Time Value of Money

Accumulation and Amount Functions

Lenders are willing to lend money because the borrower pays them a fee called interest. The fundamental insight of financial mathematics is that money has different values at different times. A dollar today is worth more than a dollar in the future because today’s dollar can be invested to earn interest. We describe this time value using two closely related functions.

The accumulation function, denoted \(a(t)\), gives the accumulated value at time \(t\) of an investment of $1 made at time 0. By definition, \(a(0) \equiv 1\), and for positive interest rates the function is increasing. The amount function, \(A(t)\), gives the accumulated value at time \(t\) of an initial investment of \(K\) (called the principal):

\[ A(t) = K \cdot a(t) \]

The interest earned between times \(t_1\) and \(t_2\) is simply the change in the amount function: \(A(t_2) - A(t_1)\). In particular, the interest earned in the \(n\)-th year is \(A(n) - A(n-1)\).

Example. Given \(A(t) = 2t^2 + 3t + 10\), the principal is \(K = A(0) = 10\), so \(a(t) = A(t)/K = 0.2t^2 + 0.3t + 1\). The interest earned in the \(n\)-th year is \(A(n) - A(n-1) = 4n + 1\), a linearly increasing function — each year earns a dollar more in interest than the previous year.

Simple Interest

Under simple interest, the accumulation function is linear:

\[ a(t) = 1 + it \]

where \(i\) is the annual rate of interest. Simple interest is most often used over short time periods. When counting days, there are three conventions: the exact method uses 365 days per year; the ordinary method approximates each month as 30 days (360 per year); and the Banker’s rule counts exact days but divides by 360. In all cases, only one of the start or end date is counted, not both.

A key feature of simple interest is that the annual interest earned is constant at \(Ki\) regardless of which year you examine. This contrasts with compound interest, where interest is reinvested and grows over time.

Worked Example: Day-Count Convention. On May 6, 2019, Karen borrows $5,000 and repays $5,094 on June M, 2019, based on exact simple interest at 14.6%. Setting up the simple interest equation: \(5000(1 + 0.146 \cdot T/365) = 5094\) gives \(T = 47\) days. May has 31 days, so the days remaining in May after the 6th are \(31 - 6 = 25\) days; then \(47 - 25 = 22\) days into June. Karen repays on June 22nd.

Worked Example: Three Simple Interest Methods. Brad borrows $5,000 on October 14 at 8% simple interest, repaying on May 7. Under exact simple interest (denominator 365): count days from Oct 14 to May 7 as \(17 + 30 + 31 + 31 + 28 + 31 + 30 + 7 = 205\) days (assuming non-leap year), giving \(5000(1 + 0.08 \cdot 205/365) = \$5{,}224.66\). Under Banker’s rule (exact days, denominator 360): \(5000(1 + 0.08 \cdot 205/360) = \$5{,}227.78\). Under ordinary simple interest (each month = 30 days, denominator 360): the numerator becomes \(30 - 14 + 30 + 30 + 30 + 30 + 30 + 30 + 7 = 203\), giving \(5000(1 + 0.08 \cdot 203/360) = \$5{,}225.56\). Banker’s rule produces the highest repayment because more days of interest accrue per unit time.

Compound Interest

Under compound interest, the accumulation function is exponential:

\[ a(t) = (1+i)^t \]

Here interest is earned on both the principal and on the accumulated interest — “interest on interest.” Over short periods \((0 < t < 1)\), simple interest actually exceeds compound interest: \(1 + it > (1+i)^t\). For \(t > 1\) the relationship reverses, and compound interest grows faster.

The present value function (or discount function) is the reciprocal of the accumulation function:

\[ v(t) = \frac{1}{a(t)} \]

Under compound interest, \(v(t) = (1+i)^{-t}\). The factor \(v = (1+i)^{-1}\) is called the annual discount factor and is ubiquitous in actuarial notation. To find the present value of an amount \(X\) due at time \(t\), we compute \(X \cdot v(t)\).

Moving money from time \(t_1\) to time \(t_2\) requires multiplying by the ratio of accumulation functions. Under compound interest this simplifies to multiplying by \((1+i)^{t_2 - t_1}\).

Worked Example: Compound Interest Algebra. At a certain compound interest rate: $1 grows to $2 in \(A\) years, $2 grows to $3 in \(B\) years, and $3 grows to $15 in \(C\) years. Express the time \(N\) for $6 to grow to $10 in terms of \(A\), \(B\), \(C\).

\[\boxed{N = C - B - A}\]

. The key technique is factoring the target ratio into pieces that match the given accumulation factors.

Effective Rate of Interest and Discount

The annual effective rate of interest in the \(n\)-th year measures growth relative to the beginning-of-year balance:

\[ i_n = \frac{A(n) - A(n-1)}{A(n-1)} = \frac{a(n) - a(n-1)}{a(n-1)} \]

Under compound interest this is constant: \(i_n = i\) for all years. Under simple interest, \(i_n = i/(1 + i(n-1))\), a decreasing function — later years earn proportionally less.

The effective rate of discount, by contrast, measures the interest relative to the end-of-period balance:

\[ d_n = \frac{A(n) - A(n-1)}{A(n)} \]

Think of it this way: if you borrow $1000 for a year and the interest of $43 is collected upfront at time 0, you actually receive $957 but repay $1000. The discount rate \(d = 43/1000 = 4.3\%\), while the interest rate is \(i = 43/957 \approx 4.49\%\). Under compound interest these are related by:

\[ d = \frac{i}{1+i}, \qquad i = \frac{d}{1-d}, \qquad 1-d = \frac{1}{1+i} = v \]

The compound interest accumulation function can be written in terms of \(d\) as \(a(t) = (1-d)^{-t}\). Note that for the same underlying transaction, \(d < i\) always.

Worked Example: Finding Effective Rates from a Quadratic Accumulation Function. Suppose \(a(t) = 0.01t^2 + 0.03t + 1\). Find \(i_2\) (effective rate in year 2) and \(d_4\) (effective rate of discount in year 4).

For \(i_2\): the interest earned in year 2 is \(a(2) - a(1) = (0.04 + 0.06 + 1) - (0.01 + 0.03 + 1) = 1.10 - 1.04 = 0.06\), divided by the opening balance \(a(1) = 1.04\). So \(i_2 = 0.06/1.04 \approx 5.769\%\).

For \(d_4\): interest in year 4 is \(a(4) - a(3) = (0.16+0.12+1) - (0.09+0.09+1) = 1.28 - 1.18 = 0.10\), divided by \(a(4) = 1.28\). So \(d_4 = 0.10/1.28 \approx 7.813\%\).

Simple Discount and Compound Discount

Simple discount uses a linear present value (discount) function:

\[ v(t) = 1 - dt, \quad 0 \le t < \frac{1}{d} \]

giving the accumulation function \(a(t) = 1/(1-dt)\), which is nonlinear and only valid until \(t = 1/d\). Unlike simple interest (where the effective rate decreases over time), under simple discount the effective interest rate increases: \(i_n = d/(1-dn)\).

Compound discount uses an exponential discount function:

\[ v(t) = (1-d)^t \]

so \(a(t) = (1-d)^{-t}\). This is simply compound interest reparameterized using the effective rate of discount. There is no structural difference between compound interest and compound discount — they are two descriptions of the same exponential accumulation.

Worked Example: Simple vs. Compound Discount. Accumulate $3,500 for 5 years at discount rate \(d = 4.5\%\). Under simple discount: \(A = 3500/(1 - 0.045 \times 5) = 3500/0.775 = \$4{,}516.13\). Under compound discount: \(A = 3500(1-0.045)^{-5} = \$4{,}406.07\). Simple discount gives a higher accumulated value because the effective interest rate is increasing over time.

Comparing the effective annual rates: under simple discount, \(i_n = d/(1-dn)\), which grows with \(n\) — the rate is 4.71% in year 1, 4.95% in year 2, 5.81% in year 3, and so on. Under compound discount, the effective rate is constant at \(d/(1-d) = 0.045/0.955 \approx 4.71\%\) — identical to simple discount only for the first year.

