Pricing Asian Options in a Semimartingale Model - Department of ...

We can write
( )
d
Xt
S t
=
=
=
( ) ( ) ( ) ( )
X t
S t
− Xt−
S t−
= q t− − Xt−
S t−
1 − 1
1+∆H t
= q t− − Xt− ∆Ht
S t− 1+∆H t
.
( ) (
)
q t− − Xt−
S t−
dH t − rdt − d〈H c 〉 t +
( ) (
q t− − Xt−
S t−
dH t − rdt − d〈H c 〉 t −
( ) ( ∫ ∞
q t− − Xt−
S t−
dHt c − d〈H c 〉 t +
−∞
(
q t− − Xt−
S t−
) (
∆Ht
1+∆H t
− ∆H t
)
.
)
∆H2 t
1+∆H t
x (µ(dt, dx) − ν(dt, dx)) −
∫ ∞
−∞
)
x 2
1+x µ(dt, dx)
or
(
∫ ∞
∫ ∞
)
dZ t = (q t− − Z t− ) dHt c − d〈H c x
〉 t + x (µ(dt, dx) − ν(dt, dx)) −
2
1+x µ(dt, dx) .
−∞
−∞
Observe that Z t is a Markovian process under Q. Theorem 2.3 and the Markovian property give us the value
process
v(t, Z t ) = E Q [(Z T − K 1 ) + |F t ],
which is a martingale by definition.
Note d〈Z c 〉 t = (q t− − Z t− ) 2 d〈H c 〉 t and thus
dv(t, Z t ) = v t (t, Z t− )dt + v z (t, Z t− )dZ + 1 2 v zz(t, Z t− )d〈Z c 〉 t
+v(t, Z t ) − v(t, Z t− ) − v z (t, Z t− )∆Z t
= v t (t, Z t− )dt + v z (t, Z t− )dZ + 1 2 v zz(t, Z t− )(q t− − Z t− ) 2 d〈H c 〉 t
(
)
+v t, Z t− + (q t− − Z t− ) ∆H
1+∆H
− v(t, Z t− ) − v z (t, Z t− )(q t− − Z t− ) ∆H
1+∆H
= Local Martingale + v t (t, Z t− )dt + 1 2 v zz(t, Z t− )(q t− − Z t− ) 2 d〈H c 〉 t
∫ ∞ { (
)
+ v t, Z t− + (q t− − Z t− ) x
1+x
−∞
}
−v(t, Z t− ) − v z (t, Z t− )(q t− − Z t− ) x
1+x
ν(dt, dx).
The fact that a predictable local martingale with finite variation starting at zero is zero concludes the proof.
⋄
Corollary 3.4 In the case when H is a Lévy process, the integro-differential equation simplifies to
(3.5) v t (t, z) + c 2 v zz(t, z)(q t− − z) 2
for 0 ≤ t ≤ T and z ∈ R.
+
∫ ∞
−∞
{ (
)
}
v t, z + (q t− − z) x
1+x
− v(t, z) − v z (t, z)(q t− − z) x
1+x
K(dx) = 0
Proof. The canonical decomposition of H is
H t = rt +
∫ t
0
√ c dWs +
∫ t
0
∫ ∞
−∞
x ( µ(ds, dx) − K(dx)dt )
where W t is a standard Brownian Motion. Applying theorem 3.3, we get
(3.6) v t (t, Z t− ) + c 2 v zz(t, Z t− )(q t− − Z t− ) 2
∫ ∞ {
+ v
−∞
(
t, Z t− + (q t− − Z t− ) x
1+x
Since the support for Z t− is R, we get the above equation.
)
}
− v(t, Z t− ) − v z (t, Z t− )(q t− − Z t− ) x
1+x
K(dx) = 0.
⋄
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