Class Information

Principles of Financial Computing

Prof. Yuh-Dauh Lyuu

Dept. Computer Science & Information Engineering and

Department of Finance National Taiwan University

Introduction

You must go into finance, Amory.

— F. Scott Fitzgerald (1896–1940), This Side of Paradise (1920) The two most dangerous words in Wall Street vocabulary are “financial engineering.”

— Wilbur Ross (2007)

Class Information

• Yuh-Dauh Lyuu. Financial Engineering & Computation:

Principles, Mathematics, Algorithms. Cambridge University Press, 2002.

• Official Web page is

www.csie.ntu.edu.tw/~lyuu/finance1.html – Lecture notes will be uploaded before class.

– Homeworks and teaching assistants will also be announced there.

– Do not mistake last year’s homeworks for this year’s!

Class Information (continued)

• Check

www.csie.ntu.edu.tw/~lyuu/capitals.html for some of the software.

• Check the CBIS (Convertible Bond Information System) at

http://www.cb.idv.tw/cb/

for a convertible-bond software system.^a

aThanks to Prof. Gow-Hsing King and his strong team.

Class Information (concluded)

• Please ask many questions in class.

– This is the best way for me to remember you in a large class.^a

a“[A] science concentrator [...] said that in his eighth semester of [Harvard] college, there was not a single science professor who could identify him by name.” (New York Times, September 3, 2003.)

Grading

• Programming assignments.

• Treat each homework as an examination.

• You are expected to write your own codes and turn in your source code.

• Do not copy or collaborate with fellow students.

• Never ask your friends to write programs for you.

• Never give your code to other students or publish your code.

What This Course Is About

• Financial theories in pricing.

• Mathematical backgrounds.

• Derivative securities.

• Pricing models.

• Efficient algorithms in pricing financial instruments.

• Research problems.

• Help in finding your thesis directions.

What This Course Is Not About

• How to program.^a

– A software bug cost Knight Capital Group, Inc.

US$457.6 million on August 1, 2012.^b

• Basic calculus, probability, combinatorics, and algebra.

• Details of the financial markets.

• How to be rich.

• How the markets will perform tomorrow.

• Professional behavior.

ahttp://www.csie.ntu.edu.tw/train/

bKirilenko & Lo (2013).

Useful Journals

• Applied Mathematical Finance.

• Communications on Pure and Applied Methematics.

• European Journal of Finance.

• European Journal of Operational Research.

• Finance and Stochastics.

• Financial Analysts Journal.

• International Journal of Finance & Economics.

• International Journal of Theoretical and Applied Finance.

Useful Journals (continued)

• Journal of Computational Finance.

• Journal of Derivatives.

• Journal of Economic Dynamics & Control.

• Journal of Finance.

• Journal of Financial Economics.

• Journal of Fixed Income.

• Journal of Futures Markets.

• Journal of Financial and Quantitative Analysis.

• Journal of Portfolio Management.

• Journal of Real Estate Finance and Economics.

Useful Journals (concluded)

• Journal of Risk and Uncertainty.

• Management Science.

• Mathematical Finance.

• Quantitative Finance.

• Review of Financial Studies.

• Review of Derivatives Research.

• Risk Magazine.

• SIAM Journal on Financial Mathematics.

• Stochastics and Stochastics Reports.

The Modelers’ Hippocratic Oath

• I will remember that I didn’t make the world, and it doesn’t satisfy my equations.

• Though I will use models boldly to estimate value, I will not be overly impressed by mathematics.

• I will never sacrifice reality for elegance without explaining why I have done so.

• Nor will I give the people who use my model false comfort about its accuracy. Instead, I will make explicit its

assumptions and oversights.

• I understand that my work may have enormous effects on society and the economy, many of them beyond my

comprehension.

aEmanuel Derman & Paul Wilmott, January 7, 2009.