Current Value

The current value of a cash flow at any point in time is found by moving its value forward or backward using the accumulation or discount function. Under compound interest this is particularly clean: to move a value of \(X\) from time \(t_1\) to time \(t_2\), multiply by \((1+i)^{t_2 - t_1}\) regardless of whether \(t_2 > t_1\) (future) or \(t_2 < t_1\) (past).

Under simple interest or simple discount, the current value depends on the original investment date, so we first recover the time-0 balance \(K = A(t_1)/a(t_1)\), then compute \(K \cdot a(t_2)\).

Example. Under simple interest with \(i = 6\%\), if the balance at time 4.5 is $1350, the balance at time 7 is:

\[ 1350 \cdot \frac{1 + 0.06 \times 7}{1 + 0.06 \times 4.5} = \$1{,}509.45 \]

Nominal Rates of Interest and Discount

In practice, interest is often quoted as an annual nominal rate but compounded more frequently. The annual nominal rate of interest compounded \(m\) times per year, denoted \(i^{(m)}\), means that \(i^{(m)}/m\) is the effective rate per \(1/m\) of a year. The accumulation function becomes:

\[ a(t) = \left(1 + \frac{i^{(m)}}{m}\right)^{mt} \]

Common values are \(m = 2\) (semiannual), \(m = 4\) (quarterly), \(m = 12\) (monthly). The annual effective rate is found by:

\[ 1 + i = \left(1 + \frac{i^{(m)}}{m}\right)^m \]

To compare rates with different compounding frequencies, convert them all to annual effective rates. When converting between two nominal rates \(i^{(m)}\) and \(i^{(k)}\), use:

\[ \left(1 + \frac{i^{(m)}}{m}\right)^m = \left(1 + \frac{i^{(k)}}{k}\right)^k \]

Similarly, the nominal rate of discount compounded \(m\) times per year, \(d^{(m)}\), gives the discount function \(v(t) = (1 - d^{(m)}/m)^{mt}\).

As \(m\) increases (more frequent compounding), the nominal rates approach the force of interest \(\delta\) from above for interest and below for discount. For a given effective rate \(i\):

\[ d < d^{(2)} < d^{(4)} < \cdots < \delta < \cdots < i^{(4)} < i^{(2)} < i \]

All these rates are equivalent — they produce the same accumulation.

Worked Example: Comparing Nominal Interest vs. Nominal Discount. A bank offers two rates: (a) \(i^{(12)} = 3\%\) and (b) \(d^{(12)} = 3\%\). Which is better for an investor?

Convert both to annual effective rates. For (a): \(1 + i = (1 + 0.03/12)^{12} = 1.0304\), so \(i = 3.04\%\). For (b): \(1 + i = (1 - 0.03/12)^{-12} = 1.0305\), so \(i = 3.05\%\). The nominal discount rate of 3% compounded monthly corresponds to a higher effective rate than the nominal interest rate of 3% compounded monthly. An investor should prefer the discount rate option. This illustrates the ordering \(i^{(m)} > d^{(m)}\) for the same quoted number — the discount rate convention always represents a higher effective yield.

Force of Interest

The force of interest is the instantaneous relative rate of change in the accumulation function:

\[ \delta_t = \frac{d}{dt} \ln a(t) = \frac{a'(t)}{a(t)} \]

The subscript \(t\) reminds us that the force can vary with time. The accumulation function can be recovered from the force of interest by integration:

\[ a(t) = e^{\int_0^t \delta_s\, ds} \]

Under simple interest, \(\delta_t = i/(1+it)\), a decreasing function. Under simple discount, \(\delta_t = d/(1-dt)\), an increasing function. Under compound interest, the force of interest is constant:

\[ \delta = \ln(1+i) = i^{(\infty)} \]

This is also called the constant force of interest, or equivalently the annual nominal rate with continuous compounding. The compound interest accumulation function can then be written as \(a(t) = e^{\delta t}\). The full hierarchy of compound interest parameterizations is:

\[ a(t) = (1+i)^t = (1-d)^{-t} = \left(1 + \frac{i^{(m)}}{m}\right)^{mt} = \left(1 - \frac{d^{(m)}}{m}\right)^{-mt} = e^{\delta t} \]

Example. Given a non-constant force of interest \(\delta_t = 0.03\sqrt{t}\), the value of $100 invested at time \(t = 1\) grown to time \(t = 4\) is:

\[ 100 \cdot e^{\int_1^4 0.03\sqrt{s}\, ds} = 100 \cdot e^{0.03 \cdot \frac{2}{3}[s^{3/2}]_1^4} = 100 \cdot e^{0.02(8-1)} = 100e^{0.14} \]

Key derivatives (useful in later modules):

• \(\frac{d}{di} v = -v^2\)
• \(\frac{d}{di} \delta = e^{-\delta} = v\)
• \(\frac{d}{d\delta} d = e^{-\delta} = 1 - d\)

Worked Example: Force of Interest from a Complex Accumulation Function. Given \(a(t) = \sqrt{(1+0.001t^2)(1-0.01t)^{-1}}\) for \(0 \le t < 100\), find \(\delta_5\).

Use \(\delta_t = \frac{d}{dt}\ln a(t)\). Take the natural log: \(\ln a(t) = 0.001t^2 - \ln(1-0.01t)\) (using the square root to halve the log, and properties of logs). Differentiating: \(\delta_t = 0.002t \cdot \frac{1}{2 \cdot 0.001t} \cdot \text{[chain rule]} \cdots\). For the specific function used in lecture, the result is \(\delta_t = 0.01 + 0.002t + \frac{0.01}{1-0.01t}\), so \(\delta_5 = 0.01 + 0.01 + 0.01/0.95 = 0.0305 = 3.05\%\).

Worked Example: Recovering a Balance from the Force of Interest. Given \(\delta_t = 0.02/(1-0.01t)\) and a balance of $20,000 at time 10, find the balance at time 4.

\[ B_4 = 20{,}000 \cdot \frac{a(4)}{a(10)} = 20{,}000 \cdot \frac{(1-0.04)^{-2}}{(1-0.10)^{-2}} = 20{,}000 \cdot \left(\frac{0.9}{0.96}\right)^2 = \$17{,}578.13 \]

The answer is less than $20,000 — as expected, since we are moving backward in time.

Inflation and Real Rate of Interest

When prices change over time, we distinguish between the nominal rate of interest \(i\) and the real rate of interest \(\tilde{i}\). If prices grow according to an inflation rate \(r\), the real accumulation function is the nominal accumulation divided by the price index:

\[ \tilde{a}(t) = \frac{a(t)}{p(t)} = \frac{(1+i)^t}{(1+r)^t} \]

The real annual effective rate is therefore:

\[ \tilde{i} = \frac{i - r}{1 + r} \]

This formula applies when both rates are effective with the same units. For monthly rates, replace \(i\) and \(r\) with \(i^{(12)}/12\) and \(r^{(12)}/12\) respectively. With continuous compounding, the formula simplifies beautifully: \(\tilde{\delta} = \delta - r^{(\infty)}\).

Worked Example: Real Rate with Different Compounding Frequencies. The interest rate is 8% per annum compounded semi-annually and the inflation rate is 3% per annum compounded monthly. Find the real rate compounded monthly.

\[ \tilde{j} = \frac{j_{\text{int}} - j_{\text{inf}}}{1 + j_{\text{inf}}} = \frac{0.655820\% - 0.25\%}{1.0025} = 0.4057\% \text{ per month} \]

Annualizing: \(i^{(12)} = 12 \times 0.4057\% = 4.87\%\). This is plausible — with 8% nominal interest and 3% inflation, the real rate is roughly 5%.

Equations of Value

An equation of value states that two sets of cash flows have the same value at some common reference date. The fundamental principle: if we accumulate (or discount) all cash flows to a common point in time using the same interest rate, equivalent cash flows have equal values.

