l******o
发帖数: 28 | 1 dB*dB=dt
计算得到的平均值确实是dt，但是variance是2(dt)^2,也就是说standard deviation也
是dt级别，为什么可以直接忽略直接写成平均值呢？
求解。 | r**a
发帖数: 536 | 2 what r u talking about? I guess you mix the def of Brownian motions and the
simulation of Brownian motions.
【在 l******o 的大作中提到】
: dB*dB=dt
: 计算得到的平均值确实是dt，但是variance是2(dt)^2,也就是说standard deviation也
: 是dt级别，为什么可以直接忽略直接写成平均值呢？
: 求解。
| w******n
发帖数: 645 | 3 I'm sorry, I made a mistake. Sorry to lyonscao!
dB is normal with mean 0 variance dt by approximation.
E[dB*dB]=E[dB^2]=dt, this is 2nd moment of a normal.
THE BELOW IS WRONG, DON'T TRUST.
****************************************************************************
******
Var[dB^2]=E[dB^4]-E[dB^2]^2=dt^2-dt^2=0, recall 4th moment of normal
【在 l******o 的大作中提到】
: dB*dB=dt
: 计算得到的平均值确实是dt，但是variance是2(dt)^2,也就是说standard deviation也
: 是dt级别，为什么可以直接忽略直接写成平均值呢？
: 求解。
| f*********5
发帖数: 367 | | l******o
发帖数: 28 | 5 谢谢回复
但是E[dB^4]=3(dt)^2
【在 w******n 的大作中提到】
: I'm sorry, I made a mistake. Sorry to lyonscao!
: dB is normal with mean 0 variance dt by approximation.
: E[dB*dB]=E[dB^2]=dt, this is 2nd moment of a normal.
: THE BELOW IS WRONG, DON'T TRUST.
: ****************************************************************************
: ******
: Var[dB^2]=E[dB^4]-E[dB^2]^2=dt^2-dt^2=0, recall 4th moment of normal
| s*j
发帖数: 7 | 6 这个表达式说的是quadratic variation，不是平均值。
【在 l******o 的大作中提到】
: dB*dB=dt
: 计算得到的平均值确实是dt，但是variance是2(dt)^2,也就是说standard deviation也
: 是dt级别，为什么可以直接忽略直接写成平均值呢？
: 求解。
| n****e
发帖数: 629 | 7 这个是在积分号下才成立.
确实variance~dt^2，
积分后因为各个dB*dB是独立的，整个积分的variance是\int f(t)(dt)^2
只要f(t)不是太变态，你就可以把这项扔掉啦。
【在 l******o 的大作中提到】
: dB*dB=dt
: 计算得到的平均值确实是dt，但是variance是2(dt)^2,也就是说standard deviation也
: 是dt级别，为什么可以直接忽略直接写成平均值呢？
: 求解。
| w******n
发帖数: 645 | 8 I'm sorry, I thought it too quickly.
The actual reason is that we cannot think dB*dB=dt as it's mean. It's simply
a statement that Brownian motion accumulates quadratic variation at rate
one per unit time. It's just an informal writing and it's only meaningful
when you do integral.
In Shreve's book, section 3.4, we get the below explanations.
Assume approximation dB ~ B(t_i+1)-B(t_i), for a partition 0= t_0 < t_1 <...
....
Now we have E[(B(t_i+1)-B_(t_i))^2] = t_i+1-t_i ~dt, Var[(B(t_i+1)-B(t_i))^2
] = 2(t_i+1-t_i)^2 ~ 2*dt^2
Now the argument here is that, when dt is so small, dt^2 is very small. And
therefore (B(t_i+1)-B(t_i))^2 is with high probability near its mean,
although it's random.
So we write (B(t_i+1)-B(t_i))^2 ~ t_i+1 - t_i , when we sum the partition up
and let the dt goes to 0, the Law of Large numbers can cancel errors and we
get the correct integral. So, when we write in differential form, we can
safely write it as dB*dB = dt, since this will gives correct integral. But
again, it's not the formal writing, it's just a convenient way to derive all
the stochastic integrals and SDEs.
【在 l******o 的大作中提到】
: 谢谢回复
: 但是E[dB^4]=3(dt)^2
| l******o
发帖数: 28 | |
|