Outstanding U.S. Debts (bln)

Year Municipal Treasury

Mortgage−

related U.S.

corporate Fed agencies

Money market

Asset−

backed Total

85 859.5 1,437.7 372.1 776.5 293.9 847.0 0.9 4,587.6

86 920.4 1,619.0 534.4 959.6 307.4 877.0 7.2 5,225.0

87 1,010.4 1,724.7 672.1 1,074.9 341.4 979.8 12.9 5,816.2

88 1,082.3 1,821.3 772.4 1,195.7 381.5 1,108.5 29.3 6,391.0 89 1,135.2 1,945.4 971.5 1,292.5 411.8 1,192.3 51.3 7,000.0 90 1,184.4 2,195.8 1,333.4 1,350.4 434.7 1,156.8 89.9 7,745.4 91 1,272.2 2,471.6 1,636.9 1,454.7 442.8 1,054.3 129.9 8,462.4 92 1,302.8 2,754.1 1,937.0 1,557.0 484.0 994.2 163.7 9,192.8 93 1,377.5 2,989.5 2,144.7 1,674.7 570.7 971.8 199.9 9,928.8 94 1,341.7 3,126.0 2,251.6 1,755.6 738.9 1,034.7 257.3 10,505.8 95 1,293.5 3,307.2 2,352.1 1,937.5 844.6 1,177.3 316.3 11,228.5 96 1,296.0 3,459.7 2,486.1 2,122.2 925.8 1,393.9 404.4 12,088.1 97 1,367.5 3,456.8 2,680.2 2,346.3 1,022.6 1,692.8 535.8 13,102.0 98 1,464.3 3,355.5 2,955.2 2,666.2 1,296.5 1,978.0 731.5 14,447.2 99 1,532.5 3,281.0 3,334.2 3,022.9 1,616.5 2,338.2 900.8 16,026.4 00 1,567.8 2,966.9 3,564.7 3,372.0 1,851.9 2,661.0 1,071.8 17,056.1 01 1,688.4 2,967.5 4,125.5 3,817.4 2,143.0 2,542.4 1,281.1 18,565.3

Global OTC Derivatives Market

Standard and Poor’s (S&P) 500 Index (by Robert Shiller^a)

Returns of S&P 500 Index

Distribution of Returns of S&P 500 Index

Analysis of Algorithms

I can calculate the motions of the heavenly bodies, but not the madness of people.

— Isaac Newton (1642–1727) It is unworthy of excellent men to lose hours like slaves in the labor of computation.

— Gottfried Wilhelm Leibniz (1646–1716)

Computability and Algorithms

• Algorithms are precise procedures that can be turned into computer programs.

• Uncomputable problems.

– Does this program have infinite loops?

– Is this program bug free?

• Computable problems.

– Intractable problems.

– Tractable problems.

Complexity

• A set of basic operations are assumed to take one unit of time (+,−, ×, /, log, x^y, e^x, . . . ).

• The total number of these operations is the total work done by an algorithm (its computational complexity).

• The space complexity is the amount of memory space used by an algorithm.

• Concentrate on the abstract complexity of an algorithm instead of its detailed implementation.

• Complexity is a good guide to an algorithm’s actual running time.

Common (Asymptotic) Complexities

• Let n stand for the “size” of the problem.

– Number of elements, number of cash flows, number of time periods, etc.

• Linear time if the complexity is O(n).

• Quadratic time if the complexity is O(n²).

• Cubic time if the complexity is O(n³).

• Superpolynomial if the complexity is higher than polynomials, say 2^{O(√n )}.^a

• Exponential time if the complexity is 2^O(n).

aE.g., Dai (B82506025, R86526008, D8852600) & Lyuu (2007); Lyuu

& C. Wang (F95922018) (2011); H. Chiu (R98723059) (2012).

Basic Financial Mathematics

In the fifteenth century mathematics was mainly concerned with questions of commercial arithmetic and the problems of the architect.

— Joseph Alois Schumpeter (1883–1950) I’m more concerned about the return of my money than the return on my money.

— Will Rogers (1879–1935)

The Time Line

-

Time 0 Time 1 Time 2 Time 3 Time 4 Period 1 Period 2 Period 3 Period 4

Time Value of Money

FV = PV(1 + r)ⁿ, (1)

PV = FV × (1 + r)⁻ⁿ.

