以下是引用irvingy在2007-12-17 2:34:00的发言:第一个直接套Girsanov,Z是zero coupon bond price
dZ_t = \mu_t dt + \sigma_t dW_t
W_t是P-Brownian motion
\lambda_t = \mu_t / \sigma_t
dZ_t = \sigma_t (\lambda_t dt + dW_t)
\int_0^t \lambda_s ds + W_t is Q-Brownian motion and Z is a martingale under Q
dQ/dP = exp(-\int_0^t \lambda_s dW_s - 1/2 \int_0^t \lambda_s^2 ds)
第二个,S_2 = 5^(1/2), S_3 = 5^(1/2 + 1/4), S_4 = 5^(1/2 + 1/4 + 1/8)...
1/2 + 1/4 + 1/8 +...收敛到1,S_n收敛到5
我喜欢看到martingale的形式如下:
dM=HdW
这是stochastic representation theorem of martinagle
其中H_t 是adapted process, W_t是 Brownian motion
解答是正确的,可是还没有推到很明显的地步
here comes my solution, i hope it easier to understand
Under original measure P, we have dynamics
dB=r B dt, bank account
dZ=mu Z dt + sigma Z dW_t(P), zero bond.
Here dW_t(P) denotes the bronian motion under measure P
take B as numeraire, after some easy calculation we have
d(Z/B)=(mu-r) (Z/B)dt+sigma (Z/B) dW_t(P), (1)
now use Girsanov measure change theorem:
dW_t(P)=Lamada_t dt + dW_t(Q), Here dW_t(Q) denotes the bronian motion under measure Q
Put it into formula (1),
d(Z/B)=(mu-r+sigma Lamada_t) (Z/B)dt+sigma (Z/B) dW_t(Q) , ( 2)
By stochastic representation theorem of martingale, we should set mu-r+sigma Lamada_t=0,
in order to let Z/B be a martingale under measure Q .
Hence, Lamada_t=(r-mu)/sigma
Again by Girsanov theorem, we know Lamada_t=dQ/dP
Hence dQ=(r-mu)/sigma dP
[此贴子已经被作者于2008-1-25 17:10:43编辑过]