Finance-Theory · Unit 6 · Video 6 · Interactive Practice

Yield to Maturity: Compressing Many Spot Rates into One Number

IKey Formulas

Formula	Name	What you need
$P_0 = \dfrac{C}{1+r_{0,1}} + \dfrac{C}{(1+r_{0,2})^2} + \cdots + \dfrac{C+F}{(1+r_{0,T})^T}$	Coupon-bond price from spot rates	A spot rate for every payment date
$(1+r_{0,t})^t = (1+R_1)(1+R_2)\cdots(1+R_t)$	Spot rate as a geometric average	The one-year rates $R_1, \dots, R_t$
$P_0 = \dfrac{C}{1+y} + \dfrac{C}{(1+y)^2} + \cdots + \dfrac{C+F}{(1+y)^T}$	Yield to maturity $y$	The price $P_0$ , coupon $C$ , face $F$
$P_0 x^T - C x^{T-1} - \cdots - C x - (C+F) = 0,\quad x = 1+y$	The yield as a polynomial root	A numerical solver

Key Insight: the yield replaces all $T$ spot rates with the one constant rate that reproduces the price, a complicated average of the spot rates; for a pure discount bond it is simply $y = r_{0,T}$ .

IIFrom Price to Yield

One constant rate $y$ must discount all three payments back to the observed price of $961.36.

Yield y 6.00%

IIIOne Sign Change, One Yield

With every cash flow positive, present value falls steadily in $y$ , so it meets the price exactly once.

Cash flows Trial y 3.0%

IVA Complicated Average of the Spot Rates

The yield always lands between the spot rates of the dates that pay, and a single payment's yield is its spot rate.

Date 1 Date 2 Date 3

💡 A thirty-year mortgage quoted at 5.89% compresses thirty different one-year rates the same way; nobody expects the one-year rate to sit at 5.89% for all thirty years.

VQuiz Questions

Problem 1 · Price at the Spot Rates

Given: a three-year bond with $1,000 face value and a 4% annual coupon ( $C = 40$ ), and the spot rates $r_{0,1} = 5\%$ , $r_{0,2} = 6\%$ , $r_{0,3} = 6.5\%$ — find its price.

✅ Correct! Each payment is discounted at the spot rate for its own date:

38.10 + 35.60 + 860.96 = 934.66

❌ Not quite. That discounts every payment at

6.5\%

, the three-year rate. The coupons at dates 1 and 2 are discounted at

r_{0,1} = 5\%

and

r_{0,2} = 6\%

❌ Not quite. That discounts every payment at the one-year rate of

5\%

. Money due at date 3 is discounted at

r_{0,3} = 6.5\%

, over three years.

❌ Check the exponents. The payment at date

t

is divided by

(1+r_{0,t})^t

, not by

1+r_{0,t}

: the rate

r_{0,t}

is an annual rate compounded over

t

years.

Show solution

The bond pays $40 at dates 1 and 2 and $1,040 at date 3. Discount each payment at the spot rate for its date:

P_0 = \frac{40}{1.05} + \frac{40}{(1.06)^2} + \frac{1{,}040}{(1.065)^3} = 38.10 + 35.60 + 860.96 = 934.66

The price is $934.66. It needs three rates, one for each payment date.

Problem 2 · The Yield of a Pure Discount Bond

Given: a three-year pure discount bond with face value $1,000, and the same spot rates $r_{0,1} = 5\%$ , $r_{0,2} = 6\%$ , $r_{0,3} = 6.5\%$ . It is priced at $827.85 — find its yield to maturity.

✅ Correct! A single payment at date 3 is discounted only at

r_{0,3}

, so its yield is

y = r_{0,3} = 6.5\%

❌ Not quite.

5.83\%

is the simple average

(5 + 6 + 6.5)/3

. The yield averages only over the payments the bond actually makes, and this bond makes one, at date 3.

❌ Not quite.

6.46\%

is the yield of the 5% coupon bond, whose coupons at dates 1 and 2 pull its yield below

6.5\%

. This bond has no coupons.

❌ Not quite.

6\%

is the two-year spot rate

r_{0,2}

. The only payment arrives at date 3.

Show solution

The bond's price can be written with its yield or with the three-year spot rate:

P_0 = \frac{F}{(1+y)^3} = \frac{F}{(1+r_{0,3})^3}

The two agree only when $y = r_{0,3}$ . Numerically:

(1+y)^3 = \frac{1{,}000}{827.85} = 1.20795 \quad\Rightarrow\quad 1+y = 1.20795^{1/3} = 1.0650

so $y = 6.50\%$ , the three-year spot rate. The spot rates for dates 1 and 2 play no part, because the bond pays nothing on those dates.