• FV (future value).

• PV (present value).

• r: interest rate.

aFibonacci (1170–1240); Irving Fisher (1867–1947).

Periodic Compounding

• Suppose the annual interest rate r is compounded m times per annum.

• Then

1 → 

1 + r m

 → 

1 + r m

₂

→ 

1 + r m

₃

→ · · ·

• Hence, after n years,

FV = PV



1 + r m

nm

. (2)

Common Compounding Methods

• Annual compounding: m = 1.

• Semiannual compounding: m = 2.

• Quarterly compounding: m = 4.

• Monthly compounding: m = 12.

• Weekly compounding: m = 52.

• Daily compounding: m = 365.

• Continuous compounding: m = ∞.

Easy Translations

• An annual interest rate of r compounded m times a year is “equivalent to” an interest rate of r/m per 1/m year.

• If a loan asks for a return of 1% per month, the annual interest rate will be 12% with monthly compounding.

Example

• Annual interest rate is 10% compounded twice per annum.

• Each dollar will grow to be

[ 1 + (0.1/2) ]² = 1.1025 one year from now.

• The rate is “equivalent to” an interest rate of 10.25%

compounded once per annum,

1 + 0.1025 = 1.1025

Rule of 72

• Let the annual interest rate be r.

• How many years T will it take for your money to double?

• The identity to solve is

(1 + r)^T = 2.

• So

T = ln 2 ln(1 + r).

• Is there an easier way?

Rule of 72 (concluded)

• The rule of 72 is a heuristic to estimate T .

• It says

T ≈ 72 r (%).

• So it takes about 72/12 = 6 years to double the GDP if the annual growth rate is 12%.

• Reason:

ln 2

ln(1 + r) ≈ 0.693

r + 0.3466.

How Good Is the Rule of 72?

aTrue interest rate subtracted by the approximation (in %).

Continuous Compounding

• Let m → ∞ so that



1 + r m

m

→ e^r in Eq. (2) on p. 28.

• Then

FV = PV × e^rn, where e = 2.71828 . . . .

aJacob Bernoulli (1654–1705) in 1685.

Continuous Compounding (concluded)

• Continuous compounding is easier to work with.

• Suppose the annual interest rate is r1 for n₁ years and r₂ for the following n₂ years.

• Then the FV of one dollar will be e^r1n1+r2n2 after n₁ + n₂ years.

Conversion between Compounding Methods

• Suppose r1 is the annual rate with continuous compounding.

• Suppose r₂ is the equivalent rate compounded m times per annum.

• How are they related?

Conversion between Compounding Methods (concluded)

• Principle: Both interest rates must produce the same amount of money after one year.

• That is, 

1 + r₂ m

m

= e^r1.

• Therefore,^a

r₁ = m ln



1 + r₂ m

 , r₂ = m



e^r1/m − 1 .

a

The PV Formula

• The PV of the cash flow C1, C₂, . . . , C_n at times 1, 2, . . . , n is

PV = C₁

1 + y + C₂

(1 + y)² + · · · + C_n

(1 + y)ⁿ. (3)

• This formula and its variations are the engine behind most of financial calculations.^a

– What is y?

– What are Cⁱ? – What is n?

• It will be justified on p. 215.

aCochrane (2005), “Asset pricing theory all stems from one simple concept [...]: price equals expected discounted payoff.”

An Algorithm for Evaluating PV in Eq. (3)

1: x := 0;

2: for i = 1, 2, . . . , n do

3: x := x + C_i/(1 + y)ⁱ;

4: end for

5: return x;

• The algorithm takes time proportional to

_n

i=1 i = O(n²).^a

• Can improve it to O(n) if you apply a^b = e^{b ln a} in step 3.^b

aIf only +, −, ×, and / are allowed.

bRecall that we count x^y as taking one unit of time.

