Accumulating Deposits from a Polynomial Force of Interest

# Force of interest third degree polynomial


# Force of interest third degree polynomial












Just struggling with a question here, any help would be appreciated.

For its deposits, a bank offers a force of interest per annum over a given year, which equals 4.99% at the beginning of the year, 5.33% after three months, 5.75% after six months, and 6.53% after nine months. Calculate the accumulated amount at the end of the year of a deposit of $78,000 at the beginning of the year, assuming the force of interest per annum is a third degree polynomial function of time over the year.

Any helpful tips or advice would be great, I'm not looking for the answer just a method that I can remember.

## Answer by Alex C (score 2)

https://quant.stackexchange.com/a/33427

If the force of interest (or instantaneous interest rate) is $r(t)$ then the accumulation function is $a(t)=e^{\int_0^t r(t)dt}$.

In this case $r(t)$ is a third degree polynomial in $t$ which is not given explicitly, but is uniquely defined by the fact that it passes through 4 points: (0.0499,0),(0.0533,0.25),(0.0575,0.50),(0.0653,0.75). Using the Newton Interpolation Formula or the Lagrange Interpolation Formula you would explicitly compute what $r(t)$ is.

After integrating this polynomial you get a 4th degree polynomial, which you evaluate at t=1 and take the exponential to find $a(1)$

The final answer is $78000*(a(1)-a(0))=78000*(a(1)-1)$

Needless to say it is a LOT of computation, especially when you have to do it with pencil and paper under time pressure of an exam. [If you have access to a computer with Mathematica or similar math program then it becomes doable.]

## Answer by Bjørn Kjos-Hanssen (score 2)

https://quant.stackexchange.com/a/36608

To correct two points in @AlexC's otherwise great answer:

- The given points are of the form $(t, r(t))$: $$\begin{eqnarray*}\left(0,\frac{499}{10^4}\right),& \left(\frac14,\frac{533}{10^4}\right),\\ \left(\frac12, \frac{575}{10^4}\right),& \left(\frac34,\frac{653}{10^4}\right).\end{eqnarray*}$$ The cubic polynomial $r(t)=at^3+bt^2+ct+d$ thus is not all that bad: it has $d=499\cdot 10^{-4}$, and $$\begin{eqnarray*}a \cdot 2^{-6}+b\cdot 2^{-4}+c\cdot 2^{-2}+d=533\cdot 10^{-4}\\ a \cdot 2^{-3}+b\cdot 2^{-2}+c\cdot 2^{-1}+d=575\cdot 10^{-4}\\ a \cdot 3^32^{-6}+b\cdot 3^22^{-4}+c\cdot 3\cdot2^{-2}+d=653\cdot 10^{-4}\end{eqnarray*}$$

- The accumulated amount is $A(1)=A(0)\cdot e^{\int_0^1 r(t)\,dt}$ where $A(0)=\$78,000$.

Here are some details on finding $a$, $b$, and $c$:

$$\begin{eqnarray*}a \cdot 2^{-2}+b+c\cdot 2^{2}+d\cdot 2^4=533\cdot 5^{-4}\\ a \cdot 2^{1}+b\cdot 2^{2}+c\cdot 2^{3}+d\cdot 2^4=575\cdot 5^{-4}\\ a \cdot 3^32^{-2}+b\cdot 3^2+c\cdot 3\cdot2^{2}+d\cdot 2^4=653\cdot 5^{-4}\end{eqnarray*}$$

$$\begin{eqnarray*}\frac14a+b+4c+16d=533\cdot 5^{-4}\\ 2a+4b+8c+16d=575\cdot 5^{-4}\\ \frac{27}4a+9b+12c+16d=653\cdot 5^{-4}\end{eqnarray*}$$

Note $16d=499\cdot 5^{-4}$ so this becomes $$\begin{eqnarray*}\frac14a+b+4c=34\cdot 5^{-4}\\ 2a+4b+8c=76\cdot 5^{-4}\\ \frac{27}4a+9b+12c=154\cdot 5^{-4}\end{eqnarray*}$$

$$\begin{bmatrix}\frac14&1&4\\ 2&4&8\\ \frac{27}4&9&12\end{bmatrix}^{-1}=\begin{bmatrix}2&-2&\frac23\\ -\frac52&2&-\frac12\\ \frac34&-\frac38&\frac1{12}\end{bmatrix}$$ so $$\begin{bmatrix}a\\ b\\ c\end{bmatrix}=5^{-4}\begin{bmatrix}2&-2&\frac23\\ -\frac52&2&-\frac12\\ \frac34&-\frac38&\frac1{12}\end{bmatrix}\begin{bmatrix}34\\ 76\\ 154\end{bmatrix}$$