Unformatted text preview: CHAPTER 20. Pricing Exotic Options 213 ~
M(T)
y x=y ~
m
(B(T), M(T)) lies in here ~
b x
~
B(T) e f
Figure 20.3: Possible values of B T ; M T .
We consider only the case S 0 K L; so 0 ~ m:
b ~
The other case, K S 0 L leads to ~ 0 m and the analysis is similar.
b
~
R m R m : : :dy dx:
~ ~
We compute ~ x
b Z mZ m
~ ~ , 2
p
S 0 expf xg , K 22y , x exp , 2y2T x + x , 1 2 T dy dx
2
~ x
b
T 2T
y=m
Zm
~
2
~
1
= ,e,rT ~ S 0 expf xg , K p 1 exp , 2y , x + x , 2 2 T
dx
2T
b
2T
y=x
" 2
Z~
,rT m S 0 expf xg , K p 1 exp , x + x , 1 2 T
=e
2
~
2T
b
2T
2m , x2 + x , 1 2 T dx
~
, exp , 2T
2
~
1 e,rT S 0 Z m exp x , x2 + x , 1 2 T dx
=p
2
~
2T
2T
b
~
1 e,rT K Z m exp , x2 + x , 1 2 T dx
,p
2
~
2T
2T
b
~
1 e,rT S 0 Z m exp x , 2m , x2 + x , 1 2 T dx
~
,p
2
~
2T
2T
b
~
1 e,rT K Z m exp , 2m , x2 + x , 1 2 T dx:
~
+p
2
~
2T
2T
b v 0; S 0 = e,rT The standard method for all these integrals is to complete the square in the exponent and then
recognize a cumulative normal distribution. We carry out the details for the ﬁrst integral and just ...
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 Fall '15
 mr.somebody
 Derivative, Normal Distribution, dx

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