MATH 4050 - Mathematical Finance and Interest Theory


Jie Zhong
Department of Mathematics
California State University, Los Angeles

Chapter 1 The growth of money

Interest

Why the Theory of Interest?

  • Interest rates is the main and popular tool of monetary policy. For example, low interest rate generally leads to economic expansion, high interest rate generally leads to economic contraction.
  • Understand how money grows, and manage our own money.

Typical problem

  • Suppose you have won a lottery with two option to collect the prize.
  • Option 1: get 20 payments of $60,000 on the first day of each year as from 2024
  • Option 2: get one million on the first day of 2024.
  • Which option should you choose?

Reasons/Motivations for your choice?

  • How badly do you need the cash?
  • How much debt do you have now?
  • Do you like to spend money as you receive it?
  • How much return can you make with the cash?
  • All of the above and other factors related to interest rate.

What is interest?

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When a landowner allows a farmer to use the land he owns, the farmer has to pay “rent” to the landowner.






When a banker lets a borrower to use a certain amount of money, the banker will charge the borrower something.

Definition of Interest

Interest may be defined as the compensation paid by a borrower of money to a lender.

Thus we can view interest as the rent paid by a borrower to a lender for the loss of use of the money.

If an investment of \(K\) (called the principal, i.e., the amount of money that the borrower borrows/lender lends at time \(t=0\)) grows to \(S\) (at a later time, say, \(T\)), then the interest is the difference

\begin{align*} S - K \ge 0 \end{align*}

Question: Why do we charge interest?

  • Investment opportunities theory
  • Time preference theory
  • Risk premium

Accumulation and amount functions

Amount function

  • Let us temporarily fix the principal \(K\).
  • \(A_K(t)\): the amount function for principal \(K\), i.e., the balance at time \(t\ge 0\) (time is always measured in some agreed upon units; think "years" for now).
  • In words, \(K\) invested at time \(t =0\) "grows" to \(A_K(t)\) at time \(t\ge 0\).
  • Note: \(A_K(0) = K\).

Accumulation function

  • \(a(t)\): the accumulation function, i.e., the amount function if the principal \(K\) is one unit of money.
  • Formally, if the principal is one unit of money, we write \(a(t) = A_1(t)\).
  • Note: \(a(0) = 1\).

The relationship between \(A_K(t)\) and \(a(t)\)

What do you think the relationship should be?

Often, \(A_K(t) = K a(t)\).

  • What does this mean?
  • When is this not true?
  • However, since the above equality holds in most cases, we will assume that it is true unless it is explicitly noted otherwise.

Examples

  • Suppose you borrow \(20\) from your parents, what would \(A_K(t)\) look like?
  • Suppose you borrow \(20\) from your friend, what would \(A_K(t)\) look like?
  • Suppose you borrow \(20\) from your bank, what would \(A_K(t)\) look like?
  • Suppose you deposit \(20\) into a bank which earns \(1\) at the end of every year (but nothing else during the year), what would \(A_K(t)\) look like?

Examples of amount function

BlankAmount.png

The increase/decrease of \(A_K(t)\) and \(a(t)\)

  • It is natural to assume that both \(a\) and \(A_K\) increase in the time \(t\).
  • Such increase may be, for example,
    • continuous and linear
    • discrete (end of the year, e.g.)
    • continuous and exponential
  • However, there are risky investment which might lose money over the time.
  • Read: Example 1.3.2-4 in the textbook.

Example

Given \(A_K(t) = \frac{1000}{50 - t}\) for \(0\le t < 50\), calculate \(K\) and \(a(10)\), assuming that \(A_K(t) = K a(t)\).

Solution:

  • \(K = A_K(0) = 1000/50 = 20\).
  • \(a(10) = A_{20}(10)/20 = 1000/(50 - 10)/20 = 5/4 = 1.25\).

Effective interest rate

When \(0 \leq t_1 \leq t_2\), the effective interest rate for \([t_1,t_2]\) is

\begin{align*} i_{[t_1,t_2]} = \frac{a(t_2) - a(t_1)}{a(t_1)}, \end{align*}

and if \(A_K(t) = Ka(t)\) then

\begin{align*} i_{[t_1,t_2]} = \frac{A_K(t_2) - A_K(t_1)}{A_K(t_1)}. \end{align*}

Question: Why not use \(A_K(t)\) to define the rate?

This is because \(A_K\) might depend on the principal \(K\).