For multiple cash flows, the present value of a stream with payments \(C_k\) at times \(t_k\) is:

\[ PV = \sum_k C_k \cdot v^{t_k} = \sum_k C_k (1+i)^{-t_k} \]

Example. A loan of $2000 is repaid by $750 after 1 year, $X after 1.5 years, and $1000 after 2 years with \(i^{(12)} = 6\%\). Setting the present value of payments equal to the loan:

\[ 2000 = 750 \cdot 1.005^{-12} + X \cdot 1.005^{-18} + 1000 \cdot 1.005^{-24} \implies X = \$444.56 \]\[ PV = 200(1.0075)^{-8} + 500(1.0075)^{-16} = \$632.05 \]

Now find the semiannual nominal rate \(i^{(2)}\) such that the PV is $600. Writing \(X = v^2\) (where \(v = (1+i^{(2)}/2)^{-1}\), the equation \(500X^2 + 200X = 600\) simplifies to \(5X^2 + 2X - 6 = 0\). The positive root is \(X = 0.91355\). Since \(X = (1 + i^{(2)}/2)^{-2}\), solving gives \(i^{(2)} = 4.57\%\). Always use the full precision of \(X\) when solving for the rate — rounding intermediate results introduces error.

Worked Example: Equation of Value as a Quadratic. Brent deposits $5,000 at time 0 and $3,000 at time 9 months. After 18 months his balance is $10,726.51. Find the annual nominal rate compounded monthly. Let \(j\) be the monthly effective rate; letting \(X = (1+j)^9\), the equation of value at 18 months is \(5000X^2 + 3000X = 10{,}726.51\). The positive root is \(X = 1.195093\), so \(j = 1.195093^{1/9} - 1 = 2.00\%\) per month, giving \(i^{(12)} = 24.00\%\).

Net Present Value and Internal Rate of Return

The net present value (NPV) of an investment is the present value of all net cash flows (positive for inflows, negative for outflows):

\[ NPV(i) = C_0 + C_1 v + C_2 v^2 + \cdots + C_n v^n \]

The internal rate of return (IRR) is the interest rate \(i\) that makes the NPV equal to zero. It represents the yield earned by the investor.

For a simple loan (lend then receive repayments), the cash flows are initially negative (outflow) followed by positive. The balance at time \(t\) under the IRR is:

\[ B_t = \sum_{k=0}^{t} C_k (1+i)^{t-k} \]

Uniqueness theorem: If the balance \(B_t > 0\) for all \(t = 0, 1, \ldots, n-1\) and \(B_n = 0\), then the IRR is unique. Economically, this means the roles of borrower and lender remain fixed throughout — the investor is always in a net lending position.

Reinvestment: The IRR can only be earned if all intermediate proceeds are reinvested at the IRR rate. If reinvestment occurs at a different rate \(j\), the actual yield will differ from the IRR.

Time Value of Money Summary

The key relationships in this module are:

Parameter	Relationship
Effective interest & discount	\(d = i/(1+i)\), \(i = d/(1-d)\)
Compound interest parameterizations	\((1+i)^t = (1-d)^{-t} = e^{\delta t}\)
Force of interest	\(\delta = \ln(1+i) = i^{(\infty)}\)
Real rate	\(\tilde{i} = (i-r)/(1+r)\)
Nominal ↔ effective	\((1 + i^{(m)}/m)^m = 1+i\)

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Annuities

Geometric Progressions

A geometric progression is a sequence where each term is a constant multiple of the previous: \(\{a, ar, ar^2, \ldots\}\). The sum of the first \(n\) terms is:

\[ \sum_{t=1}^n ar^{t-1} = \frac{a(1-r^n)}{1-r}, \quad r \ne 1 \]

This formula underlies virtually every annuity formula in this module. A regular series of level payments is an annuity; its present value is the sum of a geometric progression where each term is \(v = 1/(1+i)\) times the previous.

Annuity-Immediate

An annuity-immediate (also called an ordinary annuity) consists of payments of $1 at the end of each period for \(n\) periods. The present value (valued one period before the first payment) uses actuarial notation \(a_{\overline{n}|}\):

\[ a_{\overline{n}|} = v + v^2 + \cdots + v^n = \frac{1 - v^n}{i}, \quad i \ne 0 \]

The accumulated value (at the time of the last payment) uses \(s_{\overline{n}|}\):

\[ s_{\overline{n}|} = a_{\overline{n}|} \cdot (1+i)^n = \frac{(1+i)^n - 1}{i}, \quad i \ne 0 \]

For a payment amount of $R, multiply each formula by $R. The key equation linking PV and AV is:

\[ s_{\overline{n}|} = a_{\overline{n}|} \cdot (1+i)^n \]

Worked Example: Counting Payments Carefully. On July 10, 2010, Susan buys an annuity paying $1,000 every 3 months, with the first payment on October 10, 2010, and the final payment on April 10, 2019, priced at \(i^{(4)} = 5.2\%\). How much does she pay?

\[ P = 1000 \cdot a_{\overline{35}|1.3\%} = 1000 \cdot \frac{1 - 1.013^{-35}}{0.013} = \$27{,}976.08 \]

The total paid is $35,000 but the present value is much less because of discounting.

Worked Example: Calculating Accumulated Value with Changing Rates. Heather deposits $925 at the end of each quarter for 10 years. For the first 2 years the rate is 1% compounded quarterly; starting year 3 it rises to 2% compounded quarterly. The effective quarterly rates are \(j_1 = 0.25\%\) and \(j_2 = 0.5\%\).

Step 1: Accumulated value of first 8 deposits at end of year 2: \(925 \cdot s_{\overline{8}|0.25\%} = \$7{,}465.07\). Step 2: Grow this forward 32 quarters (8 more years) at 0.5%: \(7465.07 \times 1.005^{32} = \$8{,}756.85\). Step 3: Accumulated value of the remaining 32 deposits at end of year 10: \(925 \cdot s_{\overline{32}|0.5\%} = \$32{,}012.98\). Total: \(\$8{,}756.85 + \$32{,}012.98 = \$40{,}769.83\).

Annuity-Due

An annuity-due consists of payments at the beginning of each period. The present value (at the time of the first payment) uses \(\ddot{a}_{\overline{n}|}\):

\[ \ddot{a}_{\overline{n}|} = 1 + v + v^2 + \cdots + v^{n-1} = \frac{1 - v^n}{d}, \quad i \ne 0 \]

The accumulated value (one period after the last payment) uses \(\ddot{s}_{\overline{n}|}\):

\[ \ddot{s}_{\overline{n}|} = \frac{(1+i)^n - 1}{d} \]

The annuity-due and annuity-immediate are related by exactly one period of interest:

\[ \ddot{a}_{\overline{n}|} = (1+i) \cdot a_{\overline{n}|}, \qquad \ddot{s}_{\overline{n}|} = (1+i) \cdot s_{\overline{n}|} \]

Intuitively, payments at the beginning of each period are each worth one period of interest more than the equivalent end-of-period payments.

Perpetuities

A perpetuity is an annuity with infinitely many payments. The present value of a perpetuity-immediate (first payment in one period) is:

\[ a_{\overline{\infty}|} = \lim_{n \to \infty} a_{\overline{n}|} = \frac{1}{i}, \quad i > 0 \]

For a perpetuity-due (first payment now):

\[ \ddot{a}_{\overline{\infty}|} = \frac{1+i}{i} = \frac{1}{d} \]

When the first payment is not exactly one full period away, adjust the perpetuity value by accumulating or discounting to the correct starting point. If the first payment is in \(k\) periods from now and the payment amount is $R:

\[ PV = \frac{R}{i} \cdot v^{k-1} \]

(discount the perpetuity-immediate value back \(k-1\) periods to today).

Worked Example: Endowment Fund with Several Perpetuity Variants. Emily donates $100,000 to establish annual scholarships, invested at \(i = 8\%\).

(a) First payment in one year: This is a perpetuity-immediate. The annual scholarship is \(R = 100{,}000 \times 0.08 = \$8{,}000\).

(b) First payment immediately: This is a perpetuity-due. Then \(100{,}000 = R/d = R(1+i)/i\), so \(R = 100{,}000 \times 0.08/1.08 = \$7{,}407.41\). Less than part (a) — the fund pays out immediately without first earning a full year of interest.

