MA577 – MATHEMATICS OF FINANCIAL DERIVATIVES
MATHEMATICS OF FINANCIAL DERIVATIVES
MA577 – MATHEMATICS OF FINANCIAL DERIVATIVES
Solution to the final exam review
Problem 1
Let St denote the value of one share of a stock. Consider a European derivative with maturity T and
strike K that pays ST if ST ≥ K and that pays nothing if ST < K. This is called an asset-or-nothing call
option. Assume that the risk-free interest rate is r, the volatility of the stock is σ and that the stock price evolves
according to the Black-Scholes-Merton model.
1. Compute the value V0 and the delta ∆0 of this option at time 0.
2. Let S0 = $50, T = 1/2, K = $50, σ = 30% and r = 2%. Assume that you have a short position in 940 of these
options. How many shares of stock should you buy or sell to make your portfolio as delta neutral as possible?
Solution: 1. Using the risk-neutral valuation framework, we have
V0 = E
(
e−rTST1{ST>K}
)
= e−rTE
(
S0e
(
r−σ22
)
T+σWT
1{N>−d2}
)
= S0e
−σ2T2
∫ d2
−∞
1√
2pi
e−σ
√
Txe−
x2
2 dx
= S0e
−σ2T2
∫ d2
−∞
1√
2pi
e
σ2T
2 e−
(x+σ
√
T )2
2 dx
= S0
∫ d1
−∞
1√
2pi
e−
x2
2 dx
= S0N(d1).
For the delta at time 0, we have
∆0 =
∂V0
∂S0
,
so that
∆0 = N(d1) +
S0
S0σ
√
T
N ′(d1) = N(d1) +
N ′(d1)
σ
√
T
.
2. With these values, we have d1 = 0.153206, so that N(d1) = 0.5596 and N
′(d1) = 0.394288. Hence, the delta
of a long position in one of these options is ∆0 = 2.41829. Hence, the delta of the above mentioned portfolio is
∆portfolio = −940×∆0 = −2273.19. So in order to make the portfolio delta neutral, we should buy 2,273 shares of
stock.
1
Problem 2
Assume that {St : t ≥ 0} is a geometric Brownian motion with parameters r and σ, i.e., dSt = rStdt + σStdWt.
Prove that the process
{
Xt =
er(T−t)
St
: t ≥ 0
}
is also a geometric Brownian motion and specify its parameters.
Give an explicit expression of Xt.
Solution: We apply the Ito¯ formula to f(t, St), where f(t, x) =
er(T−t)
x . We have ∂tf(t, x) = −rf(t, x),
∂xf(t, x) = − f(t,x)x and ∂xxf(t, x) = 2f(t,x)x2 . We get
er(T−t)
St
=
erT
S0
− r
∫ t
0
f(u, Su)du−
∫ t
0
f(u, Su)
Su
dSu +
1
2
∫ t
0
2f(u, Su)
S2u
d 〈Su〉
=
erT
S0
− r
∫ t
0
f(u, Su)du−
∫ t
0
f(u, Su)
Su
rSudu−
∫ t
0
f(u, Su)
Su
σSudWu + +
∫ t
0
f(u, Su)
S2u
σ2S2udu
=
erT
S0
+
∫ t
0
(σ2 − 2r)f(u, Su)du−
∫ t
0
σf(u, Su)dWu.
Recalling that f(u, Su) = Xu, we get in differential form
dXt = (σ
2 − 2r)Xtdt− σXtdWt.
This shows that X is a geometric Brownian motion with parameters σ2−2r and −σ. Hence, an explicit expression
for Xt is given by
Xt = X0e
(
σ2
2 −2r
)
t−σWt .
Problem 3
The following table gives the prices of bonds:
Bond principal
($)
Time to maturity
(years)
Annual coupon∗
($)
Bond price
($)
100 0.50 0 98
100 1.00 0 95
100 1.50 6.2 101
100 2.00 8.0 104
∗ Half the stated coupon is assumed to be paid every 6 months.
1. Calculate zero rates for maturities of 6 months, 12 months, 18 months, and 24 months.
2. What are the forward rates for the following periods: 6 months to 12 months, 12 months to 18 months, and 18
months to 24 months?
3. What are the 6-months, 12-months, 18-months, and 24-months par yields for bonds that provide semiannual
coupons?
4. What is the price of a 2-year bond providing a semiannual coupon of 7% per annum?
5. Use long and short positions in the above bonds to build a bond portfolio that is equivalent to a 1.5-year zero
coupon bond with a principal of $15,000.
Solution: 1. The zero rate for a maturity of six months, expressed with continuous compounding is 2 ln(1+2/98) =
4.0405%. The zero rate for a maturity of one year, expressed with continuous compounding is ln(1 + 5/95) =
5.1293%. The 1.5-year rate is R where
3.1e−0.040405×0.5 + 3.1e−0.051293×1 + 103.1e−R×1.5 = 101.
The solution to this equation is R = 5.4429%. Similarly, the 2-year rate is 5.8085%.
