Financial Math FM/Time Value of Money

Learning objectives

The Candidate will understand and be able to perform calculations relating to present value, current value, and accumulated value.

Learning outcomes

The Candidate will be able to:

• Define and recognize the definitions of the following terms: interest rate (rate of interest), simple interest, compound interest, accumulation function, future value, current value, present value, net present value, discount factor, discount rate (rate of discount), convertible m-thly, nominal rate, effective rate, inflation and real rate of interest, force of interest, equation of value.
• Given any three of interest rate, period of time, present value, current value, and future value, calculate the remaining item using simple or compound interest. Solve time value of money equations involving variable force of interest.
• Given any one of the effective interest rate, the nominal interest rate convertible m-thly, the effective discount rate, the nominal discount rate convertible m-thly, or the force of interest, calculate any of the other items.
• Write the equation of value given a set of cash flows and an interest rate.

Interest

Definition. (Interest) Interest is the compensation that a borrower of an asset (or capital) pays a lender of capital for its use.

Remark.

• That is, interest is the extra thing (compensation) paid to lender, in addition to the capital.

Example. Suppose a bank lends Amy $100, and one month later Amy needs to pay back $110 to the bank. Then,

• $\$100$ is the capital;
• $\$110-\$100=\$10$ is the interest;
• Amy is a borrower;
• the bank is a lender.

Exercise. Fisher A lends fisher B a fishing rod for a week, for fisher B to catch fishes. In this week, 10% of fishes caught by fisher B (decimal numbers obtained are rounded up) need to be given to fisher A in return. For simplicity, suppose all fishes are identical.

1 Who is the lender?

	Fisher A.
	Fisher B.

2 Suppose fisher B catches 10 fishes in that week. Calculate the interest.

	One fish.
	9 fishes.
	10 fishes.
	11 fishes.
	Cannot be determined.

3 Suppose fisher B can borrow $1000 from a bank for renting a fishing rod for a week instead, and the fisher needs to pay back $1050 to the bank after a week. Also suppose each fish can be sold for $100. What is the minimum number of fishes caught in the week such that the interest paid by borrowing money from the bank is lower?

	One fish.
	9 fishes.
	10 fishes.
	11 fishes.
	None of the above.

Measurement of interest

Terminologies

In the following, we will introduce some terminologies used in the measurement of interest.

Definition. (Principal) The principal is the initial amount of money invested.

Definition. (Accumulated value) The accumulated value (or future value) at time $t$ is the total amount received at time $t$.

Definition. (Interest (alternative definition)) The interest earned during a period of investment is the difference between the accumulated value and the principal.

Remark.

• This alternative definition is equivalent to the above definition of interest.

Example. Amy invests $100 into a fund that pays $200 one year later. Then,

• the principal is $\$100$;
• the accumulated value (after one year) is $\$200$;
• the interest (earned during this year) is $\$200-\$100=\$100$.

Exercise.

1 Suppose the fund pays $\$100x$ two years later instead, and the interest earned after two years is $50. Calculate $x$.

	0.5
	0.67
	1.5
	50
	150

2 Continue from previous question. Suppose Amy invests $\$y$ instead. Express $x$ in terms of $y$.

	$x={\frac {100-y}{50}}$
	$x={\frac {y-25}{50}}$
	$x={\frac {250-y}{100}}$
	$x={\frac {50+y}{100}}$
	$x={\frac {50-y}{100}}$

Definition. (Measurement period) The measurement period is the unit in which time is measured.

Remark.

• The measurement period is often years.

Definition. (Accumulation function) The accumulation function, denoted by $a(t)$, is the function that gives the ratio of the accumulated value at time $t$ to the principal.

Remark.

• $a(0)=1$, since the accumulated value at time 0 equals the principal, and thus the ratio is 1.

Definition. (Amount function) The amount function denoted by $A(t)$, is the function that gives the accumulated value at time $t$ of principal of nonnegative number $k$.

Remark.

• $A(0)=k$, since the accumulated value at time 0 equals the principal, which is $k$ by definition.
• Also, $A(t)=ka(t)$, since $a(t){\overset {\text{ def }}{=}}{\frac {A(t)}{k}}$.

We denote the interest earned during the $n$th period ($n$ is a positive integer) from the date of investment (i.e. from the beginning of $n$th period to the end of $n$th period by $I_{n}$.

By definition, $I_{n}=A(n)-A(n-1)$, in which

• $A(n)$ is the accumulated value at the end of $n$th period, and
• $A(n-1)$ is the accumulated value at the start of $n$th period.

Example. Define $a(t)=2^{t}$ for each time $t$.

• This is a valid accumulation function, since $a(0)=2^{0}=1$.
• Also, $a(1.5)=2^{1.5}\approx 2.83$, and
• the interest earned with principal of one from $t=0$ to $t=1$ is $a(1)-a(0)=2^{1}-2^{0}=1$, in which $a(1)$ is the accumulated value at time 1, and $a(0)$ is the principal.

Exercise.

