Finance Fundementals
Compound Interest
The Effective Rate of Interest
Compound interest is interest on interest. It is the result of reinvesting interest rather than paying it out, so that interest in the next period is then earned on the principle sum plus previously accumulated interest. The different types of compound interest rates include the effective, the nominal rate and the force of interest.
For compound interest calculations we will use the following notation: \[ \begin{aligned} P &= \text{the principal (the sum given from lender to borrower)}\\ I &= \text{the total interest}\\ A &= \text{the amount (the sum repaid from borrower to lender)}\\ i &= \text{the rate of compound interest per period (normally per annum)}\\ n &= \text{the time in periods (normally years)} \end{aligned} \] Suppose an individual invests a sum \(P\) at an effective compound interest rate of \(i\) per period. It follows that:
- After one period, the accumulated value is \(A_1=P(1+i)\).
- After two periods, the accumulated value is \(A_2=A_1(1+i)=P(1+i)(1+i)=P(1+i)^2\)
- \(\dots\)
- After \(n\) periods, the accumulated is \(A_n=A_{n-1}(1+i)=P(1+i)^{n-1}(1+i)=P(1+i)^n\)
More formally, for an effective rate \(i\) per period for \(n\) periods and principal payment \(P\), the accumulated value \(A\) is given by: \[ A=P(1+i)^n \] In the expression \(A=P(1+i)^n\), \(A\) is called the future value of \(P\) or the accumulated value of \(P\). The factor \((1+i)^n\) is called an accumulation factor at compound interest and the process of calculating \(A\) from \(P\) is called accumulation at compound interest.
We can also express \(P\) in terms of \(A\), \(i\) and \(n\). \[ P=\frac{A}{(1+i)^n} \] Here, \(P\) is called the present value of \(A\) or the discounted value of \(A\). The factor \((1+i)^{-n}\) is called a discount factor at compound interest and the process of calculating \(P\) from \(A\) is called discounting at compound interest.
The interest, \(I\), is calculated as the difference between \(A\) and \(P\): \[ \begin{aligned} I&=A-P\\ &=P(1+i)^n-P\\ &=P[(1+i)^n-1] \end{aligned} \] The interest rate \(i\) is usually expressed as a percentage per period; for example, 10% per annum. It is important that we measure the number of periods \(n\) in a way which corresponds with the rate of interest.
For example if we borrow at a rate of interest of 10% per annum for a duration of 6 months, the value of \(i\) would be 10% but the value of \(n\) would be \(\frac{1}{2}\) because the duration of the loan in years is half a year.
More generally, we note that \(\forall\, i \geq 0\) \[ \begin{aligned} 1+it&>(1+i)^t &&\text{for } 0<t<1\\ 1+it&=(1+i)^t &&\text{for } t=1\\ 1+it&<(1+i)^t &&\text{for } t>1 \end{aligned} \]
The Effective Rate of Discount
The difference between the amount \(A\) and the principal \(P\) has two interpretations:
- The interest \(I\) on \(P\) which when added to \(P\) gives \(A\).
- The discount \(D\) on \(A\) which when subtracted from \(A\) gives \(P\).
For compound discount calculations we will use the following notation: \[ \begin{aligned} P &= \text{the principal (the sum given from lender to borrower)}\\ I &= \text{the total interest}\\ A &= \text{the amount (the sum repaid from borrower to lender)}\\ d &= \text{the rate of compound discount per period (normally per annum)}\\ n &= \text{the time in periods (normally years)} \end{aligned} \] Suppose an individual discounts a sum \(A\) at an effective compound discount rate of \(d\) per period. It follows that:
- One period before, the discounted value is \(P_1=A(1-d)\)
- Two periods before, the discounted value is \(P_2=C_1(1-d)=A(1-d)(1-d)=A(1-d)^2\)
- \(\dots\)
- \(n\) periods before, the discounted value is is \(P_n=C_{n-1}(1-d)=A(1-d)^{n-1}(1-d)=A(1-d)^n\)
More formally, for an effective rate of discount \(d\) per period for \(n\) periods and accumulated payment \(A\), the principal value \(P\) is given by: \[ P=A(1-d)^n \] In the expression \(P=A(1-d)^n\), \(P\) is called the present value of \(A\) or the discounted value of \(A\). The factor \((1-d)^{n}\) is called a discount factor at compound discount and the process of calculating \(P\) from \(A\) is called discounting at compound discount.