Effective interest rate, alternatively

For a positive integer \(n\), the interval \([n-1, n]\) is called \(n\text{-th}\) time period.

We write

\begin{align*} i_n = i_{[n-1, n]} = \frac{a(n) - a(n-1)}{a(n-1)}, \end{align*}

and hence,

\begin{align*} a(n) = a(n-1)(1+i_n). \end{align*}

How would this simplify for \(i_1\)?

\begin{align*} i_1 = a(1) -1. \end{align*}

Simple interest / Linear \(a(t)\)

Simple interest

Assume the accumulation function \(a(t)\) is linear in \(t\), and since \(a(0) = 1\), we have

\begin{align*} a(t) = 1 + st, \end{align*}

where \(s\) is called the simple interert rate.

Note:

\begin{align*} s = i_1. \end{align*}

We call \(A_K(t) = K(1 + st)\), the amount function for \(K\) invested by simple interest at rate \(s\).

Example 1.4.1

Tonya loans Renu \(1,600\). Renu promises that in return, she will pay Tonya \(2,000\) at the end of four years. To what rate of simple interest does this correspond?

Solution:

  • \(2000 = A_{1600}(4) = 1600(1 + 4s)\)
  • \(\displaystyle s = \frac{1}{4}\left( \frac{2000}{1600} -1 \right) = 0.0625\)

Example 1.4.2

Antonio loans his brother Bob \(2,400\) for three years at \(5\%\) simple interest. The brothers agree that if Bob wishes to repay the loan early, he may do so, and the repayment amount will still be based on \(5\%\) simple interest.

Find the amount Bob would be required to pay if he makes his repayment at the end of three years. What if the repayment is after two years or after one year? Calculate \(i_1, i_2\), and \(i_3\) if the loan lasts the full three years.

Example 1.4.2 - Solution

If Bob repays the loan after

  • three years, then \(A_{2400}(3) = 2400[1 + 3\cdot 0.05] = 2760\)
  • two years, then \(A_{2400}(2) = 2400[1 + 2\cdot 0.05] = 2640\)
  • one year, then \(A_{2400}(1) = 2400[1 + 0.05] = 2520\)

Therefore,

  • \(\displaystyle i_1 = \frac{2520-2400}{2400} = 5\%\)
  • \(\displaystyle i_2 = \frac{2640-2520}{2520} \approx 4.76\%\)
  • \(\displaystyle i_3 = \frac{2760-2640}{2640} \approx 4.55\%\)

Note: The annual effective interest rates are decreasing.

On \(i_n\)

In the simple interest case:

\begin{align*} i_n = \frac{a(n) - a(n-1)}{a(n-1)} = \frac{(1+sn) - (1+s(n-1))}{1+s(n-1)}= \frac{s}{1 + s(n-1)}. \end{align*}

So, \(i_n\) is decreasing in \(n\), and moreover,

\begin{align*} i_n \to 0, \, \text{as}~ n \to \infty. \end{align*}

Therefore, simple interest is rarely used for loans of long duration; otherwise you would go into the bank, close your account, and then instantly reopen it.

Compound interest (The usual case!)

Compound interest

Define \(i = i_1 = a(1) - 1\).

Assume that an accumulation function \(a(t)\) has the associated periodic interest rates all equal, i.e., assume that

\begin{align*} i_n = i \quad \text{for every positive integer}~ n. \end{align*}

Then the accumulation function must equal to

\begin{align*} a(t) = (1 + i)^t \quad \text{for every } t \ge 0. \end{align*}

We call \(a(t)\) defined above the compound interest rate accumulation function at interest rate \(i\).

The word "compound" means that the interest earned is automatically reinvested to earn additional interest.

Simple vs. Compound interest

SimpleCompound.png

Example

An account is opened with \(12,000\) and is closed in \(6.5\) years. The account earns \(5\%\) interest. How much is withdrawn from the account if

Compound interest is paid throughout.

\begin{align*} 12000(1.05)^{6.5} = 16478.27 \end{align*}

Compound interest is paid on each whole year and then simple interest is paid on the last half year.

\begin{align*} 12000(1.05)^6[1 + 0.5 \times 0.05] = 16483.18 \end{align*}

Example - Varying interest rates

Assume that \(1,000\) is deposited into an account. The effective annual compound interest rate is \(3\%\) for the first year, \(4\%\) for the next two, and \(1\%\) for the next three.

