c**********e
发帖数: 2007 | 1 Solve dy_t = dt + y dw_t
w_t is the SBM.
该题我见到的做法是
d (e^{-W_t-0.5 t} y_t)= e^{-W_t-0.5 t} dt
可我自己做出的解答是
d (e^{W_t-0.5 t} y_t)= e^{W_t-0.5 t} dt
差别在哪里呢？在于交叉偏导项，即rho*1*Yt*(对Wt和Yt的二阶偏导）。
哪个是对的？ |
l*******l
发帖数: 248 | 2 A few things:
What's SBM, my understanding is Brownian Montion, right?
To get d(XY), you need to apply ito's lemma,
XdY+YdX+dXdY
for d (e^{-W_t-0.5 t}), just make -W_t-0.5 t = Z
then d(e^Z)=e^ZdZ+1/2*e^Zd, then plug in, you will get it.
Your solution does not make any sense.
【在 c**********e 的大作中提到】
: Solve dy_t = dt + y dw_t
: w_t is the SBM.
: 该题我见到的做法是
: d (e^{-W_t-0.5 t} y_t)= e^{-W_t-0.5 t} dt
: 可我自己做出的解答是
: d (e^{W_t-0.5 t} y_t)= e^{W_t-0.5 t} dt
: 差别在哪里呢？在于交叉偏导项，即rho*1*Yt*(对Wt和Yt的二阶偏导）。
: 哪个是对的？
|
c**********e
发帖数: 2007 | 3 SBM is standard BM.
What is your dXdY?
【在 l*******l 的大作中提到】
: A few things:
: What's SBM, my understanding is Brownian Montion, right?
: To get d(XY), you need to apply ito's lemma,
: XdY+YdX+dXdY
: for d (e^{-W_t-0.5 t}), just make -W_t-0.5 t = Z
: then d(e^Z)=e^ZdZ+1/2*e^Zd, then plug in, you will get it.
: Your solution does not make any sense.
|
l*******l
发帖数: 248 | 4 quandratic variation, 你所谓的“交叉项”？
【在 c**********e 的大作中提到】
: SBM is standard BM.
: What is your dXdY?
|
l*******l
发帖数: 248 | 5 Oh, i see, theres a typo
it should be dy_t = y dt + y dw_t,
then ppl's solution is right.
【在 c**********e 的大作中提到】
: Solve dy_t = dt + y dw_t
: w_t is the SBM.
: 该题我见到的做法是
: d (e^{-W_t-0.5 t} y_t)= e^{-W_t-0.5 t} dt
: 可我自己做出的解答是
: d (e^{W_t-0.5 t} y_t)= e^{W_t-0.5 t} dt
: 差别在哪里呢？在于交叉偏导项，即rho*1*Yt*(对Wt和Yt的二阶偏导）。
: 哪个是对的？
|
c**********e
发帖数: 2007 | 6
No, there is no y before dt. Otherwise it becomes GBM...nothing to talk
about.
Do you really understand the question?
【在 l*******l 的大作中提到】
: Oh, i see, theres a typo
: it should be dy_t = y dt + y dw_t,
: then ppl's solution is right.
|
p*****k
发帖数: 318 | 7 careerchange, i will use the notation by chimbo in the other thread:
http://www.mitbbs.com/article_t/Quant/31266383.html
consider the auxiliary GBM:
dXt = Xt dt - Xt dWt
then apply ito for Xt*Yt, which gives
d(Xt*Yt) = Yt dXt + Xt dYt + dXt dYt = Xt dt
(as all the terms with Xt*Yt as coefficients got canceled)
the problem seems to come from the general (diffusion limit of) GARCH(1,1)
model:
dYt = (a + b Yt) dt + c Yt dWt
the same technique would work by introducing a GBM:
dXt = (c^2 - b) Xt dt - c Xt dWt |
T*****w
发帖数: 802 | |
|
f*******o
发帖数: 38 | 9 Should there be a 1/2 in front of dXdY?