2. The continuously compounded forward rates calculated using the formula
RF =
R2T2 −R1T1
T2 − T1
are, respectively, 6.2181%, 6.0700%, and 6.9054%.
3. The par yields, expressed with semiannual compounding, can be calculated from the formula
c =
(100− 100d)m
A
so that the par yields are, respectively, 4.0816%, 5.1813%, 5.4986%, and 5.8620%.
4. The price of the bond is
3.5e−0.040405×0.5 + 3.5e−0.051293×1 + 3.5e−0.054429×1.5 + 103.5e−0.058085×2 = 102.13.
5.We have the first three bonds B1, B2 and B3 at our disposal to do this.
Bond B1:
0.5Y
$98
$100
Bond B2:
1Y
$95
$100
Bond B3:
0.5Y 1Y 1.5Y
$101
$3.1 $3.1 $103.1
What we want is a 1.5-year zero coupon bond with a principal of $15,000, given by the following diagram.
1.5Y
$13,824 (= 15, 000e−0.054429×1.5)
$15,000
The only bond we can use to obtain a payment of $15,000 after 1.5 years is the bond B3. B3 pays $103.1 after 1.5
years. We hence need a long position in 15000103.1 = 145.489 bonds B3. 145.489 bonds B3 yield the following diagram.
0.5Y 1Y 1.5Y
$14,694.469 (= 145.489× 101)
$451.018 $451.018 $15,000
In order to eliminate the first coupon, we have to use the bond B1. B1 pays $100 after 0.5 years. We hence need
a short position in 451.018100 = 4.510 bonds B1.
Similarly, in order to eliminate the second coupon, we have to use the bond B2. B2 pays $100 after 1
year. We hence need a short position in 451.018100 = 4.510 bonds B2. So our portfolio is
Long position Short position
145.489 B3 4.510 B1
4.510 B2
The cost of this portfolio is hence
145.489× 101− 4.510× 98− 4.510× 95 = $13, 824,
which is the price we where looking for for our 1.5-year zero coupon bond. The associated diagram is
0.5Y 1Y 1.5Y
$13,824
(Price of the portfolio)
$451.018
(Long pos. in B3)
$451.018
(Short pos. in B1)
$451.018
(Long pos. in B3)
$451.018
(Short pos. in B2)
$15,000
(Long pos. in B3)
which is equivalent to what we wanted, i.e.,
1.5Y
$13,824
$15,000
Problem 4
The 6-month, 12-month, 18-month, and 24-month zero rates are 4%, 4.5%, 4.75%, and 5%, with semian-
nual compounding.
1. What are the rates with continuous compounding?
2. What is the forward rate for the 6-month period beginning in 18-months?
3. What is the value of a FRA that promises to pay you 6% (compounded semiannually) on a principal of $1
million for the 6-month period starting in 18 months?
Solution: 1. With continuous compounding, the 6-month rate is 2 ln(1.02) = 3.9605%. The 12-month rate
is 2 ln(1.0225) = 4.4501%. The 18-month rate is 2 ln(1.02375) = 4.6945%. The 24-month rate is 2 ln(1.025) =
4.9385%.
2. The forward rate (expressed with continuous compounding) is given by
4.9385%× 2− 4.6945%× 1.5
0.5
= 5.6707%.
3. The value of a FRA that promises to pay 6% for the six month period starting in 18 months is given by the
formula
VFRA = L(RK −RF )(T2 − T1)e−R2T2 .
Be careful that here, all the rates are expressed with semiannual compounding, and in particular the forward rate
RF . So we have to convert the forward rate that we obtained in the previous part to semiannual compounding.
We get RF = 2(e
0.056707×0.5 − 1) = 5.7518%. Hence, the value of the FRA is given by
VFRA = $1, 000, 000× (0.06− 0.057518)× 0.5e−0.049385×2 = $1, 124.
Problem 5
A $100 million interest rate swap has a remaining life of 10 months. Under the terms of the swap, 6-
month LIBOR is exchanged for 7% per annum (compounded semiannually). The average of the bid-offer rate
being exchanged for 6-month LIBOR in swaps of all maturities is currently 5% per annum with continuous
compounding. The 6-month LIBOR rate was 4.6% per annum 2 months ago. What is the current value of the
swap to the party paying floating? What is its value for the party paying fixed?
Solution: In four months $3.5 million (= 0.5 × 0.07 × $100 million) will be received and $2.3 million
(= 0.5 × 0.046 × $100 million) will be paid. In 10 months $3.5 million will be received, and the LIBOR rate
prevailing in four months time will be paid.
The value of the fixed-rate bond underlying the swap is
3.5e−0.05×4/12 + 103.5e−0.05×10/12 = $102.718 million.
The value of the floating-rate bond underlying the swap is
(100 + 2.3)e−0.05×4/12 = $100.609 million.
The value of the swap to the party paying floating is $102.718− $100.609 = $2.109 million.
The value of the swap to the party paying fixed is −$2.109 million.