How much would be in the account at the end of the six years?

Solution

\begin{align*} 1000 (1+0.03)(1+0.04)^2(1.01)^3 = 1147.80 \end{align*}

Example - Unknown rate

Suppose you want to have \(1,000\) in three years. You currently have \(900\) to invest.

What interest rate (annually compounding) do you need to accomplish your goal?

Solution

\begin{align*} 1000 = 900(1 + i)^3 \end{align*}

implies

\begin{align*} i = \left( \frac{1000}{900} \right)^{1/3} - 1 = 3.57\%. \end{align*}

Example - Unknown principal

Suppose you want to have \(1,000\) in three years. If you could earn \(2\%\) annually compounding interest, how much would you need now to invest to accomplish your goal?

Solutions

\begin{align*} 1000 = K(1 + 0.02)^3, \end{align*}

so

\begin{align*} K = \frac{1000}{(1 + 0.02)^3} = 942.32 \end{align*}

Interest in advance / The effective discount rate

Discount rate

When you rent an apartment, usually you are required to pay rent for each month at the beginning of the month. In other words, you pay the rent before you have the use of the apartment.

We said that interest may be thought of as a rent for the use of the investor's money. It is therefore not surprising that there are financial arrangements in which the interest must be paid by the borrower before the borrowed money becomes available.

Two scenarios

Loan 1: \(500\) is borrowed for a year at an effective rate of interest of \(6\%\). At the end of the year, the borrower pays the lender \(530\).

Loan 2: \(500\) is borrowed for a year, but the borrower pays the interest at the loan's inception. The borrower receives \(470\) and must repay \(500\) at the end of one year.

In both cases the interest paid is \(30\).

However, in the second case the interest is paid on a loan of only \(470\).

Loan 2 is an example of computing interest as a discount and is said to have an effective rate of discount of \(6\%\).

Definition of discount rate

The effective discount rate for the interval \([t_1, t_2]\) is

\begin{align*} d_{[t_1,t_2]} = \frac{a(t_2)-a(t_1)}{a(t_2)}. \end{align*}

If \(A_K(t) = Ka(t)\), then

\begin{align*} d_{[t_1,t_2]} = \frac{A_K(t_2)-A_K(t_1)}{A_K(t_2)}. \end{align*}

Similar to \(i_n\), when \(n\) is a positive integer,

\begin{align*} d_n = d_{[n-1, n]} = \frac{a(n) - a(n-1)}{a(n)} \qquad \text{and} \qquad a(n-1) = a(n)(1-d_n). \end{align*}

Example 1.6.5

Suppose that the growth of money is governed by the accumulation function \(a(t) = (1.05)^{t/2}(1 + 0.025t)\). Find \(d_4\) and \(i_4\).

Solution

We need

\begin{align*} a(4) = (1.05)^2(1.1) = 1.21275, \quad a(3) = (1.05)^{3/2}(1.075) - 1.156624568, \end{align*}

and thus

\begin{align*} d_4 = \frac{a(4) - a(3)}{a(4)} = 0.046279474, \qquad i_4 = \frac{a(4) - a(3)}{a(3)} = 0.048525195. \end{align*}

Note: Usually, \(i_{[t_1,t_2]}\) and \(d_{[t_1, t_2]}\) are not equal.

Equivalence of interest and discount rates

A rate of interest and a rate of discount are said to be equivalent for an interval \([t_1, t_2]\) if they produce the same accumulated value at time \(t_2\) for one unit of money invested at time \(t_1\):

\begin{align*} 1 = (1 + i_{[t_1, t_2]})(1 - d_{[t_1, t_2]}), \end{align*}

which is equivalent to

\begin{align*} i_{[t_1,t_2]} = \frac{d_{[t_1,t_2]}}{1-d_{[t_1,t_2]}} \qquad \text{or} \qquad d_{[t_1,t_2]} = \frac{i_{[t_1,t_2]}}{1+i_{[t_1,t_2]}}. \end{align*}

Similarly,

\begin{align*} i_n = \frac{d_n}{1-d_n} \qquad \text{and} \qquad d_n = \frac{i_n}{1+i_n}. \end{align*}

Discount functions / The time value of money

Discount function - Motivation

\(100\) now may be worth more than \(100\) in three years.

Why?

You could invest this \(100\) today and it would grow to \(100 a(3)\).

Question: How much should you invest now (at time \(t =0\)) to have \(1\) after \(t\) years?