We can also express \(C\) in terms of \(A\), \(i\) and \(n\). \[ A=\frac{P}{(1-d)^n} \] Here, \(A\) is called the future value of \(P\) or the accumulated value of \(P\). The factor \((1-d)^{-n}\) is called an accumulation factor at compound discount and the process of calculating \(A\) from \(P\) is called accumulation at compound discount.
The discount, \(D\), is calculated as the difference between \(A\) and \(P\): \[ \begin{aligned} D&=A-P\\ &=A-A(1-d)^{-n}\\ &=A[1-(1-d)^{-n}] \end{aligned} \] The discount rate \(d\) is usually expressed as a percentage per period; for example, 10% per annum. It is important that we measure the number of periods \(n\) in a way which corresponds with the rate of discount.
For example if we borrow at a rate of discount of 10% per annum for a duration of 6 months, the value of \(d\) would be 10% but the value of \(n\) would be \(\frac{1}{2}\) because the duration of the loan in years is half a year.
More generally, we note that \(\forall\, d=i \in (0,1)\) \[ 1-d>\frac{1}{1+i} \]
Let us now find a relationship between \(d\) and \(i\) using the equations we already know: \[ \begin{aligned} A&=P(1+i)^n\\ A&=P(1-d)^{-n} \end{aligned} \] Equating these two expressions gives \[ \begin{aligned} P(1+i)^n=P(1-d)^{-n}&&\iff&&(1+i)^n&=(1-d)^{-n}\\ &&\iff&&1+i&=(1-d)^{-1}\\ &&\iff&&i&=\frac{1}{1-d}-1\\ &&\iff&&i&=\frac{1-(1-d)}{1-d}\\ &&\iff&&i&=\frac{d}{1-d} \end{aligned} \]
Similarly, we have: \[ \begin{aligned} P(1+i)^n=P(1-d)^{-n}&&\iff&&(1+i)^n&=(1-d)^{-n}\\ &&\iff&&(1+i)^{-1}&=(1-d)\\ &&\iff&&d&=1-\frac{1}{1+i}\\ &&\iff&&d&=\frac{(1+i)-1}{1+i}\\ &&\iff&&d&=\frac{i}{1+i}\\ \end{aligned} \] Notice that, unlike simple interest, the conversion between the rate of discount and the rate of interest is independent of the amount of times that has passed.
The Nominal Rate of Interest
Previously, we have considered the effective rate of compound interest. The term effective implies that compounding occurs ones per period of time. In contrast, the term nominal implies that compounding occurs more that once per period.
We define \(i^{(m)}\) as the nominal rate of interest convertible \(m\)-thly. This means that over a 1 year period, the principle amount \(P\) will accumulate by an effective rate of \(\frac{i^{(m)}}{m}\) each \(\frac{1}{m}\) of a year for a total of \(m\) times.
For example, if interest compounds monthly at a rate of 24% per annum, then we would write \(i^{(12)}=24\%\). This means that the effective compound interest rate per month would be \(\frac{i^{(12)}}{12}=\frac{24\%}{12}=2\%\) and this effective rate would compound 12 times a year.
More generally, over an \(n\) year period, the principle amount \(P\) will accumulate by an effective rate of \(\frac{i^{(m)}}{m}\) each \({1}{m}\) of a year for a total of \(mn\) times. That is, the accumulated value is \[ A=P\left(1+\frac{i^{(m)}}{m}\right)^{mn} \] In the expression above, \(A\) is called the future value of \(P\) or the accumulated value of \(P\). The factor \(\left(1+\tfrac{i^{(m)}}{m}\right)^{mn}\) is called an accumulation factor at nominal interest and the process of calculating \(A\) from \(P\) is called accumulation at nominal interest.
