l*********o
发帖数: 88 | 1 给一个随机过程 X(t), 随机微分方程：
dS = sigma(X,t) dW.
证明：the quadratic variation of X up to T 是一个常数的必要且充分条件是
sigma(X,t) = sigma(t)。也就是说sigma不是随机的。
谢谢。 | x******a
发帖数: 6336 | 2 why dS?
【在 l*********o 的大作中提到】
: 给一个随机过程 X(t), 随机微分方程：
: dS = sigma(X,t) dW.
: 证明：the quadratic variation of X up to T 是一个常数的必要且充分条件是
: sigma(X,t) = sigma(t)。也就是说sigma不是随机的。
: 谢谢。
| l*********o
发帖数: 88 | 3 Should be dX.
【在 x******a 的大作中提到】
: why dS?
| c**********e
发帖数: 2007 | 4 What is the quadratic variation of X up to T? Could you clarify it?
是一个常数 with respect to T or with respect to randomization?
【在 l*********o 的大作中提到】
: 给一个随机过程 X(t), 随机微分方程：
: dS = sigma(X,t) dW.
: 证明：the quadratic variation of X up to T 是一个常数的必要且充分条件是
: sigma(X,t) = sigma(t)。也就是说sigma不是随机的。
: 谢谢。
| l******i
发帖数: 1404 | 5 Proof:
dX(t) = sigma(X(t),t)*dW(t)
With quadratic variation of X(t)=M(X)
=integration from 0 to t of (sigma(X(s),s)^2)*ds
which is a distribution if sigma is stochastic.
So M(X) is constant if and only if Var(M(X))=0.
Var(M(X))=0 iff sigma is a deterministic function of t. | J**********g
发帖数: 213 | 6 It's pretty straightforward except the last line:
Var(M(X))=0 iff sigma is a deterministic function of t.
can you show how to prove? Thx.
【在 l******i 的大作中提到】
: Proof:
: dX(t) = sigma(X(t),t)*dW(t)
: With quadratic variation of X(t)=M(X)
: =integration from 0 to t of (sigma(X(s),s)^2)*ds
: which is a distribution if sigma is stochastic.
: So M(X) is constant if and only if Var(M(X))=0.
: Var(M(X))=0 iff sigma is a deterministic function of t.
| p******5
发帖数: 138 | 7 同问 最后一步
Var(M(X))=0 iff sigma is a deterministic function of t ??
【在 l******i 的大作中提到】
: Proof:
: dX(t) = sigma(X(t),t)*dW(t)
: With quadratic variation of X(t)=M(X)
: =integration from 0 to t of (sigma(X(s),s)^2)*ds
: which is a distribution if sigma is stochastic.
: So M(X) is constant if and only if Var(M(X))=0.
: Var(M(X))=0 iff sigma is a deterministic function of t.
| J**********g
发帖数: 213 | 8 I actually think this result is not true.. | l******i
发帖数: 1404 | 9 Integration of distributions is approximately a sum of distributions,
which cannot be deterministic.
So M(X) is deterministic iff sigma is deterministic.
For example, suppose sigma is a distribution process
suppose with some possibility, sigma can be a for all X and t,
then M(X)=(a^2)t with some possibility;
suppose sigma can also be b for all X and t with some possibility,
then M(X)=(b^2)t with some possibility.
Thus M(X) cannot be deterministic
【在 J**********g 的大作中提到】
: It's pretty straightforward except the last line:
: Var(M(X))=0 iff sigma is a deterministic function of t.
: can you show how to prove? Thx.
| c**********e
发帖数: 2007 | 10
如果sigma_t是X_t的符号函数, 如果X_t是正的，sigma_t=1，否则sigma_t=-1.
那么sigma_t是随机的, 但是这个积分却不是随机的。
这个证明的思路是有道理的，但证明本身还是不对的。
【在 l******i 的大作中提到】
: Integration of distributions is approximately a sum of distributions,
: which cannot be deterministic.
: So M(X) is deterministic iff sigma is deterministic.
: For example, suppose sigma is a distribution process
: suppose with some possibility, sigma can be a for all X and t,
: then M(X)=(a^2)t with some possibility;
: suppose sigma can also be b for all X and t with some possibility,
: then M(X)=(b^2)t with some possibility.
: Thus M(X) cannot be deterministic
| J**********g
发帖数: 213 | 11 I thought one counter example (in my opinion), but it's relatively
complicated. I will put it up when I have some time. really busy learning
things now...
Or someone can give a rigorous proof...
【在 c**********e 的大作中提到】
:
: 如果sigma_t是X_t的符号函数, 如果X_t是正的，sigma_t=1，否则sigma_t=-1.
: 那么sigma_t是随机的, 但是这个积分却不是随机的。
: 这个证明的思路是有道理的，但证明本身还是不对的。
| M****i
发帖数: 58 | 12 If I am not missing somthing here, the sufficiency is not correct. For
example, choose the function sigma(y,t) to be the constant 1, then clearly
we have sigma(X,t) is deterministic (identically 1). But the corresponding
SDE becomes dX(t)=dW(t), so that the solution is X(t)=X(0)+W(t), whose
quadratic variation [X](t)=t, this is not a constant.
Maybe the question should be changed into:
If dX(t)=sigma(X(t),t)dW(t), then the following are equivalent:
1)the quadratic variation of X is a constant;
2) sigma(X(t),t) is identically 0 a.s.;
3) X(t)=X(0) a.s.
Proof:
1) is equivalent to 2):
The quadratic variation of X is [X](t)=int_0^t(sigma(X(t),t))^2dt, this is
a constant iff its derivative w.r.t t is identically 0, i.e. sigma(X(t),t)=0.
1) is equivalent to 3):
Observe that the SDE implies X is a continuous local martingale, so X(t)=X(0
) a.s. iff its quadratic variation is identically 0. But by definition, the
quadratic variation is a constant iff it is identically 0. CQFD |
|