Definition of discount function

The discount function is defined by

\begin{align*} v(t) = \frac{1}{a(t)}. \end{align*}

In words, \(v(t)\) is the amount of money that one should invest at time \(0\) in order to have \(1\) at time \(t\).

For example,

\begin{align*} v(t) = \begin{cases} \frac{1}{1 + st}, & \mbox{simple interest}\\ \frac{1}{(1 + i)^t}, & \mbox{compound interest} \end{cases} \end{align*}

Example 1.7.2

Suppose that the growth of money for the next five years is governed by simple interest at \(5\%\).

How much money should you invest now in order that you have a balance of \(23,000\) three years from now?

Solution

  • \(\displaystyle v(3) = \frac{1}{1 + (0.05)3} = \frac{1}{1.15}\).
  • Then we should invest \(23000 v(3) = 23000(1/1.15) = 20000\).

Invested at a time not zero

Question: What if one wishes to invest \(X\) not at time \(0\) but at a later time \(t_1>0\) with the goal of receiving \(S\) at \(t_2 > t_1\)?

Let us draw the timeline:

Timeline.png

\begin{align*} X = S \frac{v(t_2)}{v(t_1)} = S v(t_2)a(t_1) = S \frac{a(t_1)}{a(t_2)}. \end{align*}

Read: Example 1.7.3, and also page 35 - 40 for very useful instructions on Cash Flow worksheet using the BA II Plus calculator.

Example 1.7.4

Suppose that the growth of money for the next five years is governed by the linear accumulation function \(a(t) = 1+0.05t\).

If you wish to invest money two years from now so as to have \(23,000\) five years from now, how much money should you invest?

Example 1.7.4 - Solution

Let \(X\) be the amount invested at time two.

Timeline2-5.png

So we have \(X v(2) a(5) = 23000\),

and thus

\begin{align*} X & = 23000 \frac{1}{a(5)} \frac{1}{v(2)} = 23000 v(5)a(2)\\ & = 23000 \cdot \frac{1}{1.25} \cdot 1.10 = 20240. \end{align*}

Warning: \(a(5)v(2) = a(5)/a(2) \neq a(3)\)! Read Example 1.7.3 & 4, and top of page 32.

Simple and compound discount

Simple discount

In Section (1.4) we considered two parties negotiating a loan with a fixed amount of interest per basic time period for each \(1\) borrowed.

Suppose that we again consider two parties negotiating a loan, but this time they agree on a fixed amount of discount \(d\) per basic time period for each \(1\) borrowed.

Then, if the loan period is \([0, t]\) and \(K\) is the loan amount, the borrower receives \(K - Ktd = K(1 - td)\).

In particular, the borrower receives \(1\) if \(K = (1 - td)^{-1}\). It follows that \(a(t) = (1 - td)^{-1}\). Then the discount function \(v(t) = 1 - td\) is linear.

Definition of simple discount

\begin{align*} A_K(t) = \frac{K}{(1 - dt)} \end{align*}

is called the amount function for K invested by simple discount at rate \(d\).

\begin{align*} a(t) = \frac{1}{1 - dt} \end{align*}

is called the simple discount accumulation function at rate \(d\).

Note: It only makes sense to talk about loan terms that are shorter than \(1/d\) in this case.

Compound discount

In analogy with the case of compound interest, here we assume that the effective discount rate \(d_n\) is constant for every unit time period, i.e., we assume that there is a constant \(d\) such that

\begin{align*} d = d_n \qquad \text{for every } n \ge 1. \end{align*}

Then the equivalent interest rate has the form

\begin{align*} i_n = i = \frac{d}{1 - d}\qquad \text{for every } n \ge 1. \end{align*}

Terminology: If "year" is our basic time unit, then we say that \(d\) is the annual effective discount rate.

Example 1.9.8

Cassandra needs to borrow money to pay her tuition. She has a choice of borrowing at an annual effective interest rate of \(5.1\%\) or at an annual effective discount rate of \(4.9\%\). Which rate should she choose?

Example 1.9.8 - Solution

Solution 1:

An annual effective discount rate of \(4.9\%\) is equivalent to an annual effective interest rate of

\begin{align*} \frac{0.049}{1-0.049} = 0.0515 > 0.051. \end{align*}

Solution 2:

An annual effective interest rate of \(5.1\%\) is equivalent to an annual effective discount rate of

\begin{align*} \frac{0.051}{1+0.051} = 0.0485 < 0.049. \end{align*}

So by both calculations, Cassandra should borrow at the \(5.1\%\) interest rate.