We can also find the present value when the interest rate is expressed as a nominal rate convertible \(m\)-thly by rearranging the formula above \[ P=A\left(1+\frac{i^{(m)}}{m}\right)^{-mn} \] In the expression above, \(P\) is called the present value of \(A\) or the discounted value of \(P\). The factor \(\left(1+\tfrac{i^{(m)}}{m}\right)^{-mn}\) is called a discount factor at nominal interest and the process of calculating \(P\) from \(A\) is called discounting at nominal interest.
To find a relationship between the effective rate of interest \(i\) and the nominal rate of interest \(i^{(m)}\), we equate their respective accumulation factors over a 1 year period. This yields: \[ \begin{aligned} 1+i=\left(1+\frac{i^{(m)}}{m}\right)^{m} &&\iff&& (1+i)^{\frac{1}{m}}&=1+\frac{i^{(m)}}{m}\\ &&\iff&& (1+i)^{\frac{1}{m}}-1&=\frac{i^{(m)}}{m}\\ &&\iff&& i^{(m)}&=m\left[(1+i)^{\frac{1}{m}}-1\right]\\ \end{aligned} \] To find a relationship between the effective rate of discount \(d\) and the nominal rate of interest \(i^{(m)}\), we equate their respective accumulation factors over a 1 year period. This yields: \[ \begin{aligned} \frac{1}{1-d}=\left(1+\frac{i^{(m)}}{m}\right)^{m} &&\iff&& (1-d)^{-\frac{1}{m}}&=1+\frac{i^{(m)}}{m}\\ &&\iff&& (1-d)^{-\frac{1}{m}}-1&=\frac{i^{(m)}}{m}\\ &&\iff&& i^{(m)}&=m\left[(1-d)^{-\frac{1}{m}}-1\right]\\ \end{aligned} \]
The Nominal Rate of Discount
Previously, we have considered the effective rate of compound discount. The term effective implies that discounting occurs ones per period of time. In contrast, the term nominal implies that discounting occurs more that once per period.
We define \(d^{(m)}\) as the nominal rate of discount convertible \(m\)-thly. This means that over a 1 year period, the accumulated value \(A\) will discount by an effective rate of \(\frac{d^{(m)}}{m}\) each \(\frac{1}{m}\) of a year for a total of \(m\) times.
For example, if discounting compounds monthly at a rate of 24% per annum, then we would write \(d^{(12)}=24\%\). This means that the effective compound discount rate per month would be \(\frac{d^{(12)}}{12}=\frac{24\%}{12}=2\%\) and this effective rate would compound 12 times a year.
More generally, over an \(n\) year period, the accumulated amount \(A\) will discount by an effective rate of \(\frac{d^{(m)}}{m}\) each \({1}{m}\) of a year for a total of \(mn\) times. That is, the principal value is \[ P=A\left(1-\frac{d^{(m)}}{m}\right)^{mn} \] In the expression above, \(P\) is called the present value of \(A\) or the discounted value of \(A\). The factor \(\left(1-\frac{d^{(m)}}{m}\right)^{mn}\) is called a discount factor at nominal discount and the process of calculating \(P\) from \(A\) is called discounting at nominal discount.
We can also find the accumulated value when the discount rate is expressed as a nominal rate convertible \(m\)-thly by rearranging the formula above \[ A=P\left(1-\frac{d^{(m)}}{m}\right)^{-mn} \] In the expression above, \(A\) is called the future value of \(P\) or the accumulated value of \(P\). The factor \(\left(1-\frac{d^{(m)}}{m}\right)^{-mn}\) is called an accumulation factor at nominal discount and the process of calculating \(A\) from \(P\) is called accumulation at nominal discount.