Another Algorithm for Evaluating PV

1: x := 0;

2: d := 1 + y;

3: for i = n, n − 1, . . . , 1 do

4: x := (x + C_i)/d;

5: end for

6: return x;

Horner’s Rule: The Idea Behind p. 41

• This idea is



· · ·

 Cn

1 +y + C_n−1

 1

1 + y + C_n−2

 1

1 +y + · · ·

 1

1 +y.

– Due to Horner (1786–1837) in 1819.

• The algorithm takes O(n) time.

• It is the most efficient possible in terms of the absolute number of arithmetic operations.^a

aA. Borodin & Munro (1975).

Annuities

(Certain)

• An annuity pays out the same C dollars at the end of each year for n years.

• With a rate of r, the FV at the end of the nth year is

n−1

i=0

C(1 + r)ⁱ = C (1 + r)ⁿ − 1

r . (4)

aJan de Witt (1625–1672) in 1671; Nicholas Bernoulli (1687–1759) in 1709.

General Annuities

• If m payments of C dollars each are received per year (the general annuity), then Eq. (4) becomes

C

1 + _m^r nm

− 1

mr

.

• The PV of a general annuity is PV =

nm i=1

C



1 + r m

_−i

= C 1 − 

1 + _m^r _−nm

mr

.

(5)

Amortization

• It is a method of repaying a loan through regular payments of interest and principal.

• The size of the loan (the original balance) is reduced by the principal part of each payment.

• The interest part of each payment pays the interest incurred on the remaining principal balance.

• As the principal gets paid down over the term of the loan, the interest part of the payment diminishes.

• From Eq. (5) on p. 44, the monthly payment equals C = loan amount ×

mr

1 − 

1 + _m^r _−nm. (6)

Example: Home Mortgage

• By paying down the principal consistently, the risk to the lender is lowered.

• When the borrower sells the house, only the remaining principal is due the lender.

• Consider the equal-payment case, i.e., fixed-rate, level-payment, fully amortized mortgages.

– They are called traditional mortgages in the U.S.^a

aThe Economist (2016), “In most countries banks minimize their risk by offering short-term or floating-rate mortgages. American borrowers get a better deal: cheap 30-year fixed-rate mortgages that can be repaid early free.”

A Numerical Example

• Consider a 15-year, $250,000 loan at 8.0% interest rate.

• Solve Eq. (5) on p. 44 with PV = 250000, n = 15, m = 12, and r = 0.08.

• That is,

250000 = C 1 − 

1 + ^0.08₁₂ _−15×12

0.0812

.

• This gives a monthly payment of C = 2389.13.

The Amortization Schedule

Month Payment Interest Principal

Remaining principal 250,000.000 1 2,389.13 1,666.667 722.464 249,277.536 2 2,389.13 1,661.850 727.280 248,550.256 3 2,389.13 1,657.002 732.129 247,818.128

· · ·

178 2,389.13 47.153 2,341.980 4,730.899 179 2,389.13 31.539 2,357.591 2,373.308

180 2,389.13 15.822 2,373.308 0.000

Total 430,043.438 180,043.438 250,000.000

A Numerical Example (continued)

In every month:

• The principal and interest parts add up to $2,389.13.

• The remaining principal is reduced by the amount indicated under the Principal heading.^a

– The Principal column forms a geometric sequence.^b

• The interest is computed by multiplying the remaining balance of the previous month by 0.08/12.

aThis column varies with r. Thanks to a lively class discussion on Feb 24, 2010. In fact, every column varies with r. Contributed by Ms.

Wu, Japie (R01921056) on February 20, 2013.

bSee p. 1255. Contributed by Mr. Sun, Ao (R05922147) on February 22, 2017.

A Numerical Example (concluded)

Note that:

• The Principal column adds up to $250,000.

• The Payment column adds up to $430,043.438!

• If the borrower plans to pay off the mortgage right after the 178th month’s monthly payment, he needs to pay another $4,730.899.^a

aContributed by Ms. Wu, Japie (R01921056) on February 20, 2013.

Method 1 of Calculating the Remaining Principal

• A month’s principal payment = monthly payment − (previous month’s remaining principal) × (monthly interest rate).

• A month’s remaining principal = previous month’s remaining principal − principal payment calculated above.

