x*****i
发帖数: 287 | 1 $d X_t = X_t^{1.5} dB_t$
Any idea? Thanks a lot. |
r**a
发帖数: 536 | 2 Let Y=X_t^{-1/2}. Investigate d(e^{-3t/4}Y). |
x*****i
发帖数: 287 | 3 I still cannot get the answer by this way.
I tried e^{a t} x^b to solve this equ, but cannot find
good a and b's. Yours seems not working. Maybe I made some
mistakes. Still thank you so much. |
L**********u
发帖数: 194 | 4 他的那个解好像不对。
我验算过了。
呵呵
【在 x*****i 的大作中提到】
: I still cannot get the answer by this way.
: I tried e^{a t} x^b to solve this equ, but cannot find
: good a and b's. Yours seems not working. Maybe I made some
: mistakes. Still thank you so much.
|
r**a
发帖数: 536 | 5 Sorry for confusion. I made a stupid mistake.
【在 x*****i 的大作中提到】
: I still cannot get the answer by this way.
: I tried e^{a t} x^b to solve this equ, but cannot find
: good a and b's. Yours seems not working. Maybe I made some
: mistakes. Still thank you so much.
|
L**********u
发帖数: 194 | 6 这个方程应该没有closed form的解。
用变量替换 Y=1/X，
方程可以化为CIR model：
dY=dt-sqrt(Y)dB_t.
【在 x*****i 的大作中提到】
: $d X_t = X_t^{1.5} dB_t$
: Any idea? Thanks a lot.
|
x*****i
发帖数: 287 | 7 thanks，seems very good :)
but, it isn't a CIR model, since it doesn't have the mean reversion
property. |
r**a
发帖数: 536 | 8 I got another solution.
Let Y=X^{-1/2}. Then dY=3/8*Y^{-1}dt-1/2*dB
Now let Z=e^Y. Then
$$
dZ=[Z(\frac{1}{8}+\frac{3}{8\log(Z)})]dt-\frac{Z}{2}dB
$$
Notice that the above eq. regarding Z is in the form of
$$
dZ=f(Z)dt+\frac{Z}{2}dB
$$
Then we may use the result of exersice 5.16 in Oksendal's book "stochastic
differential equations" to solve it.
【在 L**********u 的大作中提到】
: 这个方程应该没有closed form的解。
: 用变量替换 Y=1/X，
: 方程可以化为CIR model：
: dY=dt-sqrt(Y)dB_t.
|
L**********u
发帖数: 194 | 9 这本书很不错，决定把第5章看看。
【在 r**a 的大作中提到】
: I got another solution.
: Let Y=X^{-1/2}. Then dY=3/8*Y^{-1}dt-1/2*dB
: Now let Z=e^Y. Then
: $$
: dZ=[Z(\frac{1}{8}+\frac{3}{8\log(Z)})]dt-\frac{Z}{2}dB
: $$
: Notice that the above eq. regarding Z is in the form of
: $$
: dZ=f(Z)dt+\frac{Z}{2}dB
: $$
|