1 Suppose the principal is $100. Calculate $A(3.5)$.

	1.13
	1.23
	1131.37
	1200
	1225

2 Calculate the interest earned from $t=3$ to $t=3.5$ with the amount of principal in the previous question.

	325
	331.37
	565.69
	800
	1131.37

3 Which of the following is (are) valid amount functions?

	$t\mapsto 2t$
	$t\mapsto 2^{t}$
	$t\mapsto t^{2}$
	$t\mapsto {\frac {1}{1+t}}$
	$t\mapsto e^{t}$

Effective interest rate

Definition. (Effective interest rate) The effective interest rate, denoted by $i$, is the ratio of the amount of interest earned during the period to the principal.

Remark.

• It follows that the effective interest rate during the $n$th period from the investment date, denoted by $i_{n}$, is

$$i_{n}={\frac {A(n)-A(n-1)}{A(n-1)}}$$

in which $A(n)-A(n-1)$ is sometimes denoted by $I_{n}$, meaning the amount of interest.

• Unless otherwise stated, rates are expressed as annual rates..

Example. $i_{n}={\frac {a(n)-a(n-1)}{a(n-1)}}$.

Proof. Suppose the principal is $k$. Then, $$i_{n}{\overset {\text{ def }}{=}}{\frac {A(n)-A(n-1)}{A(n-1)}}={\frac {ka(n)-ka(n-1)}{ka(n-1)}}={\frac {a(n)-a(n-1)}{a(n-1)}},$$ as desired.

$\Box$

Example. $A(n)=A(0)(1+i_{1})\dotsb (1+i_{n})$ for each nonnegative integer $n$.

Proof. For each nonnegative integer $n$, $$A(0)(1+i_{1})\dotsb (1+i_{n})=A(0)\cdot {\frac {A(1)}{A(0)}}\cdot \dotsb \cdot {\frac {A(n)}{A(n-1)}}=A(n),$$ as desired.

$\Box$

Exercise. Define $i_{n}=2i_{n-1}$ for each positive integer $n$, and $i_{1}=1\%$. Suppose $A(0)=100$.

1 Calculate $A(4)$.

	107.14
	115.71
	116.88
	134.225

2 Express $A(t)$ in terms of $t$.

	$A(t)=100\prod _{k=0}^{t}(1+2^{k-1}i_{1})$.
	$A(t)=100\prod _{k=0}^{t}(1+2^{k}i_{1})$.
	$A(t)=100\prod _{k=0}^{t}(1+2^{k+1}i_{1})$.
	$A(t)=100\prod _{k=0}^{t}(1+2ki_{1})$.
	$A(t)=100(1+i_{1})^{t}$.

Simple interest

For simple interest, under the simple interest rate $i$, the interest earned during each period is calculated according to the principal (and so is constant), i.e. the interest earned is $iA(0)$, i.e. $A(n)-A(n-1)=iA(0)$ for each positive integer $n$. So, $A(n)=A(0)+{\big (}A(1)-A(0){\big )}+\dotsb +{\big (}A(n)-A(n-1){\big )}=A(0)+inA(0)=A(0)(1+in).$ Since $A(n)=A(0)a(n)$, $a(n)=1+in$ for each nonnegative integer $n$. Intuitively, we may expect that the same form of accumulation function also holds for other nonnegative numbers.

Definition. (Simple interest) The accumulation of interest according to the accumulation function $$a(t)=1+it,\quad t\geq 0$$ is the simple interest, in which $i$ is the simple interest rate.

Remark.

• Since $i_{n}={\frac {a(n)-a(n-1)}{a(n-1)}}={\frac {i}{1+(n-1)i}}$ for simple interest, $i_{n}$ does not equal the simple interest rate in general.

Example. Ivan invests $10000 into a bank account which pays simple interest with an annual rate of 5%. The balance in Ivan's account after two years is $\$1000(1+2(5\%))=\$11000$.

Exercise.

1 Suppose the balance after two years is $10000 instead. How much does Ivan invest initially?

	9000
	9070.29
	9090.91
	9523.81
	11111.11

2 Calculate the effective interest rate during the second year.

	0.045
	0.048
	0.05
	0.1

Example. For simple interest, $$a(s+t)=a(s)+a(t)-1,\quad s{\text{ and }}t\geq 0.$$

Proof. $$a(s+t)=1+(s+t)i=\underbrace {1+si} _{=a(s)}+\underbrace {1+ti} _{=a(t)}-1,$$ as desired.

$\Box$

Compound interest

For compound interest, the interest earned for each period is calculated according to the accumulated value at the beginning of that period.

To be more precise, with principal of $k$ and compound interest rate $i$, at the end of first year, the interest earned is $ki$, and thus the accumulated value is $k+ki=k(1+i)$.

Thus, at the end of second year, the interest earned is $k(1+i)i$, and so the accumulated value is $k(1+i)+k(1+i)i=k(1+i)^{2}$

Using the same argument, at the end of $n$th year, the interest received is $ki(1+i)^{n-1}$, and the accumulated value is $k(1+i)^{n}$. We obtain the accumulation function with nonnegative integer $n$ as input here, namely $a(t)=(1+i)^{n}$.

Intuitively, we may expect that the same form of accumulation function also holds for other nonnegative numbers. This motivates the definition of compound interest.

