PRACTICE PROBLEMS
CHAPTER 9: MODELS OF YIELD CURVE AND
THE TERM STRUCTURE

Exercise 9.1: Suppose one bond has P = 100, u = 1.08, R = 4%, d = 0.99. A second
bond will have a price of $105 at the end of the period in an “up” state, or $90
in the “down” state. Find the current price of the second bond, and the market
prices of risk for each bond. Assume that the actual probability of an “up” state
is q = 0.60. What happens if R = 5%? Explain.

Exercise 9.2: See slide 9-56 of the Ch. 9 lecture notes. Also, find the par bond yield
curve for this problem.

Exercise 9.3: Consider the lognormal interest rate process model. Suppose the
current one-period rate is 6%, and let u = 1.2, d = 0.9, and p = 0.5.
(a) Find the term structure, i.e., the zero yield curve: y1, y2, and y3.
(b) Find y1 and y2 if p = 0.6 (rather than 0.5). How has the term structure changed?
(c) Returning to the case where p = 0.5, suppose a security pays $1 at date t = 2 in
the “uu” state but pays $0 otherwise. Find the price of this security. Can you
interpret this security in terms of options? [Hint: it might help to assume that
bonds have a face value of $100.]

SOLUTIONS
Exercise 9.1: The risk-neutral probability is p =
( )1 

R d
u d
= 0.5555. The second
bond has price
P* = [p105 + (1  p)90]/(1.04) = 94.5513.
Assuming that the actual probability is q = 0.6, we have, for bond 1,
Eq[PT/P] = qu + (1 – q)d = 1.044, Var(R1) = q(1 – q)(u – d)2 = 0.001944,
1 = 0.04409, 1 =
1
)1(]/[

RPPE T  = 0.09072.
For bond 2, first note that u* =
5513.94
105 = 1.11051, d* =
5513.94
90 = 0.95186.
Eq[PT*/P*] = qu* + (1 – q)d* = 1.047051, Var(R2) = q(1 – q)(u* – d*)2 = 0.00604,
2
2 = 0.0777, 2 =
2
** )1(]/[

RPPE T  = 0.09071.
[Note: if you do this in EXCEL, say, to avoid round-off error, the answer is
0.09072.]
If R = 5%, the risk-neutral probability is p =
( )1 

R d
u d
= 0.6667 > q = 0.60! This
creates problems, as we’ll see. The second bond has a new price:
P* = [p105 + (1  p)90]/(1.05) = 95.2381.
Assuming that the actual probability is q = 0.6, we have, for bond 1,
E[PT/P] = 1.044, Var(R1) = 0.001944, 1 = 0.04409, (these don’t change, however
note that E[PT/P] < 1 + R!). So, 1 =
1
)1(]/[

RPPE T  = 0.136 < 0!
For bond 2, first note that u* =
2381.95
105 = 1.1025, d* =
2381.95
90 = 0.945.
E[PT*/P*] = qu* + (1 – q)d* = 1.0395 < 1 + R (again),
Var(R2) = q(1 – q)(u* – d*)2 = 0.005954,
2 = 0.077159, 2 = 0.136.
So, in each case, the actual expected return (for the risky assets) is less than the risk-
free rate. This doesn’t make sense. The market prices of risk are negative ( =
0.136.)

Exercise 9.2: For b(0, 2), we have
bu(1, 2) =
ur1
1 =
0875.1
1 = 0.9195;
bd(1, 2) =
dr1
1 =
056.1
1 = 0.9470;
b(0, 2) = )]2,1()1()2,1([
1
1 du bppb
r



