Derivatives: Week 1
Harjoat S. Bhamra
Table 1: US Government Bond Prices on December 31, 1993
maturity XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX
coupon XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX125
id price XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX
ask price XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX XXXXXXXXXX
1. Using the data in Table 1, find both the bid and ask discount factors via bootstrapping. Assume the cu
ent date t is
December 31, 1993. Observe that all coupon payments are made 6 months apart. You can use MATLAB, Octave, R,
Python, or any other programming language of your choice.
Now answer the following questions
(a) The 5 year ask discount factor Zask(0, 5) is equal to
A 0.7708
B 0.7709
C 0.7703
D 0.7944
E 0.7950
(b) The 2-year bid implied USD risk-free rate
id(0, 2) is equal to
A 4.23%
B 4.20%
C 4.21%
D 4.26%
E 4.27%
(c) The bid price of a 3 year US Government bond paying a coupon of 5% semiannually is equal to
A XXXXXXXXXX
B XXXXXXXXXX
C XXXXXXXXXX
D XXXXXXXXXX
E XXXXXXXXXX
(d) You are working on the swaps trading desk of the well known bank Silberfrau Schwert. A salesperson is interested
in selling a 3 year USD LIBOR swap to a client and wants to know what the 3 year USD LIBOR swap rate would
e.
Which rate would you like to quote?
A a rate less than or equal to XXXXXXXXXX%
B a rate greater than XXXXXXXXXX%
C a rate less than or equal to XXXXXXXXXX%
D XXXXXXXXXX%
E XXXXXXXXXX%
1
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2. You now decide to estimate discount factors using the Nelson-Siegel model. You can use any programming language of
your choice.
(a) Using your new set of DF’s, what is the the bid price of a 3 year US Government bond paying a coupon of 5%
semiannually?
A XXXXXXXXXX
B XXXXXXXXXX
C XXXXXXXXXX
D XXXXXXXXXX
E XXXXXXXXXX
(b) The head of your desk has noticed that you used bootstrapping to estimate DF’s to price a 3 year USD LIBOR
swap for a client. She asks you to use the Nelson-Siegel approach to find the 3 year USD LIBOR swap rate. Based
on the Nelson-Siegel model, which rate would you like to quote?
A a rate less than or equal to XXXXXXXXXX%
B a rate greater than XXXXXXXXXX%
C a rate less than or equal to XXXXXXXXXX%
D XXXXXXXXXX%
E XXXXXXXXXX%
3. You will have learned how to use Ito’s Lemma to derive df(xt), given some function f(·) and and a stochastic process
xt. You should be able to do the following questions:
(a) Suppose
dxt = µdt+ σdZt,
where Z = (Zt)t∈[0,∞) is a standard Brownian motion under the probability measure P.
i. For α > 2, find
d(xαt )
A αxα−2t
[(
µxt +
1
2σ
2(α− 1)
)
dt+ xtσdZt
]
B αxα−2t
[(
µxt +
1
2σ
2α
)
dt+ xtσdZt
]
C αxα−2t
[(
µxt + σ
2(α− 1)
)
dt+ xtσdZt
]
D αxα−2t
[(
µxt +
1
2σ
2(α− 1)
)
dt+ σdZt
]
E αxα−2t
[(
µxt + σ
2(α− 1)
)
dt+ x2tσdZt
]
ii. Find dyt, where yt = e
xt
A dytyt = yt
(
µ+ 12σ
2
)
dt+ σdZt
B dytyt =
(
µ− 12σ
2
)
dt+ σdZt
C dytyt =
(
µ+ 12σ
2
)
dt+ σdZt
D dytyt = µdt+ σdZt
E dyt =
(
µ+ 12σ
2
)
ytdt+ σdZt
(b) Suppose
dSt
St
= µSdt+ σSdZt.