(c) First payment in 6 months: The payments are at times \(0.5, 1.5, 2.5, \ldots\). Factoring out \(v^{-1/2}\) from the perpetuity sum yields \(PV = R(1+i)^{1/2}/i\). Setting this to $100,000: \(R = 100{,}000 \times 0.08/\sqrt{1.08} = \$7{,}698.00\). This falls between parts (a) and (b).

(d) First payment in T years, scholarship $10,000: The payments start at time \(T\). The present value is \(10{,}000 \cdot v^{T-1}/i = 100{,}000\), giving \(1.08^{T-1} = 1.35\), so \(T = 1 + \ln(1.35)/\ln(1.08) = 4.899\) years.

Deferred Annuities

An \(m\)-year deferred, \(n\)-year annuity-immediate pays $R at the end of years \(m+1, m+2, \ldots, m+n\). The present value (at time 0) is:

\[ PV = v^m \cdot a_{\overline{n}|} = a_{\overline{m+n}|} - a_{\overline{m}|} \quad (m \text{ a positive integer}) \]

The second formula interprets the deferred annuity as the difference of two annuities starting now. A similarly useful identity: the value of an annuity \(m\) periods after its last payment is \((1+i)^m s_{\overline{n}|} = s_{\overline{m+n}|} - s_{\overline{m}|}\).

Nonlevel Annuities

When payment amounts vary, we decompose them into a sum of level annuities or handle each period’s rate separately. Two common techniques:

1. Group by payment level: Write the cash flows as a combination of level annuities and deferred annuities. For example, payments of $75/month for year 1, $50/month for year 2, and $25/month for year 3 can be written as three deferred annuities or as \(25a_{\overline{12}|j} + 25a_{\overline{24}|j} + 25a_{\overline{36}|j}\).

2. Separate by period: When the interest rate changes midway, calculate the accumulated value of the first batch, then grow it to the end under the new rate before adding the second batch’s value.

Worked Example: Nonlevel Payments — Two Methods. An annuity pays $200/month for 2 years, $300/month for 1 year, then $400/month for 2 years at \(i^{(12)} = 10\%\). Let \(j = 10\%/12 = 0.8\overline{3}\%\) per month.

\[ PV = 200a_{\overline{24}|j} + 300v^{24}a_{\overline{12}|j} + 400v^{36}a_{\overline{24}|j} = \$13{,}559.94 \]\[ PV = 400a_{\overline{60}|j} - 100a_{\overline{36}|j} - 100a_{\overline{24}|j} = \$13{,}559.94 \]

Method 2 avoids deferred annuities and is often faster on a financial calculator.

Payments in Geometric Progression

When payments grow at a constant rate \(g\) per period, the first payment being $P, the present value (for a finite \(n\)-period annuity-immediate) is:

\[ PV = \frac{P}{1+g} \cdot a_{\overline{n}|i^*}, \quad \text{where } i^* = \frac{i - g}{1 + g} \]

This is the real rate of interest applied to an ordinary annuity of amount \(P/(1+g)\). Alternatively, write it directly as a geometric series:

\[ PV = P \cdot v + P(1+g) \cdot v^2 + \cdots + P(1+g)^{n-1} \cdot v^n = \frac{P}{1+i} \cdot \frac{1 - \left(\frac{1+g}{1+i}\right)^n}{1 - \frac{1+g}{1+i}} \]

When \(n \to \infty\) and \(g < i\), this converges to the Gordon growth model: \(PV = P/(i-g)\), which reappears in Module 5 for stock valuation.

\[ PV = \frac{20000v\left(1-(1.05v)^{10}\right)}{1-1.05v} = \$180{,}867.50 \]

Alternatively, observe that the real rate \(j^* = (i-g)/(1+g) = 0.01/1.05 \approx 0.9524\%\), and rewrite as \(PV = (20000/1.05) \cdot a_{\overline{10}|j^*}\) — this converts the growing annuity into a level annuity at the real rate, which is especially convenient with a financial calculator.

Payments in Arithmetic Progression (P–Q Formula)

For an annuity-immediate where the first payment is $P and each subsequent payment increases by $Q, the present value is:

\[ PV = P \cdot a_{\overline{n}|} + Q \cdot \frac{a_{\overline{n}|} - n v^n}{i} \]

This is called the P–Q formula. For an annuity-due, multiply by \((1+i)\). Special case when \(P = Q = 1\) defines the increasing annuity \((Ia)_{\overline{n}|}\):

\[ (Ia)_{\overline{n}|} = \frac{\ddot{a}_{\overline{n}|} - nv^n}{i} \]\[ X = 2 a_{\overline{60}|j} + \frac{2}{j}\left(a_{\overline{60}|j} - 60v^{60}\right) = \$2{,}729.21 \]

Rounding \(j\) to fewer decimal places at an intermediate step will produce a noticeably wrong final answer because \(j\) appears raised to the 60th power.

Worked Example: Arithmetic Increasing Perpetuity. A perpetuity-immediate has first payment $3 and each subsequent payment $2 larger. The PV is $406.81. Find \(i\).

\[ 406.81i^2 - 3i - 2 = 0 \]

The positive root is \(i = (3 + \sqrt{9 + 4 \times 406.81 \times 2})/(2 \times 406.81) = 7.39\%\). The negative root is discarded since a perpetuity requires \(i > 0\).

Example (ladder payments). An annuity paying $n at end of year \(n\) for \(n = 1, 2, \ldots, 50\) and $(100−n) for \(n = 51, \ldots, 99\) has a neat closed form. Multiplying the PV by \(v\) and subtracting gives \(PV(1-v) = a_{\overline{50}|} - v^{50} a_{\overline{50}|}\), so \(PV = (1+i)(a_{\overline{50}|})^2\).

Determining the Interest Rate

Given a present or accumulated value and a set of payments, finding the interest rate requires solving a polynomial equation. The guess-and-check (bisection) method works as follows: note whether the target is above or below the first guess, narrow the interval, and iterate. A financial calculator or spreadsheet solver handles this in practice.

For a loan with payment \(R\) and present value \(L\):

\[ L = R \cdot a_{\overline{n}|i} \implies a_{\overline{n}|i} = L/R \]

Solve for \(i\) numerically.

Term of an Annuity and Final Payment

When solving for the number of payments \(n\), the solution is typically non-integer. Two methods handle this:

1. Smaller final payment: Make \(\lfloor n \rfloor\) full payments of $R plus a smaller final payment at the same payment date.
2. Full final payment: Make \(\lceil n \rceil\) payments of $R, with the last one reduced below $R.

For an annuity-immediate with target value \(FV\):

\[ n = \frac{\ln(FV \cdot i / R + 1)}{\ln(1+i)} \quad \text{(accumulated value)} \]\[ n = \frac{-\ln(1 - L \cdot i / R)}{\ln(1+i)} \quad \text{(present value)} \]

Worked Example: Finding the Term and Smaller Final Payment. You borrow $10,000 repaid by level end-of-year payments of $1,000 at \(i = 4\%\). Find the number of full payments and the smaller final payment.

\[ x = (10{,}000 - 1000 a_{\overline{13}|4\%}) \times 1.04^{14} = \$24.85 \]

A quick sanity check: \(n = 13.02\) is just 0.02 beyond 13, so we expect the final payment to be roughly \(0.02 \times 1000 = \$20\), and indeed $24.85 is in that vicinity.

\[ 20\ddot{s}_{\overline{6}|} \cdot 1.1^5 + X\ddot{s}_{\overline{3}|} \cdot 1.1^4 = 500 \]

(The annuity-due accumulated value is placed one period after the last payment, then grown to time 10.) Solving: \(X = \$51.89\).

Annuities Using a General Accumulation Function

For a general accumulation function \(a(t)\) (not necessarily compound interest), the annuity formulas generalize as:

\[ a_{\overline{n}|} = v(1) + v(2) + \cdots + v(n), \quad s_{\overline{n}|} = a(n) \cdot a_{\overline{n}|} = \frac{a(n)}{a(1)} + \frac{a(n)}{a(2)} + \cdots + 1 \]

Example. A fund with year-by-year returns 5.2%, 4.8%, 3.8%, 6.1%, 5.5% and deposits of $100 at the start of each year: the balance at end of year 5 is computed by growing each deposit forward using the product of the relevant year-by-year accumulation factors.

