Assumptions in converting between nominal/effective interest/discount

Question

$$ \left(1+\frac{i^{(m)}}{m}\right)^m = 1+i = \frac{1}{1-d} = \left(1-\frac{d^{(n)}}{n}\right)^{-n} $$ where $i=$effective interest rate, $d=$ effective discount rate, $i^{(m)}=$nominal interest rate , $d^{(m)}=$nominal discount rate

First, the assumption for all the equalities is that we are dealing with compound interest/discount.

Am I correct in that the first and third equalities only hold assuming that the nominal and effective rates are equivalent(ie, if I invested $1, then one year from now the accumulated value due to nominal and effective interest would be the same)?

Likewise, the second equality $ 1+i = \frac{1}{1-d}$ is something we can use to convert a given $i$ to an equivalent discount rate $d$. So implicit in this assumption is that $i, d$ are also equivalent rates?

So for example, if a question asks

If $i^{(8)} = 0.16$, calculate $d^{(1/2)}$

Then when setting up $$ \left(1+\frac{i^{(8)}}{8}\right)^8 = \left(1-\frac{d^{(1/2)}}{1/2}\right)^{-1/2} $$ and solving, implicit is the assumption that $d^{(1/2)}$ that the problem wants us to find is a rate equivalent to that of $i^{(8)}$?

Answer 1

Assume we have the present value $A_0$ and future value $A_1$. Using compounded interest we can write the equation $$ \big(1+\frac{i^{(m)}}{m}\big)^mA_0=(1+i)A_0=A_1 $$ but to discount, switch the exponent $m$ to $-m$ so that $$ \big(1+\frac{i^{(m)}}{m}\big)^{-m}A_1=(1-d)A_1=A_0. $$ Notice that I still use $i^{(m)}$ and for this construction to be valid it must hold that $$ d=1-\big(1+\frac{i^{(m)}}{m}\big)^{-m} $$ to which we will try to match a defined compounded discount. Define the compounded discount by $$ \big(1-\frac{d^{(m)}}{m}\big)^mA_1=(1-d)A_1=A_0. $$ From your question, you have interest to couple nominal interest $i^{(m)}$ with nominal discount $d^{(m)}$ which is, we have to solve $$ \big(1+\frac{i^{(m)}}{m}\big)^{-m}=\big(1-\frac{d^{(m)}}{m}\big)^m $$ which gives us $$ (1-\frac{d^{(m)}}{m})(1+\frac{i^{(m)}}{m})=1 $$ and we can easily see that $d^{(m)}\neq i^{(m)}$ since if $d^{(m)}=i^{(m)}$ then $(1-\frac{i^{(m)}}{m})(1+\frac{i^{(m)}}{m})=1-(\frac{i^{(m)}}{m})^2<1$.

How to find with discount rate if interest and discount have different compundings? Let interest be compunded $m_i$ times and discount be compunded $m_d$ times. Then $$ (1-\frac{d^{(m_d)}}{m_d})^{m_d}(1+\frac{i^{(m_i)}}{m_i})^{m_i}=1 $$ has the solution $$ d^{(m_d)}=m_d(1-\frac{1}{(1+\frac{i^{(m_i)}}{m_i})^{\frac{m_i}{m_d}}}). $$ We choose total time to be 2 years since your choice was $m_d=0.5$. In 2 years interest is compounded 16 times an we discount only one time(!) during 2 years. To validate this, assume you have 100 dollars that you keep for 2 years, that is 16 periods in which each period gives 2% interest, so the future value becomes 137.28 dollars. We actually have $m_i=16$ and $m_d=1$ so $$ d^{(1)}=1-\frac{1}{1.02^{16}}\approx 0.27155. $$ Then $137.28\cdot(1-0.27155)=100$.