I'm reading Marcel B. Finan's A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation for the Actuarial Exam FM/2 and have difficulty with two of his questions.
Problem 9.6 (page 73) is (paraphrased):
Given that $$1+\frac{i^{(n)}}n=\frac{1+\frac{i^{(4)}}4}{1+\frac{i^{(5)}}5}$$ find $n$.
(Note that $i^{(k)}$ denotes the nominal annual interest rate, which is convertible $k$ times per year for an effective annual interest rate $>i^{(k)}$ (when $k>1$). I assume that the three nominal annual interest rates in this question are the same, although the question doesn't so specify, since if they're not then there's obviously no way to figure out this problem.)
My only idea of how to solve this problem was that the right-hand side is the accumulation of $1+\frac{i^{(5)}}5$ over the next $\frac14-\frac15=\frac1{20}$ years, which would make $n=20$. But while that makes sense, I'm not sure it's correct and would welcome any feedback or ideas on how to solve this.
Problem 9.9 (page 74) is:
Eric deposits $X$ into a savings account at time 0, which pays interest at a nominal rate of $i$, compounded semiannually. Mike deposits $2X$ into a different savings account at time 0, which pays simple interest at an annual rate of $i$. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate $i$.
I've got $$X\left((1+\frac i2)^{16}-(1+\frac i2)^{15}\right)=2X(1+\frac i2)$$ and thus $$i(1+\frac i2)^{14}=4$$ but have no idea how to proceed from there. Any help would be much appreciatred.
Best Answer
Your reasoning is correct. I offer you an alternative, one you might find helpful in problems similar to, but more complicated than, the one given.
Let $i^{(k)}$ be a nominal annual interest rate convertible $k$ times per year and $i$ an effective annual interest rate. If $i^{(k)}$ and $i$ are equivalent, then $${\left(1 + \frac{i^{(k)}}{k}\right)}^k = 1 + i.\tag{1}$$
Using equation $(1)$, replace each occurrence of a nominal annual interest rate by the equivalent effective annual interest rate, resulting in one equation in one variable.
$$ \begin{align*} 1 + \frac{i^{(n)}}{n} = \frac{1 + \frac{i^{(4)}}{4}}{1 + \frac{i^{(5)}}{5}} & \implies {(1 + i)}^{\frac{1}{n}} = \frac{{(1 + i)}^{\frac{1}{4}}}{{(1 + i)}^{\frac{1}{5}}} \\ & \implies {(1 + i)}^{\frac{1}{n}} = {(1 + i)}^{\frac{1}{4} - \frac{1}{5}} \\ & \implies {(1 + i)}^{\frac{1}{n}} = {(1 + i)}^{\frac{1}{20}}. \end{align*} $$ Equating exponents, $\frac{1}{n} = \frac{1}{20} \implies n = 20$.
The value of Mike's investment at time $t$ is given by $$2 X (1 + ti),$$ where $t$ is measured in years. Therefore, the interest Mike earns in the last 6 months of the 8th year is given by $$2 X (1 + 8i) - 2 X (1 + 7.5i) = X i,$$ not $2 X (1 + \frac{1}{2} i)$, as covered by Ross Millikan in his answer.
As a final note, I would like to address the following:
You may assume $i^{(n)}$, $i^{(4)}$, and $i^{(5)}$ are the "same," unless stated otherwise, because the variable is the same throughout. You should not assume $i^{(4)}$ and $j^{(5)}$ are the "same" because the variable changes. Note the interest rates $i^{(n)}$, $i^{(4)}$, and $i^{(5)}$ are the "same" in the sense each is equivalent to the same effective interest rate $i$ according to equation $(1)$ above, but they are not the "same" in the sense $i^{(n)} = i^{(4})$ or $i^{(4)} = i^{(5)}$ for example.
I assume you are studying for the Financial Mathematics exam. Although it has been a number of years since I sat for the actuarial exams, my fondness for actuarial science hasn't waned. I wish you the best of luck in your actuarial studies!