Worked Example: Deposits Every 4 Years. Catherine deposits $100 at the beginning of each 4-year period for 40 years at annual effective rate \(i\). The amount at year 40 is 5 times the amount at year 20. Find the amount at year 40.

\[ \frac{100\ddot{s}_{\overline{10}|j}}{100\ddot{s}_{\overline{5}|j}} = 5 \]\[ X = 100\ddot{s}_{\overline{10}|j} = 100 \cdot \frac{(1+j)^{10}-1}{1-(1+j)^{-1}} = 100 \cdot \frac{4^2 - 1}{1 - 4^{-1/5}} = \$6{,}194.72 \]

Payment Frequency

When the payment frequency does not match the interest conversion frequency, always convert the interest rate to match the payment frequency. If payments are \(k\) times per year and the given rate is \(i^{(m)}\), the effective rate per payment period is:

\[ j = \left(1 + \frac{i^{(m)}}{m}\right)^{m/k} - 1 \]

Then proceed with the standard annuity formulas using \(j\) as the rate per period and the total number of payments as \(n\).

Continuous Annuities

As the payment frequency \(m \to \infty\), the present value of an annuity converges to the continuous annuity formula. An \(n\)-year annuity paying at a continuous rate of $1 per year has present value:

\[ \bar{a}_{\overline{n}|} = \frac{1 - v^n}{\delta} \]

where \(\delta = \ln(1+i)\) is the force of interest. The corresponding accumulated value is \(\bar{s}_{\overline{n}|} = (e^{\delta n} - 1)/\delta\).

Continuous annuities can also be interpreted via integration: for a fund paying continuously at rate \(1\) per year:

\[ \bar{a}_{\overline{n}|} = \int_0^n v^t\, dt = \int_0^n e^{-\delta t}\, dt \]

This integral representation is especially useful when the discount function is non-standard (e.g., simple discount).

\[ 43{,}000 = X \bar{a}_{\overline{15.5}|} = X \cdot \frac{1 - v^{15.5}}{\delta} \]

where \(\delta = \ln(1.04)\). Solving: \(X = 43{,}000\delta/(1 - 1.04^{-15.5}) = \$3{,}702.35\) per year.

Worked Example: Time to Exhaust a Fund. A fund of $1,600 accumulates at \(\delta = 5.5\%\) (continuously compounded) with continuous withdrawals of $150/year. Find when the fund is exhausted.

\[ 1600 = 150 \cdot \frac{1 - e^{-0.055n}}{0.055} \implies e^{-0.055n} = 1 - \frac{1600 \times 0.055}{150} = 0.4133 \implies n = 16.064 \text{ years} \]

Continuously Varying Continuous Annuities

For an annuity paying continuously at a varying rate \(r(t)\) per year at time \(t\), the present value is:

\[ PV = \int_0^n r(t) \cdot v(t)\, dt \]

where \(v(t) = 1/a(t)\) is the discount function. Under compound interest, \(v(t) = e^{-\delta t}\).

An important special case is the increasing continuous annuity paying at rate \(r(t) = t\):

\[ \bar{(I\bar{a})}_{\overline{n}|} = \int_0^n t e^{-\delta t}\, dt = \frac{\bar{a}_{\overline{n}|} - nv^n}{\delta} \]

Worked Example: Varying Payment Rate with Non-Constant Force of Interest. Deposits flow into an account at rate \((7k + tk)\) dollars per year at time \(t\), with force of interest \(\delta_t = 1/(7+t)\). After 10 years the account is worth $20,000. Find \(k\).

\[ PV = \int_0^{10} k(7+t) \cdot \frac{7}{7+t}\,dt = \int_0^{10} 7k\,dt = 70k \]

The accumulated value at time 10 is \(70k \cdot a(10) = 70k \cdot 17/7 = 170k = 20{,}000\), so \(k = \$117.65\).

Key derivatives of annuity factors (useful in immunization, Module 6):

\[ \frac{d}{di} a_{\overline{n}|} = -v\left(a_{\overline{n}|} + \frac{nv^n}{i}\right), \qquad \frac{d}{d\delta} \bar{a}_{\overline{n}|} = -\bar{(I\bar{a})}_{\overline{n}|}, \qquad \frac{d}{dn} \bar{a}_{\overline{n}|} = v^n, \qquad \frac{d}{dn} \bar{s}_{\overline{n}|} = e^{\delta n} \]

Annuities Summary

Annuity Type	Present Value	Accumulated Value
Immediate	(a_{\overline{n}	} = (1-v^n)/i)
Due	(\ddot{a}_{\overline{n}	} = (1-v^n)/d)
Continuous	(\bar{a}_{\overline{n}	} = (1-v^n)/\delta)
Perpetuity-immediate	\(1/i\)	—
Perpetuity-due	\((1+i)/i = 1/d\)	—
\(m\)-deferred \(n\)-immediate	(v^m a_{\overline{n}	} = a_{\overline{m+n}
P–Q (arith. progression)	(Pa_{\overline{n}	} + Q(a_{\overline{n}
Geom. growth (\(P, g\)	\(\frac{P}{i-g}(1-((1+g)/(1+i))^n)\)	—

Key relationship between due and immediate: \(\ddot{a}_{\overline{n}|} = (1+i) a_{\overline{n}|}\) and \(\ddot{a}_{\overline{n}|} = 1 + a_{\overline{n-1}|}\).

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Loans

Outstanding Loan Balance

Consider a loan of amount \(L\) repaid by \(n\) level payments of \(R\) at the end of each period with effective rate \(i\) per period. The initial condition is:

\[ L = R \cdot a_{\overline{n}|} \]

At any time \(t\) during repayment, the outstanding loan balance can be computed by two equivalent methods:

\[ B_t = L(1+i)^t - R \cdot s_{\overline{t}|} \]\[ B_t = R \cdot a_{\overline{n-t}|} \]

Both give the same answer when payments are exactly level. If payments are rounded (e.g., to the nearest cent), the retrospective method is more accurate for intermediate balances; the prospective method gives the correct value only if we adjust the final payment. With a slightly smaller final payment \(R^*\), the prospective formula is:

\[ B_t = R \cdot a_{\overline{n-t-1}|} + R^* (1+i)^{-(n-t)} \]

Mortgage example. A $480,000 mortgage at \(i^{(2)} = 7.6\%\) over 25 years has a bi-weekly effective rate of \(j = (1 + 0.076/2)^{1/13} - 1 = 0.2873\%\) and bi-weekly payment \(R = \$1{,}631.88\). Choosing a shorter amortization period (20 or 15 years) dramatically reduces total interest paid — from $580,722 (25yr) to $318,736 (15yr).

Worked Example: Retrospective vs. Prospective Balance. A $20,000 loan at \(i = 8\%\) is repaid by $2,500/year plus a smaller final payment. First find \(n\): solving \(a_{\overline{n}|8\%} = 8\) gives \(v^n = 0.36\), so \(n = -\ln(0.36)/\ln(1.08) = 13.27\). Thus 13 full payments plus a smaller 14th payment \(R^* = (20000 - 2500a_{\overline{13}|8\%}) \times 1.08^{14} = \$706.57\) (about 0.27 × $2,500, as expected from the fractional part).

The outstanding balance at end of year 6 via the retrospective method: \(B_6 = 20000(1.08)^6 - 2500s_{\overline{6}|8\%} = \$13{,}397.66\). Via the prospective method: \(B_6 = 2500a_{\overline{7}|8\%} + 706.57 \cdot 1.08^{-8} = \$13{,}397.66\). Both methods agree exactly.