Definition. (Compound interest) The accumulation of interest according to the accumulation function $$a(t)=(1+i)^{t},\quad t\geq 0$$ is the compound interest, in which $i$ is the compound interest rate.

Example. Ivan invests $10000 into a bank account which pays compound interest with an annual rate of 5%. Then, the balance in Ivan's account after two years is $\$10000(1.05)^{2}=\$11025$.

Exercise.

1 After two years, the bank account pays simple interest with the same rate for three years, and then pays compound interest with rate 10%. Calculate the balance after seven years.

	$13781.25
	$13946.63
	$15341.29
	$16875.42
	$17755.87

2 With simple interest rate $i$, the balance after two years is 12000. Calculate the difference between the balance after two years with compound interest rate $i$ and that with simple interest rate $i$.

	0
	0.01
	25
	91.1
	100

Example. For compound interest rate $i$, the effective interest rate during the $n$th period $i_{n}=i$.

Proof. $$i_{n}={\frac {(1+i)^{n}-(1+i)^{n-1}}{(1+i)^{n-1}}}={\frac {(1+i)^{n-1}{\big (}(1+i)-1{\big )}}{(1+i)^{n-1}}}=i.$$

$\Box$

Remark.

• Because of this nice result, from now on, every interest (rate) is assumed to be compound interest (rate) unless stated otherwise.

Effective discount rate

The effective interest rate was defined as a measure of interest paid at the end of the period. However, there are also discount rates, denoted by $d$, which is a measure of the interest paid at the beginning of the period.

Example.

• If Amy borrows $100 from a bank for a year at an effective interest rate of 5%, then the bank will give Amy $100 at the beginning of the year, and at the end of the year, Amy will repay the bank $100 plus the interest of $5, a total of $105.
• On the other hand, if Amy is charged by an effective discount rate of 5%, then Amy will need to pay the interest of $5 at the beginning of the year, so the bank will only give Amy $\$100-\$5=\$95$ at the beginning, and Amy will repay the bank $100 at the end (interest is paid already).

We can see from this example that the effective interest rate is a percentage of the principal, while the effective discount rate is a percentage of the balance at the end of the year. Thus, we can define effective discount rate more precisely as follows:

Definition. (Effective discount rate) The effective discount rate, denoted by $d$, is the ratio of the amount of interest earned during the period to the amount invested at the end of the period.

Remark.

• It follows that the effective discount rate during the $n$th period, denoted by $d_{n}$, is

$$d_{n}={\frac {A(n)-A(n-1)}{A(n)}}.$$

Example. In the above example, the effective discount rate is 5% since ${\frac {\$100-\$95}{\$100}}=5\%$.

Example. $d_{n}={\frac {a(n)-a(n-1)}{a(n)}}$.

Proof. $$d_{n}{\overset {\text{ def }}{=}}{\frac {A(n)-A(n-1)}{A(n)}}={\frac {ka(n)-ka(n-1)}{ka(n)}}={\frac {a(n)-a(n-1)}{a(n)}}.$$

$\Box$

Example. $A(0)=A(n)(1-d_{n})\dotsb (1-d_{1})$.

Proof. For each positive integer $n$, $$A(n)(1-d_{n})\dotsb (1-d_{1})=A(n)\cdot {\frac {A(n-1)}{A(n)}}\cdot \dotsb \cdot {\frac {A(0)}{A(1)}}=A(0).$$

$\Box$

Example. Given that $a(t)=e^{t}$, then $$d_{3}={\frac {e^{3}-e^{2}}{e^{3}}}\approx 0.632.$$ In contrast, $$i_{3}={\frac {e^{3}-e^{2}}{e^{2}}}\approx 1.718.$$

Exercise.

Select all correct expression(s) for $d_{k}-i_{k+1}$ in terms of $k$, for each positive integer $k$.

	0
	$2-{\frac {a(k-1)+a(k+1)}{a(k)}}$
	$1-{\frac {a(k-1)+a(k+1)}{a(k)}}$
	${\frac {a(k+1)-a(k-1)}{a(k)}}$
	$2\cdot {\frac {a(k+1)-a(k-1)}{a(k)}}$

Simple discount rate

For simple discount, the interest paid is calculated according to the accumulated value at the end of $n$th period. That is, the interest paid at the beginning of each period is $dA(n)$ (constant), i.e. $A(m)-A(m-1)=dA(n)$ for each positive integer $m$.

So, $$A(0)=A(n)-{\big (}A(n)-A(n-1){\big )}-\dotsb -{\big (}A(1)-A(0){\big )}=A(n)-dnA(n)=A(n)(1-dn).$$ Since $A(n)=A(0)a(n)\Leftrightarrow A(0)=A(n)a^{-1}(n)$ , $a^{-1}(n)=1-dn\Rightarrow a(n)={\frac {1}{1-dn}}$ for each nonnegative integer $n$ such that $dn<1\Leftrightarrow n<1/d$ (so that the accumulation function is defined). Similarly, we may intuitively expect that the same form of accumulation function holds for other nonnegative numbers, which motivates the following definition.

Definition. (Simple discount) The accumulation of discount according to the accumulation function $$a(t)={\frac {1}{1-dt}},\quad 0\leq t<1/d$$ is the simple discount, in which $d$ is the simple discount rate.