Problem 3 · Price First, Then Yield

Given: a two-year bond with $1,000 face value and a 6% annual coupon ( $C = 60$ ), with spot rates $r_{0,1} = 5\%$ and $r_{0,2} = 6\%$ — find its price, then its yield to maturity.

What is the price?

What is the yield to maturity?

✅ Correct! Price $1,000.54 at the spot rates; the one constant rate that reproduces it is

y = 5.97\%

, between the two spot rates and close to

6\%

, the rate on the date that carries most of the value.

❌ Not quite. That is the price if both payments were discounted at

6\%

. The coupon at date 1 is discounted at

r_{0,1} = 5\%

, which makes it worth a little more.

❌ Not quite. That discounts both payments at

5\%

. The payment at date 2 is discounted at

r_{0,2} = 6\%

, over two years.

❌ Check the exponent. The date-2 payment is divided by

(1.06)^2

, not by

1.06

❌ Not quite.

5\%

is the one-year spot rate. The $60 at date 1 is only about 6% of the price, so it barely moves the yield away from

6\%

❌ Not quite.

5.50\%

is the simple average of the two spot rates. The yield is not a simple average: date 2 carries about 94% of the value, so the yield sits close to its

6\%

rate.

❌ Close, but…

6\%

would be the yield of a two-year pure discount bond. Discounting the date-1 coupon at

6\%

instead of

5\%

undervalues it, so the constant rate that reproduces $1,000.54 must sit slightly below

6\%

Show solution

Step 1: Price at the spot rates. The bond pays $60 at date 1 and $1,060 at date 2:

P_0 = \frac{60}{1.05} + \frac{1{,}060}{(1.06)^2} = 57.14 + 943.40 = 1{,}000.54

Step 2: Set up the yield equation. With $x = 1 + y$ :

1{,}000.54 = \frac{60}{x} + \frac{1{,}060}{x^2} \quad\Rightarrow\quad 1{,}000.54\,x^2 - 60x - 1{,}060 = 0

Step 3: Solve. With $T = 2$ the polynomial is a quadratic; take its positive root:

x = \frac{60 + \sqrt{60^2 + 4(1{,}000.54)(1{,}060)}}{2(1{,}000.54)} = \frac{60 + 2{,}060.56}{2{,}001.08} = \frac{2{,}120.56}{2{,}001.08} = 1.0597

so $y = 5.97\%$ . Check: $\frac{60}{1.0597} + \frac{1{,}060}{(1.0597)^2} = 56.62 + 943.93 = 1{,}000.55$ , one cent off the price because $y$ was rounded to $5.97\%$ .

Problem 4 · When the Signs Change Twice

Given: an investment costs $100 today, pays $250 at date 1, and requires a further payment of $150 at date 2, so its yield solves $100 = \dfrac{250}{1+y} - \dfrac{150}{(1+y)^2}$ — find every yield above $-100\%$ that fits.

✅ Correct! The coefficients

+,\,-,\,+

change sign twice, and both

x = 1

and

x = 1.5

solve the quadratic: two yields fit the same price.

❌ Not quite.

y = 0\%

does fit, since

250 - 150 = 100

, but with two sign changes a second positive root is possible. Try

x = 1.5

❌ Not quite.

y = 50\%

fits, but so does

y = 0\%

: at

x = 1

100 - 250 + 150 = 0

. One sign change guarantees one root; two sign changes allow two.

❌ Not quite. The quadratic's discriminant is

250^2 - 4(100)(150) = 2{,}500 > 0

, so both roots are real.

Show solution

Write $x = 1 + y$ and multiply both sides by $x^2$ :

100x^2 - 250x + 150 = 0 \quad\Rightarrow\quad 2x^2 - 5x + 3 = 0 \quad\Rightarrow\quad (x - 1)(2x - 3) = 0

The roots are $x = 1$ and $x = 1.5$ , so $y = 0\%$ and $y = 50\%$ . Check $y = 50\%$ : $\frac{250}{1.5} - \frac{150}{(1.5)^2} = 166.67 - 66.67 = 100$ .

The coefficient signs $+,\,-,\,+$ change twice, so Descartes' rule allows two positive roots (or none), and here there are two. Cash flows that change sign more than once can fit the same price with several yields, and it becomes hard to say which one is the right one.

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