exchange rate call option.pdf
TRANSCRIPT
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Exchange Rate Call OptionMauricio Bedoya
September 2014
To understand this blog, we must know:
1. Girsanov Theorem.
2. Probability Theory.
Definition: Girsanov Theorem (P. Glasserman Book)Let Y be as
Y(t) = e12
t0 ||(u)||
2du+t0 dw(u) (1)
witht
0||(u)||2du
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In the blog Exchange Rate Process and Interest Rate Parity , we found that X(t) is charac-terized in a risk neutral world
X(t) = X(0)
e(r(d)r(f)
12
2)T+w(T) (5)
with r(d), r(f), constant and w(T) N[0,dt]. Operating in equation 4, we found
C(0) = er(d)T E[X(T) I{X(T)K}]
Part I
er(d)T K E[I{X(T)K}] Part II
(6)
We split the solution of equation 6, because the procedure to solve part I use Girsanov Theo-rem. Girsanov Theorem is not used to solve part II.
1. Solution Part IReplacing equation 5 in equation 6, we get
er(d)T E[X(T) I{X(T)K}] =er(d)T E[X(0) e(r(d)r(f)12
2)T+w(T) I{X(T)K}]=er(f)T X(0) E[e 122T+w(T) I{X(T)K}]
Girsanov Here
=er(f)T
X(0)
E[I{X(T)K}]
=er(f)T X(0) E[I{X(0)e
(r(d)r(f)+12 2)T+ w(T)K}]
=er(f)T X(0) E[I{(r(d)r(f)+ 122)T+ w(T)ln KX(0) }]=er(f)T X(0) E[I{ w(T)Ln( KX(0) )(r(d)r(f)+ 122)T}]=er(f)T X(0) E[I
{ w(T)Ln( K
X(0))(r(d)r(f)+122)T
}
]
=er(f)T X(0) E[I{ Z(T)
Ln( KX(0)
)(r(d)r(f)+12 2)T
T
}
]
(7)
Now lets explain a little bit what happen. In the second line of equation 7, we cancelr(d). What we get next, is the Radom Nikodym Derivative times an indicator function.Using Girsanov, we change the measure w(T) by w(T). Where we have w(T) we will insert
w(T)+ T. If we multiply by the previous expression, we get
w(T)+ 2 T = w(T)
operating with 1
2 2
T, we get the fourth line. In the fifth line, we take logarithms andleave w(T)alone. Next, we use the fact that w(T) N[0,T], and proceed to standardize,
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leaving the expression
Z(T)Ln( K
X(0)
)
(r(d)
r(f)+
12
2)
T
T (8)
finally, we know that Z(T) N[0, 1], and due to the symmetry of Standard Normal Dis-tribution, equation 8 can be rewritten
Z(T) Ln(
X(0)
K ) + (r(d) r(f)+ 12 2) T
T d1
(9)
Equation 7 can be rewritten, using result of equation 9, to obtain
er(d)T E[X(T) I{X(T)K}] =er(f)T X(0) N[d1] (10)
with N[d1] characterizing the CDF till d1 of a Standard Normal Distribution.
2. Solution of Part IIPart II is easy because the expectation of an indicator function, is equal to the probabilityof the event, then
er(d)T K E[I{X(T)K}] =er(d)T K E[I{X(0)e(r(d)r(f)12 2)T+w(T)K}]=er(d)T K E[I{(r(d)r(f) 122)T+w(T)Ln KX(0) }]=er(d)T K E[I{w(T)Ln KX(0)(r(d)r(f) 122)T}]=er(d)T K E[I
{ Z(T)Ln K
X(0)(r(d)r(f) 12 2)T
T
}
]
(11)
Knowing that Z(T) N[0, 1], we can use the symmetry property of Z(T), to get
E[I{ Z(T)
Ln KX(0)
(r(d)r(f) 122)T
T
}
] =E[I
{ Z(T)Ln
X(0)
K + (r(d) r(f) 12 2) T
T d2
}
] (12)
using the result of equation 12, we can rewrite equation 11
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er(d)T K E[I{X(T)K}] =er(d)T K N[d2] (13)
with N[d2] characterizing the CDF till d2 of a Standard Normal Distribution. The solution
of equation 6 (Exchange Rate Call Option) is the difference between equation 10 and 13.
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