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Quant版 - 一道新的布朗题
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1 (共1页)
l******i
发帖数: 134
1
X(t) = a t + b W(t) , where W(t) is standard Brownian motion.
T = Min(t, X(t) = x ) where x > 0;
What is E(T)
What is Var(T)
e***x
发帖数: 13
2
I use Girsanov to make Xt/b to be a Brownian motion under measure Q, and use
Doob's OST to find the Laplace transform of T, the rest of problem is
trivial.

【在 l******i 的大作中提到】
: X(t) = a t + b W(t) , where W(t) is standard Brownian motion.
: T = Min(t, X(t) = x ) where x > 0;
: What is E(T)
: What is Var(T)

c********y
发帖数: 30813
3
Actually, we don't need to change the measure. Directly apply the Laplace
transform on stopping time T.
since [X(t)-at]/b = W(t) is a brownian motion,
exp(sigma*[X(t)-at]/b-sigma^2 * t/2)) is an exponential martingale.
therefore,
E[exp(sigma*[X(t)-aT]/b-sigma^2 * T/2))]=1
E[exp(lamda*T)]=exp(-sigma*x/b), where lamda = -(a*sigma/b + sigma^2 /2)
E[T]=d[exp(-sigma*x/b)]/d(lamda) @ (lamda=0) = x/a
E[T^2]=dd[exp(-sigma*x/b)]/d (lamda^2) @ (lamda=0) = x^2/a^2+x*b^2/a^3
Var[T]=x*b^2/a^3
l******i
发帖数: 134
4
解的很好啊。下面是我的解法。
1. since [X(t)-at]/b = W(t) is a martingale, E([X(T)- a T ]/b) = 0 , so E(T)
= x/a
2. since Y(t) = W(t)^2 - t is also a martingale, E( ([X(T)- a T ]/b)^2 - T)
= 0
E(T^2) = x b^2/a^3 + x^2/a^2
so, Var(T) = x b^2/a^3
Two concept is used,
1. stopped martingale is martingale,so you can replace t with T, the
stopping time.
2. E(W(t)^2) = t , the basic property of Brownian motion.

【在 c********y 的大作中提到】
: Actually, we don't need to change the measure. Directly apply the Laplace
: transform on stopping time T.
: since [X(t)-at]/b = W(t) is a brownian motion,
: exp(sigma*[X(t)-at]/b-sigma^2 * t/2)) is an exponential martingale.
: therefore,
: E[exp(sigma*[X(t)-aT]/b-sigma^2 * T/2))]=1
: E[exp(lamda*T)]=exp(-sigma*x/b), where lamda = -(a*sigma/b + sigma^2 /2)
: E[T]=d[exp(-sigma*x/b)]/d(lamda) @ (lamda=0) = x/a
: E[T^2]=dd[exp(-sigma*x/b)]/d (lamda^2) @ (lamda=0) = x^2/a^2+x*b^2/a^3
: Var[T]=x*b^2/a^3

1 (共1页)
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