To find a relationship between the effective rate of interest \(i\) and the nominal rate of discount \(d^{(m)}\), we equate their respective accumulation factors over a 1 year period. This yields: \[ \begin{aligned} 1+i=\left(1-\frac{d^{(m)}}{m}\right)^{-m} &&\iff&& (1+i)^{-\frac{1}{m}}&=1-\frac{d^{(m)}}{m}\\ &&\iff&& 1-(1+i)^{-\frac{1}{m}}&=\frac{d^{(m)}}{m}\\ &&\iff&& d^{(m)}&=m\left[1-(1+i)^{-\frac{1}{m}}\right]\\ \end{aligned} \] To find a relationship between the effective rate of discount \(d\) and the nominal rate of discount \(d^{(m)}\), we equate their respective discount factors over a 1 year period. This yields: \[ \begin{aligned} 1-d=\left(1-\frac{d^{(m)}}{m}\right)^{m} &&\iff&& (1-d)^{\frac{1}{m}}&=1-\frac{d^{(m)}}{m}\\ &&\iff&& 1-(1-d)^{\frac{1}{m}}&=\frac{d^{(m)}}{m}\\ &&\iff&& d^{(m)}&=m\left[1-(1-d)^{\frac{1}{m}}\right]\\ \end{aligned} \]
To find a relationship between the effective rate of interest \(i\) and the nominal rate of discount \(d^{(m)}\), we equate their respective accumulation factors over a 1 year period. This yields: \[ \begin{aligned} \left(1+\frac{i^{(m)}}{m}\right)^{m}=\left(1-\frac{d^{(m)}}{m}\right)^{-m} &&\iff&& \left(1+\frac{i^{(m)}}{m}\right)^{-1}&=1-\frac{d^{(m)}}{m}\\ &&\iff&& \frac{m}{m+i^{(m)}}&=\frac{m-d^{(m)}}{m}\\ &&\iff&& \frac{m^2}{m+i^{(m)}}&=m-d^{(m)}\\ &&\iff&& d^{(m)}&=m-\frac{m^2}{m+i^{(m)}}\\ &&\iff&& d^{(m)}&\frac{m^2+m\,i^{(m)}-m^2}{m+i^{(m)}}\\ &&\iff&& d^{(m)}&\frac{m\,i^{(m)}}{m+i^{(m)}}\\ \end{aligned} \]
The Force of Interest
Previously, we have considered the nominal rate of interest convertible \(m\)-thly. We defined \(i^{(m)}\) as the nominal rate of interest convertible \(m\)-thly. This means that over a 1 year period, the principle amount \(P\) will accumulate by an effective rate of \(\frac{i^{(m)}}{m}\) each \(\frac{1}{m}\) of a year for a total of \(m\) times.
We define the force of interest,\(\delta\), as the limit of \(i^{(m)}\) as \(m\to \infty\) while maintaining a constant accumulation factor over a one year period. That is \[ \delta = \lim_{m\to \infty} i^{(m)} \] where \(i^{(m)}=m\left[(1+i)^{\frac{1}{m}}-1\right]\) and \(1+i\) is kept constant. This means that compounding occurs continuously. This definition results in \[ \delta=\log(1+i) \]
To prove that \(\delta=\log(1+i)\) we must first make use of the Taylor expansion for the exponential function. Recall that \[ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \] It follows that \[ \begin{aligned} \delta & =\lim _{m \rightarrow \infty} i^{(m)} \\ & =\lim _{m \rightarrow \infty} m\left\{(1+i)^{1 / m}-1\right\} \\ & =\lim _{m \rightarrow \infty} m\left\{e^{D / m}-1\right\} \quad \text { where } D=\log (1+i) \\ & =\lim _{m \rightarrow \infty} m\left\{1+\frac{D}{m}+\frac{D^{2}}{2 m^{2}}+\frac{D^{3}}{6 m^{3}}+\ldots-1\right\} \\ & =\lim _{m \rightarrow \infty}\left\{D+\frac{D^{2}}{2 m}+\frac{D^{3}}{6 m^{2}}+\ldots\right\} \\ & =D \\ & =\log (1+i)\end{aligned} \]
Thus, over an \(n\) year period, the principle amount \(P\) will accumulate by a force of interest \(\delta\) resulting in an accumulated value of \[ A=Pe^{\delta n} \] In the expression above, \(A\) is called the future value of \(P\) or the accumulated value of \(P\). The factor \(e^{\delta n}\) is called an accumulation factor at constant force of interest and the process of calculating \(A\) from \(P\) is called accumulation at continuously compounding interest.