Amortization of a Debt

Under the amortization method, each payment \(R\) splits into interest and principal portions. At time \(t\):

• Interest paid: \(I_t = i \cdot B_{t-1} = R(1 - v^{n-t+1})\)
• Principal repaid: \(P_t = R - I_t = R \cdot v^{n-t+1}\)
• Updated balance: \(B_t = B_{t-1} - P_t = R \cdot a_{\overline{n-t}|}\)

The principal repaid grows geometrically: each period the principal portion increases by a factor of \((1+i)\):

\[ P_{t+1} = (1+i) P_t \]

The complete amortization schedule:

Time	Payment	Interest	Principal	Balance
0	—	—	—	(Ra_{\overline{n}
1	\(R\)	\(R(1-v^n)\)	\(Rv^n\)	(Ra_{\overline{n-1}
\(t\)	\(R\)	\(R(1-v^{n-t+1})\)	\(Rv^{n-t+1}\)	(Ra_{\overline{n-t}
\(n\)	\(R\)	\(R(1-v)\)	\(Rv\)	0

Total interest paid = \(nR - L\).

Amortization Examples

Given any two adjacent rows of an amortization table, the interest rate can be recovered. Since \(P_{t+1} = (1+i) P_t\):

\[ i = \frac{P_{t+1}}{P_t} - 1 \]

Given the interest paid at time \(t+1\), the balance at time \(t\) is:

\[ B_t = \frac{I_{t+1}}{i} \]

These relationships allow reconstruction of a full amortization table from partial information. For mortgages, the principal in the 25^th payment and the 37^th payment differ by 12 periods of compounding: \(P_{37} = P_{25}(1+i)^{12}\), allowing us to solve for the effective rate.

Worked Example: Recovering Amortization Data from Two Balances. A 4-year loan at \(i = 8\%\) with level end-of-year payments has outstanding balance $1,076.82 at end of year 2 and $559.12 at end of year 3. Since principal portions grow geometrically, the interest in year 3 is \(I_3 = 0.08 \times 1076.82 = 86.15\). The principal in year 3 is \(P_3 = B_2 - B_3 = 1076.82 - 559.12 = 517.70\). The payment is \(R = I_3 + P_3 = \$603.85\). Since \(P_t/(1+i) = P_{t-1}\), the principal at time 1 is \(P_1 = 517.70/1.08^2 = \$443.84\).

Refinancing a Loan

When interest rates fall, it may be worth refinancing — paying a penalty to break the existing loan and take out a new loan at the lower rate. The analysis compares:

1. The outstanding balance plus any prepayment penalty as the new “amount borrowed”
2. The new payment under the new rate and remaining term

If the new payment is lower, refinancing saves money. The total savings depend on the penalty size and how much lower the new rate is.

Canadian mortgages are typically renegotiated every 5 years when the rate guarantee period expires. The penalty for breaking early is often 3 months’ interest on the outstanding balance.

Worked Example: Should Kenny Refinance? Kenny buys $5,000 of furniture, pays $500 down, and finances $4,500 over 5 years at \(i^{(12)} = 12\%\) (\(j = 1\%\)/month). Monthly payment: \(R = 4500/a_{\overline{60}|1\%} = \$100.10\). After 24 months, the retrospective balance is \(B_{24} = 4500(1.01)^{24} - 100.10 \cdot s_{\overline{24}|1\%} = \$3{,}013.76\).

Kenny can refinance at \(i^{(12)} = 8.4\%\) (\(j^* = 0.7\%\)/month) but must pay a 3-month payment penalty. New loan: \(L^* = 3013.76 + 3 \times 100.10 = \$3{,}314.06\). New payment over 36 remaining months: \(R^* = 3314.06/a_{\overline{36}|0.7\%} = \$104.46\). Since $104.46 > $100.10, refinancing is not worthwhile — the penalty outweighs the interest savings.

Sinking Funds

With the sinking fund method, the borrower pays only interest each period to the lender, then repays all principal in one lump sum at the end. To fund the final payment, the borrower makes regular deposits into a separate sinking fund earning rate \(j\):

\[ D = \frac{L}{s_{\overline{n}|j}} \]

The total cost per period is the interest payment plus the sinking fund deposit:

\[ iL + D = iL + \frac{L}{s_{\overline{n}|j}} \]

When \(j = i\), this equals \(L/a_{\overline{n}|}\) — exactly the same as the amortization method payment. When the sinking fund earns less than the loan rate \((j < i)\), the sinking fund method costs more. The outstanding “principal” at time \(t\) is the original loan minus the sinking fund balance: \(L - Ds_{\overline{t}|j}\).

Worked Example: Sinking Fund with Partial Amortization. A 12-year, $8,000 loan at 8% charges interest of $640/year, but the borrower pays $800/year to the lender (partly reducing principal). At the end of 12 years, the remaining balance — computed retrospectively as \(8000(1.08)^{12} - 800s_{\overline{12}|8\%} = \$4{,}963.66\) — must be repaid via a sinking fund earning 4%. The required annual sinking fund deposit is \(X = 4963.66/s_{\overline{12}|4\%} = \$330.34\).

Worked Example: Mutual Fund Interest into a Bank Account. Betty invests $10,000 in a mutual fund earning 8%. She withdraws the $800 annual interest and deposits it in a bank account earning 4%. The mutual fund balance stays at $10,000 throughout (interest is removed, not reinvested). After 10 years, the bank account holds \(800 \cdot s_{\overline{10}|4\%} = \$9{,}604.89\). Total wealth: $10,000 + $9,604 = $19,604.

Loans Summary

Principal repaid between times \(t_1\) and \(t_2\): \(B_{t_1} - B_{t_2}\)

Interest paid between times \(t_1\) and \(t_2\): \(R(t_2 - t_1) - (B_{t_1} - B_{t_2})\)

Sinking fund deposit: \(D = L/s_{\overline{n}|j}\), where \(j\) is the sinking fund rate

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Bonds

Terminology and Basic Price Formula

A bond is a loan where the borrower (issuer) promises to pay:

• Periodic coupons of amount \(Fr\), where \(F\) is the face value (par value) and \(r\) is the coupon rate per period
• A redemption value \(C\) at maturity (usually \(C = F\), called redeemable at par)

The yield to maturity \(j\) is the effective rate per coupon period. The bond price is the present value of all payments at this yield:

\[ P = Fr \cdot a_{\overline{n}|j} + C v^n \]

This basic bond price formula contains five quantities: \(P, Fr, n, C, j\). Given any four, the fifth can be found. For semiannual bonds, the annual yield is quoted as \(i^{(2)} = 2j\).

\[ P = 40a_{\overline{10}|j} + 45v^{10}a_{\overline{10}|j} + 50v^{20}a_{\overline{20}|j} + 1000v^{40} = \$968.72 \]

Since the weighted average coupon rate (about 9.25%) is close to the yield (9.2%), the price is near par.

Premium and Discount Pricing Formula

Using the identity \(v^n = 1 - ja_{\overline{n}|}\), the basic formula rewrites as the premium/discount formula:

\[ P = C + (Fr - Cj) a_{\overline{n}|} \]

• Premium bond: \(Fr > Cj \implies P > C\). For par bonds (\(C = F\): \(r > j \implies P > F\)
• Discount bond: \(Fr < Cj \implies P < C\). For par bonds: \(r < j \implies P < F\)
• Par bond: \(Fr = Cj \implies P = C\). For par bonds: \(r = j \implies P = F\)

The amount of premium is \(P - C = (Fr - Cj)a_{\overline{n}|}\). The amount of discount is \(C - P\).

Worked Example: Using the Premium/Discount Formula. A 28-year $1,200 par bond pays annual coupons at rate \(r = 2i\) (double the yield). Bart pays $1,968. Find the sale price after 7 years.

\[ 1968 = 1200 + 1200(1 - v^{28}) \implies v^{28} = 0.36 \]

At time 7 with 21 coupons remaining: \(B_7 = 1200 + 1200(1 - v^{21}) = 2400 - 1200 \times 0.36^{21/28} = \$1{,}842.29\). The key trick: the premium/discount formula absorbs the interest rate into \(v^n\), letting us solve without explicitly finding \(i\).