Compound discount rate

For compound discount, the interest paid at the beginning of each period is calculated according to the balance at the end of that period.

To be more precise, suppose $A(n)=k$ and the compound discount rate is $d$. At the beginning of $n$th year, the interest paid is $kd$, and so the balance at the beginning of $n$th year is $k-kd=k(1-d)$.

Since the balance at the end of $n-1$th year (which is the same as that at the beginning of $n$th year) is $k(1-d)$, the interest paid at the beginning of $n-1$th year is $k(1-d)d$, and thus the balance at the beginning is $k(1-d)-k(1-d)d=k(1-d)^{2}$.

Using the same argument, the balance at the beginning of first year is $k(1-d)^{n}$, i.e. $A(0)=k(1-d)^{n}=A(n)(1-d)^{n}$, and we can see that $a^{-1}(n)=(1-d)^{n}\Rightarrow a(n)={\frac {1}{(1-d)^{n}}}$ similarly for each nonnegative integer $n$. This motivates the following generalized definition similarly.

Definition. (Compound discount) The accumulation of interest according to the accumulation function $$a(t)={\frac {1}{(1-d)^{t}}},\quad t\geq 0$$ is the compound discount, in which $d$ is the compound discount rate.

Example.

• Amy invests 1000 into a fund with compound discount rate $d$, and she receives 2000 after ten years.
• Then, we know that

$$1000(1-d)^{-10}=2000\Rightarrow (1-d)^{-10}=2.$$

• Then, we can calculate that the accumulated value after 100 years is

$$A(100)=1000(1-d)^{-100}=1000{\big (}(1-d)^{-10}{\big )}^{10}=1000(2^{10})=1024000.$$

Exercise.

1 Suppose the fund value accumulates at simple interest rate $i$ instead. Calculate $i$ such that the amount received by Amy is the same.

	0.05
	0.07
	0.1
	0.93
	13.93

2 Suppose the fund value accumulates at compound interest rate $i$ instead. Calculate $i$ such that the amount received by Amy is the same.

	0.05
	0.07
	0.1
	0.93
	13.93

3 Suppose the fund value accumulates at simple discount rate $d$ instead. Calculate $d$ such that the amount received by Amy is the same.

	0.05
	0.07
	0.1
	0.93
	13.93

Suppose the fund value accumulates at different rates according to the following pattern: for each nonnegative integer $n$,

• during $4n$th year, the fund values accumulate at simple interest rate of 5%;
• during $4n+1$th year, the fund values accumulate at simple discount rate of 5%;
• during $4n+2$th year, the fund values accumulate at compound interest rate of 5%;
• during $4n+3$th year, the fund values accumulate at compound discount rate of 5%.

Calculate the amount received by Amy after ten years with the same amount of initial investment.

	1649.41
	1652.76
	1666.67
	1826.74
	2231.55

Equivalent rates

Definition. (Equivalent rates) Two interest or discount rates are equivalent if a given amount of principal invested for the same length of time at each of the rates produces the same accumulated value.

Remark.

• It means that the corresponding accumulation functions have the same output with the same input.
• The "$i$" and "$d$" in the exam FM questions are assumed to be equivalent, since it is assumed that $d={\frac {i}{1+i}}$, which will be shown below to be equivalent to the equivalence of $i$ and $d$.

Example. Suppose $i$ is an effective interest rate and $d$ is an effective discount rate that is equivalent to $i$. Then, $d={\frac {i}{1+i}}$.

Proof. Since they are equivalent, their accumulation functions are equal in value with the same input. In particular, with input $t=1$, $$1+i={\frac {1}{1-d}}\Leftrightarrow 1-d={\frac {1}{1+i}}\Leftrightarrow d={\frac {1+i-1}{1+i}}={\frac {i}{1+i}}.$$

$\Box$

Remark.

• It follows that $d=iv$ if and only if $i$ and $d$ are equivalent, since $v={\frac {1}{1+i}}$.
• Also, $v={\frac {1}{1+i}}=1-d$.

Example. Suppose $i$ is an effective interest rate and $d$ is an effective discount rate that is equivalent to $i$. Then, $i={\frac {d}{1-d}}$.

Proof. Since they are equivalent, $$1+i={\frac {1}{1-d}}\Leftrightarrow i={\frac {1-(1-d)}{1-d}}={\frac {d}{1-d}}.$$

$\Box$

Exercise.

1 Suppose the compound interest rate $i$ and the compound discount rate 3% are equivalent. Calculate $i$.

	0.029
	0.031
	0.942
	0.97

2 Select all correct expression(s) of $i$.

	$i={\frac {1-v}{v}}$
	$i={\frac {1+v}{v}}$
	$i={\frac {v}{1+v}}$
	$i={\frac {v}{1-v}}$
	$i={\frac {1-d}{v^{2}}}$

3 Select all possible value(s) of $i$ such that $i=d$ ($i$ and $d$ are equivalent).

	0
	0.5
	0.62
	1.62
	-1.62

4 Select all possible value(s) of $i$ such that $i=v$.

	0
	0.5
	0.62
	1.62
	-1.62

Nominal rates

We have discussed effective interest and discount rates. For those effective rates, the interest is paid exactly once per measurement period (either at the beginning (for discount rates) or at the end (for interest rates)).