• Generate the amortization schedule until you reach the particular month you are interested in.

Method 1 of Calculating the Remaining Principal (concluded)

• This method is relatively slow but is universal in its applicability.

• It can, for example, accommodate prepayments and variable interest rates.

Method 2 of Calculating the Remaining Principal

• Right after the kth payment, the remaining principal is the PV of the future nm − k cash flows,

nm−k

i=1

C



1 + r m

_−i

= C 1 − 

1 + _m^r _−nm+k

mr

. (7)

• This method is much faster.

• But it is more limited in applications because it makes more assumptions.

Yields

• The term yield denotes the return of investment.

• Two widely used yields are the bond equivalent yield (BEY) and the mortgage equivalent yield (MEY).

• Recall Eq. (2) on p. 28: FV = PV

1 + _m^r nm

.

• BEY corresponds to the r above that equates PV with FV when m = 2.

• MEY corresponds to the r above that equates PV with FV when m = 12.

• Again, although BEY and MEY might be different, they nevertheless mean the same yield.

Internal Rate of Return (IRR)

• It is the yield y which equates an investment’s PV with its price P ,

P = C₁

(1 + y) + C₂

(1 + y)² + · · · + C_n (1 + y)ⁿ.

• IRR assumes all cash flows are reinvested at the same rate as the internal rate of return because:

FV = C₁(1 + y)ⁿ⁻¹ + C₂(1 + y)ⁿ⁻² + · · · + Cⁿ.

• So it must be used with extreme caution.

Numerical Methods for IRRs

• Define

f (y) =^Δ

n t=1

C_t

(1 + y)^t − P.

– P is the market price.

• Solve f(y) = 0 for a real y ≥ −1.^a

• f(y) is monotonically decreasing in y if Ct > 0 for all t.

• So a unique real-number solution exists for this f(y).

aNegative interest rates became a reality for German and Swiss bonds in 2015. In 2016, Sweden, Denmark, and Japan imposed negative interest rates on excess reserves. As many as 355 corporate bonds were issued with negative yields as of June of 2016. Even so, −100% should be a natural lower bound because why would anyone or financial institution

The Bisection Method

• Start with a and b where a < b and f(a) f(b) < 0.

• Then f(ξ) must be zero for some ξ ∈ [ a, b ].

• If we evaluate f at the midpoint c = (a + b)/2, either^Δ (1) f (c) = 0, (2) f (a) f (c) < 0, or (3) f (c) f (b) < 0.

• In the first case we are done, in the second case we

continue the process with the new bracket [ a, c ], and in the third case we continue with [ c, b ].

• The bracket is halved in the latter two cases.

• After n steps, we will have confined ξ within a bracket of length (b − a)/2ⁿ.

, . -

The Newton-Raphson Method

• It converges faster than the bisection method.

• Start with a first approximation x0 to a root of f (x) = 0.

• Then

x_k+1 = x^Δ _k − f (x_k) f^(x_k).

• When computing yields, f^(x) = −

n t=1

tC_t

(1 + x)^t+1. (8)

x_k x_k

x y

f x 

The Secant Method

• A variant of the Newton-Raphson method.

• Replace differentiation with difference.

• Start with two approximations x0 and x₁.

• Then compute the (k + 1)st approximation with x_k+1 = x_k − f (x_k)(x_k − x_k−1)

f (x_k) − f(xk−1) .

• Note that it is easier to calculate

[ f (x_k) − f(xk−1) ]/(x_k − xk−1) than the f^(x_k) on p. 59.

The Secant Method (concluded)

• Its convergence rate is 1.618.

• This is slightly worse than the Newton-Raphson method’s 2.

• But the secant method does not need to evaluate f^(x_k) needed by the Newton-Raphson method.

• This saves about 50% in computation efforts per iteration.

• The convergence rate of the bisection method is 1.

Solving Systems of Nonlinear Equations

• It is not easy to extend the bisection method to higher dimensions.

• But the Newton-Raphson method can be extended to higher dimensions.