We can also find the present value when the interest rate is expressed as a constant force of interest by rearranging the formula above \[ P=Ae^{-\delta n} \] In the expression above, \(P\) is called the present value of \(A\) or the discounted value of \(P\). The factor \(e^{-\delta n}\) is called a discount factor at constant force of interest and the process of calculating \(P\) from \(A\) is called discounting at continuously compounding interest.
We can also express \(\delta\) in terms of \(d^{(m)}\). Recall that we define \(d^{(m)}\) as the nominal rate of discount convertible \(m\)-thly. This means that over a 1 year period, the accumulated value \(A\) will discount by an effective rate of \(\frac{d^{(m)}}{m}\) each \(\frac{1}{m}\) of a year for a total of \(m\) times.
We can define the force of interest,\(\delta\), as the limit of \(d^{(m)}\) as \(m\to \infty\) while maintaining a constant accumulation factor over a one year period. That is \[ \delta = \lim_{m\to \infty} d^{(m)} \] where \(d^{(m)}=m\left[1-(1+i)^{-\frac{1}{m}}\right]\) and \(1+i\) is kept constant. This means that compounding occurs continuously. This definition results also in \[ \delta=\log(1+i) \]
To prove that \(\delta=\log(1+i)\) we must again make use of the Taylor expansion for the exponential function. Recall that \[ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \] It follows that \[ \begin{aligned} \delta & =\lim _{m \rightarrow \infty} d^{(m)} \\ & =\lim _{m \rightarrow \infty} m\left[1-(1+i)^{-\frac{1}{m}}\right] \\ & =\lim _{m \rightarrow \infty} m\left[1-e^{D/m}\right] \quad \text { where } D=-\log (1+i) \\ & =\lim _{m \rightarrow \infty} m\left\{1-1-\frac{D}{m}-\frac{D^{2}}{2 m^{2}}-\frac{D^{3}}{6 m^{3}}-\ldots\right\} \\ & =\lim _{m \rightarrow \infty}\left\{-D-\frac{D^{2}}{2 m}-\frac{D^{3}}{6 m^{2}}-\ldots\right\} \\ & =-D \\ & =\log (1+i) \end{aligned} \]
To find a relationship between the force of interest \(\delta\) and the effective rate of discount \(d\), we equate their respective accumulation factors over a 1 year period. This yields: \[ e^{\delta}=\left(1-d\right)^{-1} \implies \delta = -\log \left(1-d\right) \text{ and } d=1-e^{-\delta} \] To find a relationship between the force of interest \(\delta\) and the nominal rate of interest \(i^{(m)}\), we equate their respective accumulation factors over a 1 year period. This yields: \[ e^{\delta}=\left(1+\frac{i^{(m)}}{m}\right)^{m} \implies \delta = m\log \left(1+\frac{i^{(m)}}{m}\right) \text{ and } i^{(m)}=m\left[e^{\delta/m}-1\right] \] To find a relationship between the force of interest \(\delta\) and the nominal rate of discount \(d^{(m)}\), we equate their respective accumulation factors over a 1 year period. This yields: \[ e^{\delta}=\left(1-\frac{d^{(m)}}{m}\right)^{-m} \implies \delta = -m\log \left(1-\frac{d^{(m)}}{m}\right) \text{ and } i^{(m)}=m\left[1-e^{-\delta/m}\right] \]
Equations of Value
\[ \onslide<2->{Y=X(1+i)^n \quad \textrm{or} \quad X=\cfrac{Y}{(1+i)^n}} \]