i. Find d lnSt
A d lnSt =
(
µS − 12Stσ
2
S
)
dt+ σSdZt
B d lnSt =
(
µS − 12Stσ
2
S
)
dt− σSdZt
C d lnSt = µSdt+ σSdZt
D d lnSt =
(
µS +
1
2σ
2
S
)
dt+ σSdZt
E d lnSt =
(
µS − 12σ
2
S
)
dt+ σSdZt
2
ii. Find d(Sαt )/S
α
t
A α
[(
µS − 12 (1− α)σ
2
S
)
dt+ ασSdZt
]
B
(
µS − 12 (1− α)σ
2
S
)
dt+ StσSdZt
C α
[(
µS +
1
2 (1− α)σ
2
S
)
dt+ σSdZt
]
D α
[(
µS − 12 (1− α)σ
2
S
)
dt+ σSdZt
]
E α
[(
µS − 12σ
2
S
)
dt+ σSdZt
]
4. You may not be fully comfortable with indicator functions and may find the use of a martingale to define a new
probability measure to be somewhat exotic, so let’s explore these ideas in some examples.
(a) Consider a fair coin which can land on either heads or tails. Label the event of heads via H and the event of tails
via T . Now define an indicator function 1H via
1H =
{
1 event H occurs
0 event H does not occu
i. What is E[1H ]?
A 0.4
B 1
C 0
D 0.52
E 0.5
ii. Now suppose someone tampers with the coin so that the probability of it landing on heads is now 0.51. What
is E[1H ]?
A 0.4
B 1
C 0
D 0.51
E 0.5
(b) Now consider an economy with two dates t ∈ {0, 1} = T . The state of the economy at date-1 is not known at
date-0. What is known is that there are K possible states, where K is some strictly positive integer. We denote
the K-states by ω1, . . . , ωK . When we collect them all together in one set, we have the state space (sometimes
called the sample space) Ω = {ω1, . . . , ωK}. The physical probability of being in the state ωk at date-1 is denoted
y Pk, where k ∈ {1, . . . ,K}. Of course, ∀ k ∈ {1, . . . ,K}, 0 ≤ Pk < 1 and
∑K
k=1 PK = 1. Define a set of K
securities, where security k has date-1 payoff given by the indicator function 1ωk .
i. Find E0[1ωk ].
A Pk−1
B Pk
C 1
D Pk + Pk+1
E Pk+Pk+12
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ii. Define the discrete-time stochastic process M = {Mt}t∈T = {M0,M1}. At date-0 we know that M0 = 1, but
M1 is a random variable which takes the value M1(ωk) if the economy ends up in state ωk at date-1. We shall
also impose the conditions that ∀ k ∈ {1, . . . ,K}, M1(ωk) > 0 and E0[M1] = M0. Hopefully, it is obvious that
M is a martingale with respect to the physical probabilities and the state space we have described above. M
is also strictly positive.
We now use M1 to define Q1, Q2, . . . , QK via
Qk = E0[M11ωk ], ∀ k ∈ {1, . . . ,K}.
A. What is
∑K
k=1Qk?
A Q2
B 0
C 1
D 0.5
E Pk+Pk+12
When you think about this question, you will see that we used a martingale which was strictly positive to define
a new set of numbers, Q1, Q2, . . . , QK , which represent probabilities – do you see why?. These probabilities
are different from the physical probabilities – they are a new set of probabilities.
iii. Which one of the following statements is true?
A M1(ωk) =
Pk
Qk
B M1(ωk) =
Qk
Pk
C M1(ωk) =
Q2k
Pk
D M1(ωk) = 2
Qk
P 2k
E M1(ωk) =
Qk
P 2k
iv. Where EQ0 [·] denotes a date-0 conditional expectation using the new probabilities, Q1, . . . , QK , find E
Q
0 [M1]
A
∑K
k=1
Qk
2Pk
B 1
C
∑K
k=1
Q2k
Pk
D
∑K
k=1
Q2k
P 2k
E
∑K
k=1
Q2k
2Pk
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