The discount factor

In the compound discount case, i.e., the discount rate is a constant, we have

\begin{align*} d = \frac{i}{1 + i}. \end{align*}

Using the notation for the discount factor \(\displaystyle v = \frac{1}{1 + i}\), we can rewrite the above as

\begin{align*} d = iv. \end{align*}

Note:

\begin{align*} d + v = 1. \end{align*}

Compound discount accumulation function

In the compound discount case, we can rewrite the accumulation function

\begin{align*} a(t) = (1 + i)^t = (1 - d)^{-t}. \end{align*}

and call it compound discount accumulation function at discount rate \(d\).

Read: Example 1.9.13 in the textbook.

Nominal rates of interest and discount

Nominal interest rate

Assume that the bank credits the interest more than once per year, say \(m\) times in a single year.

We denote by \(i^{(m)}\), the nominal (annual) interest rate compounded (convertible, payable) m times per year.

The word "nominal" means that the interest rate \(i^{(m)}\) is annual in name only, i.e., the mechanism is such that the bank pays interest at the rate of

\begin{align*} \frac{i^{(m)}}{m}\quad \text{after each $m^{\text{th}}$ of a year} \end{align*}

which then gets reinvested in the account.

Connection between \(i^{m}\) and \(i\)

Suppose that we have an investment governed by compound interest. Let \(i\) be the annual effective interest rate. It is also known as the annual percentage yield (APY). Then

\begin{align*} \frac{i^{(m)}}{m} = (1 + i)^{1/m} - 1, \end{align*}

which is equivalent to

\begin{align*} i^{(m)} = m[(1 + i)^{1/m} - 1], \quad i = \left( 1 + \frac{i^{(m)}}{m} \right)^m -1. \end{align*}

Question: Why?

Connection between \(i^{m}\) and \(i\) - Continued

In words, for every invested unit of money, one gets

\begin{align*} 1 + i \end{align*}

at the end of the year.

On the other hand, after every \(m^{\text{th}}\) of the year the money currently on the account grows by a factor of \((1 + i^{(m)}/m)\); there are \(m\) such compoundings in a single year, so the final amount of money equals

\begin{align*} \left( 1 + \frac{i^{(m)}}{m} \right)^m. \end{align*}

Finally, the two values displayed above must be equal.

Example

Find the accumulated value of \(500\) invested for five years at \(0.08\) nominal interest compounded quarterly.

Solution

\begin{align*} 500 \left( 1 + \frac{0.08}{4} \right)^{4\times 5} = 500 \cdot 1.02^{20} = 742.97. \end{align*}

Note: The above investment scheme is equivalent to the one in which one invests \(500\) at \(2\%\) for \(20\) years.

Read: Example 1.10.5 and 1.10.6 in the textbook.

Example 1.10.7 - \(i^{(m)}\) for \(m\) not an integer

Jolene invests money in a fund for which interest is paid once every two years. The rate per two-year period is \(14\%\). Find the nominal interest rate convertible biennially and the annual effective interest rate governing the fund.

Solution

  • In this case, \(m = 1/2\).
  • The nominal (annual) interest rate \(i^{(1/2)} = 14\% / 2 = 7\%\).
  • The annual effective interest rate \(i = (1 + i^{(m)}/m)^m -1 = (1.14)^{1/2} -1 = 6.77\%\).

Nominal discount rate

Similar facts exist for discount rate.

With \(d\) as the annual discount rate, we have

\begin{align*} 1 - d = \left( 1 - \frac{d^{(m)}}{m} \right)^m, \end{align*}

or, equivalently,

\begin{align*} d = 1 - \left( 1 - \frac{d^{(m)}}{m} \right)^m, \quad \text{or}\quad d^{(m)} = m \left[ 1 - (1 -d)^{1/m} \right], \end{align*}

where \(d^{(m)}\) is called the nominal discount rate compounded (convertible, payable) m times per year.

