4
Interest Rate Derivative Securities
243
with
A
(
T
) =
B
(
T
) = 0 and determine the system of ordinary diFerential
equations the functions
A
(
t
) and
B
(
t
) should satisfy.
Solution
:
Substituting
V
(
r, t
) =
e
A
(
t
)

rB
(
t
)
into the equation yields
dA
dt

r
dB
dt
+ (
a
0
+
a
1
r
)
B
2

(
b
0
+
b
1
r
)
B

r
= 0
.
Thus
dA
dt
+
a
0
B
2

b
0
B
= 0
,
dB
dt

a
1
B
2
+
b
1
B
+ 1 = 0
.
This system of ordinary diFerential equations with the conditions
A
(
T
) =
B
(
T
) = 0 has a unique solution. Therefore the original equation has a
solution in the form
V
(
r, t
) =
e
A
(
t
)

rB
(
t
)
.
Because
A
(
T
) =
B
(
T
) = 0, we have
V
(
r, T
) = 1
.
Consequently, the solution of the system of ordinary diFerential equations
with conditions
A
(
T
) =
B
(
T
) = 0 gives the solution of the bond problem.
6. In the Vasicek model, the spot interest rate is assumed to satisfy
dr
= (¯
μ

γr
)
dt
+
p

βdX,
β <
0
,
γ >
0
,
where
¯
μ, γ
and
β
are constants and
dX
is a Wiener process. Let the
market price of risk
λ
(
r, t
) =
¯
λ
√

β
. Then the price
V
(
r, t
;
T
) of a zero
coupon bond maturing at time
T
with a face value
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 Spring '10
 Zhu
 ORDINARY DIFFERENTIAL EQUATIONS, Interest rate derivative, Zea, Rate Derivative Securities

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