Bond Amortization

A bond is analogous to a loan from the bondholder’s perspective. The book value at time \(t\) is the present value of remaining payments at the original yield:

\[ B_t = Fr \cdot a_{\overline{n-t}|j} + C v^{n-t} = C + (Fr - Cj) a_{\overline{n-t}|j} \]

At each coupon date, the interest earned is \(I_t = j B_{t-1} = Fr - (Fr - Cj)v^{n-t+1}\), and the book value adjustment (amount written down or up) is:

\[ P_t = Fr - I_t = (Fr - Cj) v^{n-t+1} \]

Book value adjustments grow geometrically: \(P_{t+1} = (1+j) P_t\). For a premium bond, book values decrease toward \(C\) (writing down the premium). For a discount bond, book values increase toward \(C\) (writing up the discount).

\[ B_6 = 800a_{\overline{4}|6\%} + 10000(1.06)^{-4} = \$10{,}693.02 \]

The interest portion of the 7th coupon is \(I_7 = 0.06 \times 10{,}693.02 = \$641.58\). The remaining \(800 - 641.58 = \$158.42\) is the premium writedown (principal adjustment).

Book Value Between Coupon Dates

Between coupon dates, two price concepts apply:

Dirty price (actual purchase price): accumulate the most recent coupon-date price for fraction \(f\) of a period:

\[ P_f = (1+j)^f P_0 \]

where \(P_0\) is the price at the last coupon date (using current market yield \(j\) and \(f = \text{days since last coupon}/\text{days in coupon period}\).

Semi-practical clean price (used for quoting): subtract the accrued coupon:

\[ \text{Clean price} = P_f - f \cdot Fr \]

Bond quotes are the clean price per $100 of face value. The clean price removes the effect of where we are in the coupon cycle, making bonds with the same yield but different coupon dates more comparable.

\[ B_{t+f} = (1-f) B_t + f B_{t+1} \]

Worked Example: Dirty and Clean Price. A $1,000 par bond redeemable December 1, 2021, with 7% semi-annual coupons. Find the dirty and clean price on August 8, 2010, to yield \(i^{(2)} = 6\%\) (\(j = 3\%\). Coupon = $35.

The last coupon date is June 1, 2010 (time 0). From June 1 to December 1, 2021, there are 23 remaining coupons. The price at June 1: \(B_0 = 35a_{\overline{23}|3\%} + 1000(1.03)^{-23} = \$1{,}082.22\).

Count days: June 1 to August 8 = 68 days; June 1 to December 1 = 183 days. The fraction is \(f = 68/183\).

Dirty price: \(P_f = 1082.22 \times 1.03^{68/183} = \$1{,}094.17\).

Clean price: \(1094.17 - (68/183) \times 35 = \$1{,}081.16\).

Determining the Yield Rate

When a bond is bought and then sold before maturity, the investor’s yield over the holding period is found by solving for the rate that equates the purchase price to the present value of received cash flows (coupons plus sale proceeds). Use a financial calculator: enter PV (negative), payments, FV, and compute the rate.

Callable Bonds

A callable bond gives the issuer the right to redeem the bond early on specified dates, usually at slightly higher redemption values. This is analogous to a mortgage prepayment privilege.

When purchasing a callable bond to guarantee a minimum yield:

• Calculate the price at each possible call date assuming the desired yield rate
• The minimum of all these prices guarantees the desired minimum yield

Reasoning: if the issuer calls early when it is in their interest (when rates have fallen), the investor gets a higher return. If the issuer calls at the date giving the lowest price, the investor just earns the desired yield. Any other call date yields more.

For a premium bond (\(Fr > Cj\), the earlier call date typically produces the lower price (shorter annuity of the premium). For a discount bond, the later maturity date produces the lower price. When the call redemption value differs from the maturity redemption value, compute the price at every possible call/maturity date and take the minimum.

Bonds Summary

Formula	Expression
Basic price	(P = Fra_{\overline{n}
Premium/discount	(P = C + (Fr - Cj)a_{\overline{n}
Book value at time \(t\)	(B_t = C + (Fr - Cj)a_{\overline{n-t}
Book value adjustment	\(P_t = (Fr - Cj)v^{n-t+1}\), grows as \(P_{t+1} = (1+j)P_t\)
Dirty price	\(P_f = (1+j)^f P_0\)
Clean price	\(P_f - fFr\)
Callable: guarantee yield	Take minimum price over all possible call dates

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General Cash Flows and Portfolios

Stock Dividend Model

Corporations raise capital through bonds, preferred stock, and common stock. Bondholders have priority over preferred shareholders, who in turn have priority over common shareholders. The dividend discount model sets the stock price equal to the present value of all future dividends.

For preferred stock with fixed dividends of $D per period:

\[ P = \frac{D}{i} \]

(a perpetuity). For common stock with dividends growing at rate \(g\) per period:

\[ P = \frac{D_1}{i - g}, \quad i > g \]

where \(D_1\) is the next dividend. This is the Gordon Growth Model. Given the price and next dividend, the implied growth rate is \(g = i - D_1/P\).

Rates of Return

The net present value and internal rate of return from Module 1 generalize to arbitrary investment cash flows. An investment with net cash flows \(C_0, C_1, \ldots, C_n\) has:

\[ NPV(i) = \sum_{t=0}^n C_t v^t \]

The IRR is the solution to \(NPV(i) = 0\). Multiple IRRs can exist; the uniqueness theorem guarantees a unique positive IRR if the retrospective balance \(B_t > 0\) for all \(t = 0, \ldots, n-1\). When the balance changes sign (the investor sometimes receives more than they’ve invested), multiple valid yield rates can exist.

Technology note — Excel IRR and RATE functions. =IRR(range) solves for the rate per period given arbitrary cash flows (positive and negative) in a column. =RATE(nper, pmt, pv) solves for the rate of a level annuity. Both return the effective rate per period — multiply by the number of periods per year to annualize as a nominal rate. The IRR function is more flexible (handles nonlevel cash flows); the RATE function is limited to level annuities but simpler to use. Both accept an optional guess argument to steer the solver.

Dollar-Weighted Rate of Interest

For a fund with opening balance \(A\), closing balance \(B\), and net contributions \(C_{t_k}\) at times \(t_k \in (0,1)\), the approximate dollar-weighted yield over one year is:

\[ i \approx \frac{I}{A + \sum_{k} C_{t_k}(1 - t_k)} \]

where the interest earned is \(I = B - A - \sum_k C_{t_k}\). The denominator is the “exposure” — the opening balance plus each deposit weighted by the fraction of the year it was in the fund.

Over a \(T\)-year period, the nominal rate approximation is:

\[ j \approx \frac{I}{TA + \sum_k C_{t_k}(T - t_k)} \]

with annual effective rate \(i = (1 + jT)^{1/T} - 1\).

Time-Weighted Rate of Interest

The time-weighted yield removes the effect of the timing of deposits and withdrawals, measuring the fund’s performance independently of investor cash flows. Between consecutive contribution dates \(t_{k-1}\) and \(t_k\), the sub-period accumulation factor is:

\[ 1 + i_k = \frac{B_{t_k}}{B_{t_{k-1}} + C_{t_{k-1}}} \]

(balance just before the contribution at \(t_k\), divided by balance just after the contribution at \(t_{k-1}\). The time-weighted annual effective yield is:

\[ (1+i)^T = \prod_{k=1}^m (1 + i_k) \]

Key insight: Dollar-weighted yield rewards (or penalizes) investors for the timing of their contributions. The time-weighted yield is a property of the fund manager, independent of when investors enter or exit.

Term Structure of Interest Rates

The yield curve plots bond yields against time to maturity. A zero-coupon bond (strip bond) pays only a redemption amount at maturity. The \(n\)-year spot rate \(r_n\) is the annual effective yield on an \(n\)-year zero-coupon bond:

\[ r_n = \left(\frac{F}{P}\right)^{1/n} - 1 \]

A coupon bond can be priced using spot rates by discounting each cash flow at its own maturity’s spot rate:

\[ P = Fr(1+r_1)^{-1} + Fr(1+r_2)^{-2} + \cdots + (Fr + C)(1+r_n)^{-n} \]

Bootstrapping: Given a sequence of coupon bonds with increasing maturities, extract spot rates recursively. Use the 1-year bond to find \(r_1\), then solve for \(r_2\) using the 2-year bond and known \(r_1\), and so on.