Equivalent discount and interest rate

We can derive the following relationships

\begin{align*} \left( 1 - \frac{d^{(m)}}{m} \right) \left( 1 + \frac{i^{(m)}}{m} \right) =1, \end{align*}
\begin{align*} i^{(m)} = \frac{d^{(m)}}{1 - \frac{d^{(m)}}{m}}\quad \text{and} \quad d^{(m)} = \frac{i^{(m)}}{1 + \frac{i^{(m)}}{m}}, \end{align*}

and most generally, for integers \(n\) and \(p\),

\begin{align*} \left( 1 + \frac{i^{(n)}}{n} \right)^n = 1 + i = (1 - d)^{-1} = \left( 1 - \frac{d^{(p)}}{p} \right)^{-p}. \end{align*}

Example

If you invest \(100\) today and it grows to \(115\) in one year, what is

annual simple interest rate:

\begin{align*} 100(1 + i) = 115 \Longrightarrow i = 0.15. \end{align*}

annual compound interest rate:

\begin{align*} 100(1 + i) = 115 \Longrightarrow i = 0.15. \end{align*}

nominal interest compounded monthly:

\begin{align*} 100 \left( 1 + \frac{i^{(12)}}{12} \right)^{12} = 115 \Longrightarrow i^{(12)} = 0.1406. \end{align*}

nominal discount compounded monthly:

\begin{align*} \left( 1 - \frac{d^{(12)}}{12} \right) \left( 1 + \frac{i^{(12)}}{12} \right) =1 \Longrightarrow d^{(12)} = 0.1389. \end{align*}

Force of interest

Continuous compounding

What happens as \(m\) increases?

\begin{align*} \lim_{m \to \infty} i^{(m)} = \lim_{m\to \infty} m \left[ (1+i)^{1/m} -1 \right] = \ln (1 + i). \end{align*}

We call

\begin{align*} \delta = \lim_{m\to\infty} i^{(m)} = \ln (1 + i) \end{align*}

the force of interest.

Equivalently, \(i = e^{\delta} - 1, e^{\delta} = 1 + i\).

Continuous compounding - Continued

With such a notation, the accumulation function takes the form

\begin{align*} a(t) = e^{\delta t}. \end{align*}

One should imagine that the continuous compounding occurs at infinitesimally short time intervals …

Rates comparison

We also have

\begin{align*} \lim_{m \to \infty} d^{(m)} = \lim_{m\to\infty} \frac{i^{(m)}}{1 + \frac{i^{(m)}}{m}} = \lim_{m\to \infty} i^{(m)}=\delta. \end{align*}

If \(i>0\) and \(m>1\), then

\begin{align*} i > i^{(m)} \searrow \delta \nwarrow d^{(m)} > d. \end{align*}

Question: Can you prove that \(i^{(m)}\) is decreasing and \(d^{(m)}\) is increasing in \(m\)?

Force of interest at time \(t\)

Assuming that the interest rate is variable, you may be interested in looking at the interest rate over short periods of time. That interest rate is:

\begin{align*} i_{[t, t+1/m]} = \frac{a(t+1/m) - a(t)}{a(t)}, \end{align*}

and the nominal interest rate is

\begin{align*} \frac{\left(\frac{a(t + 1/m ) - a(t)}{a(t)}\right)}{1/m} = \frac{\left(\frac{a(t + 1/m ) - a(t)}{1/m}\right)}{a(t)}, \end{align*}

which as \(m \rightarrow \infty\) tends to

\begin{align*} \delta_t = \frac{a'(t)}{a(t)} = \frac{d}{dt} \ln a(t). \end{align*}

Force of interest at time \(t\) - Continued

We define the function of time

\begin{align*} \delta_t = \frac{a'(t)}{a(t)} = \frac{d}{dt} \ln a(t) \end{align*}

as the force of interest at time \(t\).

Special cases:

  • Simple interest: \(a(t) = 1 + st\), \(\displaystyle \delta_t = \frac{s}{1 + st}\).
  • Simple discount: \(a(t) = (1 - dt)^{-1}\), \(\displaystyle \delta_t = \frac{d}{1 - dt}\).
  • Compound interest: \(a(t) = (1 + i)^t\), \(\delta_t = \ln ( 1 + i ) = \delta\).

From \(\delta_t\) to \(a(t)\)

Recalling the definition of \(\delta_t\), and using the Fundamental Theorem of Calculus (simply put - we integrate both sides of the equality), we have

\begin{align*} a(t) = e^{\int_0^t \delta_r\, dr} = \exp \left( \int_0^t \delta_r\, dr \right). \end{align*}

In particular, if \(\delta_t = \delta\), then

\begin{align*} a(t) = \exp\left(\int_0^t \delta dt\right) = e^{t \delta}. \end{align*}