However, the interest can be paid more than once per measurement period, and the interest and discount rates, for which the interest is paid more than once per measurement period, are called nominal, rather than effective, rates.

Definition. (Nominal interest and discount rates) Nominal interest and discount rates are rates for which the interest is paid more than once per measurement period.

The reason for calling those rates as "nominal" is that the notation for the nominal interest (discount) rate payable (or "convertible" or "compounded") $m$thly per measurement period is $i^{(m)}$ ($d^{(m)}$), and its value is a nominal value only, in the sense that the actual rate used in the calculation for each payment is $i^{(m)}/m$ ($d^{(m)}/m$) by definition, rather than $i^{(m)}$ ($d^{(m)}$), and this is the $m$thly rate.

Example. Amy deposits 1000 into a bank account with 24% interest compounded monthly (i.e. "12thly", compounded 12 times a year). Then,

• the nominal interest rate $i^{(12)}=0.24$;
• the accumulated value after a year (or 12 months) is $1000\left(1+{\frac {i^{(12)}}{12}}\right)^{12}\approx 1268.24$,

by regarding "months" as the measurement period, since the monthly rate is ${\frac {i^{(12)}}{12}}$, and there are 12 measurement periods, and thus the result follows from the definition of compound interest.

• We can see that the effective interest rate is approximately ${\frac {1268.24-1000}{1000}}\approx 27\%$, which is different from the nominal rate of 24%, showing how the nominal rate is "nominal" again.

Exercise.

1 Calculate the accumulated value after a year if the rate is discount rate payable monthly instead.

	784.72
	788.49
	1268.24
	1274.35
	26930.03

2 Calculate the accumulated value after a year if the rate is interest rate payable daily instead. (Treat one year as 365 days.)

	1007.92
	1271.15
	1271.35
	1377408.29

3 Calculate the accumulated value after a year if the rate is discount rate payable per second instead. (Treat one year as 31536000 seconds.)

	999.99722222605
	1000.0027777817
	1020.2013389814
	1271.2491512355
	1271.2491535574

Example. The nominal interest rate convertible monthly, denoted by $i^{(12)}$, that is equivalent to $d^{(6)}=18\%$ is calculated by $${\begin{aligned}&&\left(1+{\frac {i^{(12)}}{12}}\right)^{12}&=\left(1-{\frac {d^{(6)}}{6}}\right)^{-6}\\&\Rightarrow &1+{\frac {i^{(12)}}{12}}&={\big (}(1-0.03)^{-6}{\big )}^{1/12}{\text{ or }}-{\big (}(1-0.03)^{-6}{\big )}^{1/12}({\text{rejected}})\\&\Rightarrow &i^{(12)}&=12(0.97^{-6/12}-1)\approx 0.184.\end{aligned}}$$

Exercise.

Calculate the effective interest rate $i$ that is equivalent to $d^{(6)}=18\%$.

	0.184
	0.2
	2.29
	0.22
	0.03

Force of interest

We have discussed nominal interest rates, and in this subsection, we will discuss what will happen if the compounding frequency gets higher and higher, i.e. the $m$ in "compounded $m$thly" becomes larger and larger, to the infinity. We call this "compounded continuously".

To be more precise, we would like to know the value of $\lim _{m\to \infty }i^{(m)}$ during the "infinitesimal" time interval $[t,t+1/m]$ which tends to be simply the time point $t$, and we call this force of interest at time $t$, denoted by $\delta _{t}$. Now, we would like to develop a formula for $\delta _{t}$.

For nominal interest rate $i^{(m)}$, we have the following relationship between $A(t)$ and $A(t+1/m)$ by definition (treating $1/m$ of year as measurement period, then the effective interest rate during the period $[t,t+1/m]$ is $i^{(m)}/m$ by definition): $$A(t+1/m)=A(t)(1+i^{(m)}/m)\Rightarrow i^{(m)}={\frac {m{\big (}A(t+1/m)-A(t){\big )}}{A(t)}}={\frac {A(t+1/m)-A(t)}{1/m}}\cdot {\frac {1}{A(t)}}$$ So, taking limit, $$\underbrace {\lim _{m\to \infty }i^{(m)}} _{=\delta _{t}}={\frac {1}{A(t)}}\cdot \lim _{m\to \infty }{\frac {A(t+1/m)-A(t)}{1/m}}={\frac {1}{A(t)}}\cdot \underbrace {\lim _{1/m\to 0}{\frac {A(t+1/m)-A(t)}{1/m}}} _{{\overset {\text{ def }}{=}}A'(t)}={\frac {A'(t)}{A(t)}}.$$ This motivates the definition of force of interest.

Definition. (Force of interest) The force of interest at time $t$, denoted by $\delta _{t}$, is defined by $\delta _{t}={\frac {A'(t)}{A(t)}}$.

Remark.

• It follows that $\delta _{t}={\frac {a'(t)}{a(t)}}$, since $A(t)=A(0)a(t)$ and thus $A'(t)=A(0)a'(t)$.
• The expression ${\frac {A'(t)}{A(t)}}$ may be interpreted as relative rate of change of the amount function at time $t$, in the sense that it tells the rate of change at time $t$ equals what portion of $A(t)$, such that $\delta _{t}$ is independent from the value of $A(t)$.