• Let (x^k, y_k) be the kth approximation to the solution of the two simultaneous equations,

f (x, y) = 0, g(x, y) = 0.

Solving Systems of Nonlinear Equations (continued)

• The (k + 1)st approximation (x_k+1, y_k+1) satisfies the following linear equations,

⎡

⎣ ^{∂f(x∂xk,yk) ∂f(x∂yk,yk)}

∂g(xk,y_k)

∂x ∂g(xk,y_k)

∂y

⎤

⎦

⎡

⎣ Δx_k+1 Δy_k+1

⎤

⎦ = −

⎡

⎣ f (x_k, y_k) g(x_k, y_k)

⎤

⎦ ,

with unknowns

Δx_k+1 =^Δ x_k+1 − xk, Δy_k+1 =^Δ y_k+1 − yk.

Solving Systems of Nonlinear Equations (concluded)

• The above has a unique solution for (Δx_k+1, Δy_k+1) when the 2 × 2 matrix is invertible.

• Finally, set

x_k+1 = x_k + Δx_k+1, y_k+1 = y_k + Δy_k+1.

Zero-Coupon Bonds (Pure Discount Bonds)

• By Eq. (1) on p. 27, the price of a zero-coupon bond that pays F dollars in n periods is

F/(1 + r)ⁿ, (9)

where r is the interest rate per period.

• Can be used to meet future obligations as there is no reinvestment risk.^a

aRecall p. 55.

Example

• The interest rate is 8% compounded semiannually.

• A zero-coupon bond that pays the par value 20 years from now will be priced at 1/(1.04)⁴⁰, or 20.83%, of its par value.

• It will be quoted as 20.83.^a

• If the bond matures in 10 years instead of 20, its price would be 45.64.

aOnly one fifth of the par value!

Level-Coupon Bonds

• Coupon rate.

• Par value, paid at maturity.

• F denotes the par value, and C denotes the coupon.

• Cash flow:

6 6 6 6 -

1 2 3 n

C C C

· · ·

C + F

• Coupon bonds can be thought of as a matching package of zero-coupon bonds, at least theoretically.^a

a“You see, Daddy didn’t bake the cake, and Daddy isn’t the one who gets to eat it. But he gets to slice the cake and hand it out. And when he does, little golden crumbs fall off the cake. And Daddy gets to eat

Pricing Formula

P =

n i=1

 C

1 + _m^r _i +  F

1 + _m^r _n

= C 1 − 

1 + _m^r _−n

mr

+  F

1 + _m^r _n . (10)

• n: number of cash flows.

• m: number of payments per year.

• r: annual rate compounded m times per annum.

• Note C = F c/m when c is the annual coupon rate.

• Price P can be computed in O(1) time.

Yields to Maturity

• It is the r that satisfies Eq. (10) on p. 69 with P being the bond price.

• For a 15% BEY, a 10-year bond with a coupon rate of 10% paid semiannually sells for

5 × 1 − [ 1 + (0.15/2) ]^−2×10

0.15/2 + 100

[ 1 + (0.15/2) ]^2×10

= 74.5138 percent of par.

• So 15% is the yield to maturity if the bond sells for 74.5138.^a

Price Behavior (1)

• Bond prices fall when interest rates rise, and vice versa.

• “Only 24 percent answered the question correctly.”^a

aCNN, December 21, 2001.

Price Behavior (2)

• A level-coupon bond sells

– at a premium (above its par value) when its coupon rate c is above the market interest rate r;

– at par (at its par value) when its coupon rate is equal to the market interest rate;

– at a discount (below its par value) when its coupon rate is below the market interest rate.

aConsult the text for proofs.

9% Coupon Bond

Yield (%) Price (% of par)

7.5 113.37

8.0 108.65

8.5 104.19

9.0 100.00

9.5 96.04

10.0 92.31

10.5 88.79

Terminology

• Bonds selling at par are called par bonds.

• Bonds selling at a premium are called premium bonds.

• Bonds selling at a discount are called discount bonds.

