y**********0
发帖数: 425 | 1 从Qd转换到Qf
likelihood ratio is d(Ld)=Ld*ud(t)*dWp(dWp is Brownian Motion under
P(real)
martingale)
d(Lf)=Lf*uf(t)*dWp
从Qd到Qf的
L=Lf/Ld
那么 dL=L*(uf(t)-ud(t))*dWp
但是这里就有问题了，从Qd转换到Qf时上面的dWp应该是dWd吧(under Qd martingale)
从Qd到Qf的likelihood ratio的kernel 应该是uf(t)-ud(t)对吧，书上是这么说的，但
是得到的公式却出现了dWp而不是dWd。问题在哪里呢？谢谢。 | Q***5
发帖数: 994 | 2 It seems that the author has omitted the drifting terms in all the equations
, only showed the volatility part -- which is the same under all the
equivalent processes. | y**********0
发帖数: 425 | 3
equations
hope you read more books about martingale.
Here we are talking about likelihood ratio, not the martingale dynamics of
asset.
【在 Q***5 的大作中提到】
: It seems that the author has omitted the drifting terms in all the equations
: , only showed the volatility part -- which is the same under all the
: equivalent processes.
| Q***5
发帖数: 994 | 4
not a very good attitude -- but I will ignore it this time :-)
asset.>
These two are closely related, think about numeraries. | r**a
发帖数: 536 | 5 我也觉得有问题。最后的方程少了一个drift。
假设real prob is Q_p, the two martingale measures are Q_f, Q_d. According to
the equations of likelihood process, we have
dQ_d/dQ_p=L_d and dQ_f/dQ_p=L_f
So L_f/L_d=dQ_f/dQ_d, this implies that there exists a kernel \phi(t) such
that under the measure Q_d, we have
d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d
Now if we change the measure Q_d to Q_p, the dW_d should be replaced by (dW_
p-u_d*dt).Thus there is a drift term appearing in the final result.
Another way to think about this is to use the Ito's formula on L_f/L_d
directly when you calculate d(L_f/L_d). Finally, you'll find the answer
gotten from above two ways are consistant. | y**********0
发帖数: 425 | 6
to
dW_
right, there should be a drift,
but it's like this d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_p+(...)dt. (based on the
math deduction d(L_f/L_d)=.....(using dL_f=L_f*uf*dW_p,dL_d=L_d*ud*dW_p)
how you get this d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d, how you get dW_d instead
of dW_p, and the \phi(t)here should be uf-ud, right? Anyway, can you give
the way to get d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d,
【在 r**a 的大作中提到】
: 我也觉得有问题。最后的方程少了一个drift。
: 假设real prob is Q_p, the two martingale measures are Q_f, Q_d. According to
: the equations of likelihood process, we have
: dQ_d/dQ_p=L_d and dQ_f/dQ_p=L_f
: So L_f/L_d=dQ_f/dQ_d, this implies that there exists a kernel \phi(t) such
: that under the measure Q_d, we have
: d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d
: Now if we change the measure Q_d to Q_p, the dW_d should be replaced by (dW_
: p-u_d*dt).Thus there is a drift term appearing in the final result.
: Another way to think about this is to use the Ito's formula on L_f/L_d
| r**a
发帖数: 536 | 7
the
instead
You can check that L_f/L_d is the likelihood process corresponding changing
measure from Q_d to Q_f. We all know that the likelihood process is a
martingale, so there should exist \phi(t) such that d(L_f/L_d)=(L_f/L_d)*\
phi(t)*dW_d. The key is what this \phi is.
and the \phi(t)here should be uf-ud, right?
Yes, \phi(t)=uf-ud. I got this from Ito formula calculation. I do not know
how to get it directly from d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d. The point
here is that if you plug \phi(t)=uf-ud into d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_
d and change dW_d to dW_p-u_d*dt, then you can recover the result gotten
from applying Ito's formula.
Anyway, can you give
【在 y**********0 的大作中提到】
:
: to
: dW_
: right, there should be a drift,
: but it's like this d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_p+(...)dt. (based on the
: math deduction d(L_f/L_d)=.....(using dL_f=L_f*uf*dW_p,dL_d=L_d*ud*dW_p)
: how you get this d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d, how you get dW_d instead
: of dW_p, and the \phi(t)here should be uf-ud, right? Anyway, can you give
: the way to get d(L_f/L_d)=(L_f/L_d)*\phi(t)*dW_d,
| y**********0
发帖数: 425 | 8 well, thank you.
i finally found that i made a big mistake when calculating the derivatation
of likelihood process which shuold NOT be like this: d(L_f/L_d)={d(L_f)*L_d-
L_f*d(L_d)}/square(L_f). So all the problems have been solved.
changing
【在 r**a 的大作中提到】
:
: the
:
: instead
: You can check that L_f/L_d is the likelihood process corresponding changing
: measure from Q_d to Q_f. We all know that the likelihood process is a
: martingale, so there should exist \phi(t) such that d(L_f/L_d)=(L_f/L_d)*\
: phi(t)*dW_d. The key is what this \phi is.
: and the \phi(t)here should be uf-ud, right?
: Yes, \phi(t)=uf-ud. I got this from Ito formula calculation. I do not know
| r**a
发帖数: 536 | 9 This mistake is not rare :)
derivatation
d-
【在 y**********0 的大作中提到】
: well, thank you.
: i finally found that i made a big mistake when calculating the derivatation
: of likelihood process which shuold NOT be like this: d(L_f/L_d)={d(L_f)*L_d-
: L_f*d(L_d)}/square(L_f). So all the problems have been solved.
:
: changing
|
|