• Thus, the force of interest measure the "intensity" of interest at time $t$, since the only factor that change $A(t)$ is interest.

Proposition. (Expression of accumulation function in terms of force of interest) $$a(t)=\exp \left(\int _{0}^{t}\delta _{s}\,ds\right).$$

Proof. $$\delta _{s}={\frac {a'(s)}{a(s)}}\Leftrightarrow \int _{0}^{t}\delta _{s}\,ds=\int _{0}^{t}{\frac {a'(s)}{a(s)}}\,ds\Leftrightarrow \int _{0}^{t}\delta _{s}\,ds=\int _{0}^{t}{\frac {1}{a(s)}}\,d{\big (}a(s){\big )}\Leftrightarrow \int _{0}^{t}\delta _{s}\,ds=\ln a(t)-\underbrace {\ln a(0)} _{=\ln 1=0}\Leftrightarrow a(t)=\exp \left(\int _{0}^{t}\delta _{s}\,ds\right).$$

$\Box$

Remark.

• A corollary is that ${\frac {a(t_{2})}{a(t_{1})}}=\exp \left(\int _{0}^{t_{2}}\delta _{s}\,ds-\int _{0}^{t_{1}}\delta _{s}\,ds\right)=\exp \left(\int _{t_{1}}^{t_{2}}\delta _{s}\,ds\right)$, which is sometimes useful.

Proposition. (Constant force of interest) If and only if the force of interest over a measurement period is constant, denoted by $\delta$, and the effective interest rate during the time interval is $i$, $\delta =\ln(1+i)$.

Proof.

• Without loss of generality, suppose the measurement period is $[t,t+1]$. Then,

$${\frac {a(t+1)}{a(t)}}=\exp \left(\int _{t}^{t+1}\delta \,ds\right)\Leftrightarrow {\frac {(1+i)^{t+1}}{(1+i)^{t}}}=\exp \left(\delta \right)\Leftrightarrow \delta =\ln(1+i).$$

$\Box$

Remark.

• In general, if and only if the force of interest over $n$ measurement periods, say time interval $[0,n]$, is constant, the accumulation function

$$a(n)=\exp \left(\int _{0}^{n}\delta \,ds\right)\Leftrightarrow a(n)=e^{n\delta }.$$

Example. (Equivalency between forces of interest and discount) The force of discount can be analogously defined by $\lim _{m\to \infty }d^{(m)}$. Then, the force of discount $\lim _{m\to \infty }d^{(m)}=\delta _{t}$.

Proof.

• Similar to the motivation for force of interest, we have

$$A(t)=A(t+1/m)(1-d^{(m)}/m)\Rightarrow d^{(m)}={\frac {-m{\big (}A(t)-A(t+1/m{\big )})}{A(t+1/m)}}={\frac {A(t+1/m)-A(t)}{1/m}}\cdot {\frac {1}{A(t+1/m)}}.$$

• Taking limit,

$$\lim _{m\to \infty }d^{(m)}=\lim _{1/m\to 0}\left({\frac {A(t+1/m)-A(t)}{1/m}}\cdot {\frac {1}{A(t+1/m)}}\right)=A'(t)\cdot {\frac {1}{A(t)}}=\delta _{t}.$$

$\Box$

Remark.

• Thus, it suffices to discuss force of interest, but not force of discount.

Example. Value of a fund accumulates at 10% (annual, which is assumed) force of interest, i.e. $\delta =0.1$. Amy invests 10000 into the fund. Then, the accumulated value of the investment is $10000e^{2\cdot (0.1)}\approx 12214.03$.

Exercise.

1 Suppose Amy invests $P$ into the fund instead, and the accumulated value is 10000 instead. Express the constant force of interest $\delta$ at which the fund value accumulates in terms of $P$.

	$\delta ={\frac {1}{2}}\ln(10000/P)$
	$\delta ={\frac {1}{2}}\ln(20000/P)$
	$\delta =\ln(20000/P)$
	$\delta =\ln(10000/P)$
	$\delta =\ln(50000/P)$

2 Calculate $A'(1)$ with the same condition in the example.

	11051.71
	1105.17
	1221.4
	1.11
	0.11

3 Select all correct expression(s) of $A'(t)$ with the same condition in the example.

	$A'(t)=e^{\delta t}$
	$A'(t)=\delta e^{\delta t}$
	$A'(t)=e^{\delta t}/\delta ^{3}$
	$A'(t)=\delta (1+i)^{t}$ ($i$ is the annual effective interest rate)
	$A'(t)=\delta (1+it)$ ($i$ is the annual effective interest rate)

Example. Value of a fund accumulates at the annual force of interest $\delta _{t}=(2+t)^{-1}$ for each nonnegative $t$ (in years, i.e. $t=j$ means "at the end of $j$th year", since the time interval from $t=j-1$ to $t=j$ is the $j$th year.). Then, the accumulated value at $t=15$ with principal one is $${\begin{aligned}a(15)=\exp \left(\int _{0}^{15}(2+t)^{-1}\,dt\right)=\exp(\ln 17-\ln 2)=\exp \left(\ln(17/2)\right)=17/2.\end{aligned}}$$

Exercise.