= )]9470.0(5.0)9195.0(5.0[
07.1
1  = 0.8722;
So, y2 = 2/1)2,0(
1
b
 1 =
8722.0
1  1 = 7.076%.
The price of a 3-year 5% bond is
5b(0, 1) + 5b(0, 2) + 105b(0, 3) = 5(0.9346) + 5(0.8722) + 105(0.8130) = 94.399.
3
For the par bond yield curve, let’s use a face value of $1, so that the coupon payment,
ci, is a percent.
One-year bonds:
1 = (1 + c1)b(0, 1); c1 =
)1,0(
)1,0(1
b
b =
0.9346
0.93461 = 7.00%.
Two-year bonds:
1 = c2b(0, 1) + (1 + c2)b(0, 2);
c2 =
)2,0()1,0(
)2,0(1
bb
b

 =
8722.09346.0
8722.01

 = 7.07%.
Three-year bonds:
1 = c3b(0, 1) + c3b(0, 2) + (1 + c3)b(0, 3);
c3 =
)3,0()2,0()1,0(
)3,0(1
bbb
b

 =
8130.08722.09346.0
8130.01

 = 7.14%.

Exercise 9.3: Here’s the tree for the one-period rates:






7.2%
6%
4.86%
6.48%
8.64
t = 0 t = 2 t = 1
5.4%
4
(a) Obviously y1 = 6%. Here is the tree for the two-year bond:






For example, bu(1,2) =
ur1
1  Ep[1] =
072.1
1 = 0.9328,
and b(0, 2) =
r1
1 [pbu(1, 2) + (1  p)bd(1, 2)]
=
06.1
1 [0.5(0.9328) + 0.5(0.9488)] = 0.8875.

So, b(0,2) = (1 + y2)2 = 0.8875, which implies y2 = 6.15%.

Here is the tree for the three-year bond:





For example, bdd(2, 3) =
rd 21
1

 Ep[1] =
0486.1
1 = 0.9537,
and bd(1, 3) =
dr1
1 [pbud(2, 3) + (1  p)bdd(2, 3)]
0.9328
0.8875
1
1
1
t = 0 t = 2 t = 1
0.9488
0.8674
0.8327
0.9537
0.9391
0.9205
t = 0 t = 2 t = 1
0.8979
5
=
054.1
1 [0.5(0.9391) + 0.5(0.9537)] = 0.8979.

From the tree, b(0,3) = (1 + y3)3 = 0.8327, and thus y3 = 6.29%.

(b) First, changing p does not affect y1 = the current one-period rate = 6%. Next, if
we re-consider the tree for the two-year bond, note that the prices at date t = 1
do not depend on p, either. So, we just need to calculate b(0,2):
b(0,2) =
r1
1 [pbu(1, 2) + (1  p)bd(1, 2)] =
06.1
1 [0.6(0.9328) + 0.4(0.9488)]
= 0.8860 = (1 + y2)2; y2 = 6.23% > 6.15% = y2 when p = 0.5.
The term structure is steeper than it was when p = 0.5. Higher future interest rates
occur with a higher probability when p = 0.6.
(c) Let’s call the price of this security f. Here is the tree for the two-year bond:






Here, f u =
ur1
1 [p1 + (1  p)0] =
072.1
1 (0.5) = 0.4664
f d =
dr1
1 [p0 + (1  p)0] = 0
f =
r1
1 [pf u + (1  p)f d] =
06.1
1 (0.5)(0.4664) = 0.22.
This is an option that pays $1 in state “uu” and 0 otherwise. To interpret this, we
need to look at the tree for the three-year bond. We can say that this derivative
pays $1 if b(2,3) = buu(2,3) = 0.9205, but pays $0 in the cases where b(2,3) >
0.9205. This is like a put option on a 3-year zero coupon bond. To find the
exercise price, X, (now assuming a face value equal to $100), note that we want
the put payoff to be
f u
f
0
0
1
t = 0 t = 2 t = 1
f d
6
X – 100buu(2,3) = 1. Solving gives X = 93.05 < 100bud(2,3) = 93.91. (This
implies that the bond, 100bud(2,3) is out of the money, which we also require.)
This put will pay $1 in state “uu” and 0 otherwise.












































































































































































































































































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