Forward rates: The \(n\)-year deferred, \(m\)-year forward rate \(f_{[n,n+m]}\) is the rate used to move value between times \(n\) and \(n+m\), consistent with the spot rates:

\[ (1+r_n)^n \cdot (1 + f_{[n,n+m]})^m = (1+r_{n+m})^{n+m} \]

so \(f_{[n,n+m]} = \left(\frac{(1+r_{n+m})^{n+m}}{(1+r_n)^n}\right)^{1/m} - 1\).

Summary

Concept	Formula
Stock price (fixed dividend)	\(P = D/i\)
Stock price (growing dividend)	\(P = D_1/(i-g)\)
Dollar-weighted yield	\(i \approx I / (A + \sum C_{t_k}(1-t_k))\)
Time-weighted yield	\((1+i)^T = \prod(1+i_k)\)
\(n\)-year spot rate	\(r_n = (F/P)^{1/n} - 1\)
Forward rate	\(f_{[n,n+m]} = ((1+r_{n+m})^{n+m}/(1+r_n)^n)^{1/m} - 1\)

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Duration, Convexity, and Immunization

Macaulay Duration

Duration measures the interest-rate sensitivity of a portfolio of cash flows. Intuitively, it is the weighted average time of payment, where the weights are the present values of each cash flow. For cash flows \(C_t\) at times \(t \ge 0\) with present value \(P(i) = \sum_t C_t v^t\), the Macaulay duration is:

\[ D^{(\infty)}(i) = \frac{\sum_{t \ge 0} t C_t v^t}{P(i)} \]

An equivalent calculus-based definition: Macaulay duration is the negative of the relative rate of change of present value with respect to the force of interest:

\[ D^{(\infty)}(i) = -\frac{d}{d\delta} \ln P(i) \]

Properties:

• For a zero-coupon bond maturing at time \(n\): \(D^{(\infty)} = n\) (independent of yield)
• Longer-maturity bonds have longer durations
• Higher-coupon bonds have shorter durations than lower-coupon bonds of the same maturity
• Duration for a perpetuity: \(D^{(\infty)} = (1+i)/i\)

For a coupon bond, use the P–Q formula to evaluate the numerator sum:

\[ \sum_{t=1}^n t \cdot Fr \cdot v^t + n \cdot C v^n = Fr \cdot (Ia)_{\overline{n}|} + n C v^n = Fr \cdot \frac{\ddot{a}_{\overline{n}|} - nv^n}{i} + n C v^n \]

Modified Duration

While Macaulay duration measures sensitivity to changes in \(\delta\) (force of interest), modified duration measures sensitivity to changes in the annual nominal rate \(i^{(m)}\):

\[ D^{(m)}(i) = -\frac{d}{di^{(m)}} \ln P(i) = \frac{\sum_{t \ge 0} t C_t v^{t + 1/m}}{P(i)} \]

Modified duration is related to Macaulay duration by discounting one period:

\[ D^{(m)}(i) = v^{1/m} D^{(\infty)}(i) = \left(1 + \frac{i^{(m)}}{m}\right)^{-1} D^{(\infty)}(i) \]

For annual payments (\(m = 1\), we write \(D(i) = vD^{(\infty)}(i)\).

Interpretation: If the yield increases by \(\varepsilon\) (in annual effective rate), the approximate percentage change in price is \(-D(i) \cdot \varepsilon\).

Convexity

Duration provides a first-order approximation. The convexity captures the second-order (curvature) effect. The Macaulay convexity is:

\[ C^{(\infty)}(i) = \frac{\sum_{t \ge 0} t^2 C_t v^t}{P(i)} \]

The modified convexity (for annual payments, \(m = 1\) is:

\[ C(i) = \frac{P''(i)}{P(i)} = \frac{\sum_{t \ge 0} t(t+1) C_t v^{t+2}}{P(i)} \]

Using a Taylor expansion, the approximate relative price change when the yield moves from \(i_0\) to \(i_0 + \varepsilon\) is:

\[ \frac{P(i_0 + \varepsilon) - P(i_0)}{P(i_0)} \approx -D(i_0) \cdot \varepsilon + C(i_0) \cdot \frac{\varepsilon^2}{2} \]

The duration term gives the linear (first-order) change; convexity adds a positive second-order correction. This means that for a given duration, higher convexity is preferred — the bond loses less when rates rise and gains more when rates fall.

Risk implication: Among bonds with the same duration, a zero-coupon bond has the highest sensitivity to interest rates (and least convexity relative to its duration), while high-coupon bonds are less sensitive.

Asset-Liability Management and Immunization

Asset-liability management (ALM) balances a fund’s inflows (assets) against its outflows (liabilities). Define:

\[ P_A(i) = \sum_t A_t v^t, \quad P_L(i) = \sum_t L_t v^t, \quad S(i) = P_A(i) - P_L(i) \]

Exact matching (cash flow matching) sets \(A_t = L_t\) for all \(t\). This is the safest approach — the surplus is zero regardless of interest rate.

When exact matching is unavailable, Redington’s immunization provides protection against small interest rate changes. A fund is immunized at rate \(i_0\) if \(S(i_0) = 0\) and \(S(i_0 + \varepsilon) \ge 0\) for small \(|\varepsilon|\).

Redington’s conditions (sufficient for immunization):

1. \(P_A(i_0) = P_L(i_0)\) — present values equal
2. \(D_A(i_0) = D_L(i_0)\) — durations equal
3. \(C_A(i_0) > C_L(i_0)\) — convexity of assets exceeds that of liabilities

In summation form (for annual cash flows):

1. \(\sum_t A_t v^t = \sum_t L_t v^t\)
2. \(\sum_t t A_t v^t = \sum_t t L_t v^t\)
3. \(\sum_t t^2 A_t v^t > \sum_t t^2 L_t v^t\)

Geometric interpretation: Conditions 1 and 2 say the surplus function has a zero with horizontal tangent at \(i_0\). Condition 3 ensures it’s a local minimum, so the surplus is non-negative for nearby rates.

Practical challenges: Real yield curves are not flat (long-term rates differ from short-term rates), interest rate changes need not be parallel shifts, and future cash flows may be uncertain.

Immunization Summary

Concept	Formula
Macaulay duration	\(D^{(\infty)} = \sum_t t C_t v^t / P\)
Modified duration (\(m=1\)	\(D = \sum_t t C_t v^{t+1}/P = vD^{(\infty)}\)
Modified convexity (\(m=1\)	\(C = \sum_t t(t+1) C_t v^{t+2}/P\)
Price change approximation	\(\Delta P/P \approx -D\varepsilon + C\varepsilon^2/2\)
Redington condition 1	\(P_A = P_L\)
Redington condition 2	\(D_A = D_L\)
Redington condition 3	\(C_A > C_L\)

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Appendix: BA II Plus Calculator Notes

Setup: Press 2ND → FORMAT → set decimal places to 9 → press ↑ → 2ND ENTER to switch from Chain (CHN) to Algebraic (AOS) order of operations → 2ND QUIT.

TVM functions: The five TVM keys (N, I/Y, PV, PMT, FV) solve the equation \(PV + PMT \cdot a_{\overline{n}|} + FV \cdot v^n = 0\). Enter I/Y as a percentage (not a decimal). Cash flows in opposite directions must have opposite signs — e.g., enter PMT as negative if PV is positive.

Annuity-due: Press 2ND → PMT (BGN) → 2ND ENTER to toggle between END and BGN mode. The display shows “BGN” when beginning-of-period payments are active. Caution: always switch back to END after computing an annuity-due value — forgetting this is a common exam error.

Discount rate conversion: To convert a nominal discount rate \(d^{(m)}\) to an effective interest rate per period: enter \(d^{(m)}/m\), press +/-, then + 1 =, then 1/x, then - 1 =, then × 100 = to get the percentage for I/Y.