1 Calculate the effective interest rate $i$ during the time period from $t=3$ to $t=11$.

	1.6
	2.4
	6.5
	8.5

2 Another fund accumulates at the simple interest rate of $i$. Calculate $i$ such that the accumulated value of this fund equals that of the fund in the example at each positive $t$, with the same principal.

	0.5
	1
	1.5
	2.5
	There does not exist such $i$.

Example.

• Amy deposits 100 into a bank and the deposit of 100 accumulates according to the force of interest

$$\delta _{t}={\begin{cases}0.01,&0\leq t<7;\\(1+t)^{-1},&7\leq t<10;\\0.0009t^{2},&t\geq 10,\\\end{cases}}$$

in which $t$ is in years.

• Then, the accumulated value of the deposit at the end of 8th year is

$$100\exp \left(\int _{0}^{8}\delta _{t}\,dt\right)=100\exp \left(\int _{0}^{7}0.01\,dt+\int _{7}^{8}(1+t)^{-1}\,dt\right)=100\exp \left(0.07+\ln 9-\ln 8\right)\approx 120.66;$$

• the accumulated value at the end of 11th year is

$$100\exp \left(\int _{0}^{11}\delta _{t}\,dt\right)=100\exp \left(\int _{0}^{7}0.01\,dt+\int _{7}^{10}(1+t)^{-1}\,dt+\int _{10}^{11}0.0009t^{2}\,dt\right)=100\exp \left(0.07+\ln 11-\ln 8+0.0003(11^{3}-10^{3})\right)\approx 162.87.$$

Exercise.

Calculate the effective discount rate $d$ during the time period from the end of 3rd year to the end of 14th year.

	0.08
	0.59
	0.6
	1.49
	1.56

Present, current and future values

From previous sections, we have seen that money has a time value because of the interest, in the sense that $1 today will worth more than $1 after a period of time (assuming positive interest rate).

To be more precise, an investment of $k$ will accumulate to $k(1+i)$ at the end of one period, in which $i$ is the effective interest rate during the period. In particular, the term $1+i$ is called accumulation factor, since it accumulates the investment value at the beginning to its value at the end. Graphically, it looks like the following time diagram (a graph represents statuses at different time).

   *----------*
   |          |
   |          v
 k |            k(1+i)
---*----------*----
  beg        end

We would often like to do something "reverse" to calculating the accumulated value given the principal. That is, calculating the principal given the accumulated value. Since the principal is the investment value at initial time, which is often now (or present), the "reverse" calculation is essentially calculating the present value of the investment, given its accumulated (or future) value at the future.

To be more precise, we would like to calculate the principal (or present value, denoted by $PV$) such that it accumulates to $k$ (which is future value, denoted by $FV$) at the end of one period. Using equation to describe this situation, we have $$PV(1+i)=FV\Leftrightarrow PV=FV\cdot {\frac {1}{1+i}}$$ in which $i$ is the effective interest rate during this period. The term ${\frac {1}{1+i}}$, which is denoted by $v$ , is the discount factor, since it "discounts" the future value to the present value.

   *----------*
   |          |
   v          |
 k/1+i        | k     
---*----------*----
  beg        end

The term current value (which is "at the middle" of present and future values) is sometimes used. It means the value of the payments at a specified date, and some payments are made before that date, while some payments are made after that date.

We have discussed how to calculate the present value for one period, but we can generalize the result to more than one period. To be more precise, we would like to also calculate the present value given the future value at the end of $t$ periods. We can use the accumulation function $a(t)$ to describe this situation in general . $$PVa(t)=FV\Leftrightarrow PV=FV\cdot {\frac {1}{a(t)}}=FVa^{-1}(t)$$ in which $a^{-1}(t)$ is the inverse function of $a(t)$ .

Also, given multiple future values, we can calculate the total present value of these future values by summing up all present values corresponding to these future values.

Example. In exchange for your car, John has promised to pay you 5,000 after one year and 10,000 after three years. Using an annual effective interest rate of 5%, the present value of these payments is $$PV=5000(1.05)^{-1}+10000(1.05)^{-3}\approx 13400.28.$$ Time diagram:

    *--------------------*
    |                    |
    *-------*            |
    |       |            |
    |       |            |
    v     5000        10000
----*------*------------*---- t
    0      1            3

Exercise.

1 Calculate the present value if the given interest rate is simple interest rate.

	11337.07
	13043.48
	13457.56
	13852.81
	14285.71

2 Calculate the effective interest rate $i$ such that the present value is 14000. (Hint: you may use Newton's method with initial guess $i_{0}=0.03$ to approximate the answer.)

	1.12%
	3.01%
	3.02%
	3.03%
	3.10%

Remark.

• It is often useful to draw a rough time diagram to understand the situation given in the question.

Example. A fund provides payment of 50000 at the end of 7th year. Suppose there is annual force of interest $\delta _{t}=0.04t+0.05$ ($t$ is in years). Then, the present value of the payment is $$PV=50000e^{-{\big (}0.02(7^{2})+0.05(7)-0.02(0^{2})-0.05(0){\big )}}\approx 13223.86.$$

Proof.

