_t$ and $P_t$ to be the principal portion of $X_t$ ($\boxed{X_t = I_t + P_t}$).

Let $B_t$ be the balance of the annuity at time $t$. Recall that $I_t = iB_{t-1}$ ($t = 1, 2, \dots, 20$) and that we can use the "retrospective method" (i.e., present value of future payments) to find $B_t$ for all $t$.

Claim. $B_{t-1} = \dfrac{1-v^{20-(t-1)}}{i}$.

At time $t = 1$, we have to consider $B_0$, the present value of future payments at time $0$. This is simply $a_{\overline{20}|}$.

For $t = 2$, we consider the present value of future payments at time $1$. There are $19$ payments of $1$, simply with present value $a_{\overline{19}|}$.

For an arbitrary integer $t \in [1, 20]$, we have a present value of $B_{t-1} = a_{\overline{20-(t-1)}|}$.

This gives $$\boxed{I_t = iB_{t-1} = 1-v^{20-(t-1)}\text{, }} t \in [1, 20]\text{ an integer.}$$ Thus, the total interest payments paid throughout the duration of the annuity is $$\sum\limits_{t=1}^{20}I_t = \sum\limits_{t=1}^{20}[1-v^{20-(t-1)}] = 20 - (v^{20}+v^{19}+\cdots + v^{2}+v) = \boxed{20 - a_{\overline{20}|}\text{.}}\tag{1}$$

Level Annuity with $7^{\text{th}}$ Payment Changed

What is different in this case? Using $\prime$ to denote the values under this change, we have $X_7^{\prime} = X_7 + P_8$. Notice that $$X_t = I_t + P_t = 1 - v^{20-(t-1)}+P_t = 1$$ thus implying that $P_t = v^{20-(t-1)}$. Therefore, $P_8 = v^{20-(8-1)} = v^{13}$ so that $X^{\prime}_7 = 1 + v^{13}$.

Notice $$X^{\prime}_t = \begin{cases} 1, & t = 1, 2, \dots, 6 \\ 1+v^{13}, & t = 7 \\ 1, & t = 8, \dots, n \end{cases}$$ for some $n$.

Consider the present value of the payments of $X^{\prime}_t$ at interest rate $i$. This would be equal to $a_{\overline{n}|} + v^{13}v^{7}$, which is equal to the present value of the loan originally ($a_{\overline{20}|}$). Hence $$a_{\overline{n}|} + v^{20} = a_{\overline{20}|}\Longleftrightarrow \dfrac{1-v^{n}}{i} + v^{20}=\dfrac{1-v^{20}}{i} \Longleftrightarrow v^{n} = (1+i)v^{20} = v^{19} \implies n = 19\text{.}$$ So we have $$X^{\prime}_t = \begin{cases} 1, & t = 1, 2, \dots, 6 \\ 1+v^{13}, & t = 7 \\ 1, & t = 8, \dots, 19\text{.} \end{cases}$$ Use the idea that $B_{t-1}^{\prime}$ ($t \in [1, 19]$) is the present value of future payments. For $B_7^{\prime}, B_8^{\prime}, \dots, B_{18}^{\prime}$, this is easy: it is the remaining present values of $1$, i.e., $B_7^{\prime} = a_{\overline{12}|}$, $B_8^{\prime} = a_{\overline{11}|}$, $\dots$, $B_{17}^{\prime} = a_{\overline{2}|}$, $B_{18}^{\prime} = a_{\overline{1}|}$. Hence $B_k^{\prime} = a_{\overline{19-k}|}$, $k \in [7, 18]$ an integer.

For $B_6^{\prime}$, we have the $12$ payments of $1$ plus the payment of $1 + v^{13}$. Considering the present value only the part of the payment of $1$, we have $13$ payments of $1$. In addition, we have $v^{13}$ discounted one year, so we have $B_6^{\prime} = a_{\overline{13}|} + v^{14}$. Go back one year for $B_5^{\prime}$: we have another payment of $1$ to take into account and have to discount $v^{14}$ by another year, so we have $B_5^{\prime} = a_{\overline{14}|} + v^{15}$. In general, $B_{k}^{\prime} = a_{\overline{19-k}|}+v^{20-k}$, $k \in [0, 6]$ an integer.

Hence, $$B_k^{\prime} = \begin{cases} a_{\overline{19-k}|}+v^{20-k}, & k \in [0, 6] \\ a_{\overline{19-k}|}, & k \in [7, 18]\text{,} \end{cases}$$ or $$B_{t-1}^{\prime} = \begin{cases} a_{\overline{19-(t-1)}|}+v^{20-(t-1)}, & t \in [1, 7] \\ a_{\overline{19-(t-1)}|}, & t \in [8, 19]\text{.} \end{cases}$$ Hence $$I_t^{\prime} = iB_{t-1}^{\prime} = \begin{cases} 1-v^{19-(t-1)}+iv^{20-(t-1)}, & t \in [1, 7] \\ 1-v^{19-(t-1)}, & t \in [8, 19]\text{,} \end{cases}$$ and $$\begin{align}\sum\limits_{i=1}^{19}I_t^{\prime}&= \sum\limits_{i=1}^{19}\left[1-v^{19-(t-1)}\right]+i\sum\limits_{i=1}^{7}v^{20-(t-1)} \\ &= 19-(v^{19}+v^{18}+\cdots + v)+i(v^{20}+\cdots + v^{14})\text{.} \\ &= 19-a_{\overline{19}|}+iv^{13}(v+\cdots + v^{7}) \\ &= 19-a_{\overline{19}|}+iv^{13}a_{\overline{7}|} \\ &= 19-a_{\overline{19}|}+v^{13}(1-v^{7}) \\ &= \boxed{19-a_{\overline{19}|} + v^{13} - v^{20}\text{.}}\tag{2}\end{align}$$ The series $a_{\overline{20}|} = v+v^2 + \cdots + v^{20}$, hence $$\begin{align} \text{Interest Savings} &= (1) - (2) \\ &= \sum\limits_{i=1}^{20}I_t - \sum\limits_{i=1}^{19}I_t^{\prime} \\ &= 20 - a_{\overline{20}|} - (19-a_{\overline{19}|} + v^{13} - v^{20}) \\ &= 20 - (v+v^2 + \cdots + v^{20})-19+(v+v^2 + \cdots + v^{19})- v^{13}+v^{20} \\ &= \boxed{1 - v^{13}\text{.}} \end{align}$$

How to Do This for an Actuarial Exam?

The key relationships to know is that for a level annuity of $n$ payments of $1$ with interest rate $i$, $X_t = I_t + P_t$, $I_t = iB_{t-1}$, and $P_t = v^{n-(t-1)}$. You could also memorize the sum of the $I_t$, but it is easily derived, as I showed above. The formulas for $X_t$ and $I_t$ are true in general, but the formula for $P_t$ above is only true for the level annuity case.

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