Price Behavior (3): Convexity

0 0.05 0.1 0.15 0.2

Yield 0

250 500 750 1000 1250 1500 1750

Price

Day Count Conventions: Actual/Actual

• The first “actual” refers to the actual number of days in a month.

• The second refers to the actual number of days in a coupon period.

• The number of days between June 17, 1992, and October 1, 1992, is 106.

– 13 days in June, 31 days in July, 31 days in August, 30 days in September, and 1 day in October.

Day Count Conventions: 30/360

• Each month has 30 days and each year 360 days.

• The number of days between June 17, 1992, and October 1, 1992, is 104.

– 13 days in June, 30 days in July, 30 days in August, 30 days in September, and 1 day in October.

• In general, the number of days from date (y1, m₁, d₁) to date (y₂, m₂, d₂) is

360 × (y2 − y1) + 30 × (m2 − m1) + (d2 − d1). (11)

Day Count Conventions: 30/360 (continued)

• If d₁ or d₂ is 31, we must change it to 30 before applying formula (11).^a

• Hence:

– There are 3 days between February 28 and March 1.

– There are 2 days between February 29 and March 1.

– There are 29 days between March 1 and March 31.

aThe simplest of all the “30/360” variations, this is called the

“30E/360” convention, used mainly in the Eurobond market (Kosowski

& Neftci, 2015).

Day Count Conventions: 30/360 (concluded)

• An equivalent formula to (11) on p. 77 without any adjustment is (check it)

360 × (y₂ − y₁) + 30 × (m₂ − m₁ − 1) + max(30 − d₁, 0) + min(d₂, 30).

• There are many variations on the “30/360” convention regarding 31, February 28, and February 29.^a

aKosowski & Neftci (2015).

Full Price (Dirty Price, Invoice Price)

• In reality, the settlement date may fall on any day between two coupon payment dates.

• Let

ω =^Δ

number of days between the settlement and the next coupon payment date

number of days in the coupon period .

(12)

Full Price (continued)

- 6

coupon payment date

C(1 − ω)

coupon payment date

 _{(1 − ω)} - _ω -

Full Price (concluded)

• The price is now calculated by

PV = C

1 + _m^r ω + C

1 + _m^r _ω+1 · · ·

=

n−1

i=0

 C

1 + _m^r _ω+i + F

1 + _m^r _ω+n−1. (13)

Accrued Interest

• The quoted price in the U.S./U.K. does not include the accrued interest; it is called the clean price or flat price.

• The buyer pays the invoice price: the quoted price plus the accrued interest (AI).

• The accrued interest equals

C ×

number of days from the last

coupon payment to the settlement date

number of days in the coupon period = C × (1 − ω).

Accrued Interest (concluded)

• The yield to maturity is the r satisfying Eq. (13) on p.

82 when PV is the invoice price:

clean price + AI =

n−1

i=0

 C

1 + _m^r _ω+i + F

1 + _m^r _ω+n−1.

Example (“30/360”)

• A bond with a 10% coupon rate and paying interest semiannually, with clean price 111.2891.

• The maturity date is March 1, 1995, and the settlement date is July 1, 1993.

• There are 60 days between July 1, 1993, and the next coupon date, September 1, 1993.

• The accrued interest is (10/2) × (1 − ₁₈₀⁶⁰ ) = 3.3333 per

$100 of par value.

Example (“30/360”) (concluded)

• The yield to maturity is 3%.

• This can be verified by Eq. (13) on p. 82 with – ω = 60/180,

– n = 4, – m = 2, – F = 100, – C = 5,

– PV= 111.2891 + 3.3333, – r = 0.03.

Price Behavior (2) Revisited

• Before: A bond selling at par if the yield to maturity equals the coupon rate.

• But it assumed that the settlement date is on a coupon payment date.

• Now suppose the settlement date for a bond selling at par (i.e., the quoted price is equal to the par value) falls between two coupon payment dates.

• Then its yield to maturity is less than the coupon rate.^a – The short reason: Exponential growth to C is

replaced by linear growth, hence “overpaying.”

aSee Exercise 3.5.6 of the textbook for proof.