【在 l*******l 的大作中提到】
: A few things:
: What's SBM, my understanding is Brownian Montion, right?
: To get d(XY), you need to apply ito's lemma,
: XdY+YdX+dXdY
: for d (e^{-W_t-0.5 t}), just make -W_t-0.5 t = Z
: then d(e^Z)=e^ZdZ+1/2*e^Zd, then plug in, you will get it.
: Your solution does not make any sense.
|
f*******o
发帖数: 38 | 10 Uh, my bad. 1 is correct.
【在 f*******o 的大作中提到】
: Should there be a 1/2 in front of dXdY?
|
l*******l
发帖数: 248 | 11 No...u r confused with sth else
【在 f*******o 的大作中提到】
: Should there be a 1/2 in front of dXdY?
|
s****n
发帖数: 41 | 12 pcasnik大牛，不好意思我还是有些不明白。
你的方法解出来，Xt=X0*exp(-Wt+0.5t),这样得到的结果貌似和careerchange的正确答案Xt=X0*exp(-0.5t-Wt)还是不一样，到底哪个才是正解？
另外，我看到之前Blook给出的解法，如下
“
1. you know how to do the following:
dY + aY dt = f(t)dt
2. you know how to perform ito's lemma to
d(e^W_t)
3. follow the same logic as 1, you will know how to do
dY_t + bY_tdW_t = dt
或者麻烦一点，推导一下如下通解
dYt=aYtdt + bYtdBt
然后记住结论，这样对所有类似题就通吃了。
”
我用他的方法解，似乎最后要解一个ODE。请问你们用的是一个方法吗？
多谢！
【在 p*****k 的大作中提到】
: careerchange, i will use the notation by chimbo in the other thread:
: http://www.mitbbs.com/article_t/Quant/31266383.html
: consider the auxiliary GBM:
: dXt = Xt dt - Xt dWt
: then apply ito for Xt*Yt, which gives
: d(Xt*Yt) = Yt dXt + Xt dYt + dXt dYt = Xt dt
: (as all the terms with Xt*Yt as coefficients got canceled)
: the problem seems to come from the general (diffusion limit of) GARCH(1,1)
: model:
: dYt = (a + b Yt) dt + c Yt dWt
|
c**********e
发帖数: 2007 | 13 swchen,
Xt=X0*exp(-0.5t-Wt) is not a correct solution.
It is a solution given somebody before. The reason for 重起一贴
was that this solution is wrong.
So I made my solution, which was the same as pcasnik大牛's
solution. But when I posted it wrong (it is corrected now.)
pcasnik大牛's solution Xt=X0*exp(-Wt+0.5t) is the correct one.
答案Xt=X0*exp(-0.5t-Wt)还是不一样，到底哪个才是正解？
【在 s****n 的大作中提到】
: pcasnik大牛，不好意思我还是有些不明白。
: 你的方法解出来，Xt=X0*exp(-Wt+0.5t),这样得到的结果貌似和careerchange的正确答案Xt=X0*exp(-0.5t-Wt)还是不一样，到底哪个才是正解？
: 另外，我看到之前Blook给出的解法，如下
: “
: 1. you know how to do the following:
: dY + aY dt = f(t)dt
: 2. you know how to perform ito's lemma to
: d(e^W_t)
: 3. follow the same logic as 1, you will know how to do
: dY_t + bY_tdW_t = dt
|
s****n
发帖数: 41 | 14 Careerchange,
Thanks a lot. No wonder I cannot figure out exp(-0.5t-Wt) in the old posts.
【在 c**********e 的大作中提到】
: swchen,
: Xt=X0*exp(-0.5t-Wt) is not a correct solution.
: It is a solution given somebody before. The reason for 重起一贴
: was that this solution is wrong.
: So I made my solution, which was the same as pcasnik大牛's
: solution. But when I posted it wrong (it is corrected now.)
: pcasnik大牛's solution Xt=X0*exp(-Wt+0.5t) is the correct one.
:
: 答案Xt=X0*exp(-0.5t-Wt)还是不一样，到底哪个才是正解？
|