• First, the inverse of accumulation function corresponding to the force of interest $\delta _{t}$ at $t=7$ is

$$a^{-1}(7)=\left(\exp \left(\int _{0}^{7}0.04t+0.05\,dt\right)\right)^{-1}=\left(\exp \left([0.02t^{2}+0.05t]_{0}^{7}\right)\right)^{-1}=\exp \left(-0.02(7^{2})-0.05(7)\right).$$

• The result follows from the relationship $PV=50000a^{-1}(7)$.

$\Box$

Exercise.

Suppose after the end of 8th year, there is a simple discount of 3% instead of the force of interest, and there is an additional payment of 300000 at the end of 15th year. Calculate the current value of the payments at the end of 10th year.

	295953.38
	313348.23
	330482.32
	333030.32
	334057.07

Equations of value

For two or more payments at different time points, to compare them fairly, we need to accumulate or discount them to a common time point, so that the effects on payments from the time value of money are eliminated.

The equation which accumulates or discounts each payment as in above is called the equation of value.

Indeed, we have encountered equations of value in previous sections, since an example of equations of value is calculating present values of multiple payments ("present" is the common time point).

The concepts involved in equations of value have been discussed previously.

Inflation and real interest rate

In previous sections, we have not consider the effect from inflation, and we will introduce how interest rates changes under inflation.

Because of the inflation, there are two types of interest rates, namely nominal interest rate and real interest rate.

For nominal interest rate, it is the same as the "normal" interest rate discussed previously, and thus is denoted by $i$.

Definition. (Real interest rate) The real interest rate, denoted by $r$, is the interest rate with inflation effect eliminated. To be more precise, the real interest rate is defined by $$1+r={\frac {1+i}{1+j}}$$ with inflation rate $j$ for the corresponding year at which interest rates are calculated..

Remark.

• By definition of (annual) inflation rate, $1+j={\frac {\text{price level at the end of the year}}{\text{price level at the beginning of the year}}}$, so ${\frac {1}{1+j}}={\frac {\text{price level at the beginning of the year}}{\text{price level at the end of the year}}}$, and thus ${\frac {1}{1+j}}$ can be interpreted as adjusting the price level at which the interest is paid (i.e. at the end of the year) to the price level at the beginning, so the change in price level in this year is eliminated.
• An approximation of real interest rate is $r\approx i-j$, since $1+i=(1+r)(1+j)=1+r+j+rj\approx 1+r+j$ ($rj$ is small).

Example.

• Suppose the annual inflation rate is 5% and the nominal interest rate is 4%.
• Then, the annual real interest rate is ${\frac {1.04}{1.05}}-1\approx -0.0095.$
• In general, the (effective) nominal interest rate during $n$ years is $1.04^{n}$, and
• the inflation rate for $n$ years is $1.05^{n}$.
• So, the real (effective) interest rate during $n$ years is ${\frac {1.04^{n}}{1.05^{n}}}-1=\left({\frac {1.04}{1.05}}\right)^{n}-1$.
• As we can see here, if the nominal interest rate is compound interest rate, the resulting real interest rate is also compound interest rate.
• We can use real interest rate for various calculations, e.g. calculating present values, i.e. use the real interest rate for the "$i$" in formulas.

Proof. Let $r$ be the real interest rate. Then, by definition, $$1+r={\frac {1+4\%}{1+5\%}}={\frac {1.04}{1.05}}\Rightarrow r={\frac {1.04}{1.05}}-1\approx -0.0095.$$

$\Box$

Exercise.

1 Suppose a fund pays 1000 at the end of first year. Let $x$ and $y$ be the present value of the payment calculated using nominal interest rate and real interest rate respectively. Calculate $y-x$.

	-48.08
	-0.05
	0
	0.05
	48.08

2 Suppose the inflation rate for the $n$th year is $(4+n)\%$, and the nominal interest rate is the same during each year, and a fund pays $k^{n/2}$ ($k>0$) at the end of $n$th year, until third payment is made, and the payment stops thereafter. Select all correct expression(s) for the present value of these three payments using real interest rate.

	${\frac {1.05k^{1/2}}{1.04}}+\left({\frac {1.06k^{1/2}}{1.04}}\right)^{2}+\left({\frac {1.07k^{1/2}}{1.04}}\right)^{3}$
	$\left({\frac {1.05k^{1/2}}{1.04}}\right)^{-1}+\left({\frac {1.06k^{1/2}}{1.04}}\right)^{-2}+\left({\frac {1.07k^{1/2}}{1.04}}\right)^{-3}$
	${\frac {1}{1.04}}(1.05^{-1}k^{1/2}+1.06^{-2}k+1.07^{-3}k^{3/2})$
	$k^{1/2}\cdot {\frac {1.05}{1.04}}+k\cdot \left({\frac {1.05}{1.04}}\right)^{2}+k^{3/2}\cdot \left({\frac {1.05}{1.04}}\right)^{3}$
	${\frac {(1.05k^{1/2})(1-(1.05k^{1/2}/1.04)^{3})}{1.04-1.05k^{1/2}}}$