j**x
发帖数: 34 | 1 If Bt is standard brownian motion, t1
Bt2*Bt3)? |
x******a
发帖数: 6336 | 2 any relation between the brownian motion and the expectation?
t1*
【在 j**x 的大作中提到】
: If Bt is standard brownian motion, t1: Bt2*Bt3)?
|
j**x
发帖数: 34 | |
J******d
发帖数: 506 | 4 显然是0.
Bt1*
【在 j**x 的大作中提到】
: If Bt is standard brownian motion, t1: Bt2*Bt3)?
|
f****e
发帖数: 590 | 5 t2-t1?
【在 J******d 的大作中提到】
: 显然是0.
:
: Bt1*
|
k*******d
发帖数: 1340 | |
f****e
发帖数: 590 | 7 恩，我错了
【在 k*******d 的大作中提到】
: 0吧
|
p*****k
发帖数: 318 | 8 jpmx, as already pointed out, the answer is 0.
the rule of thumb is:
if there are odd number of normal r.v.s, the expectation of
their product (central moment) vanishes;
[of course it also applies to r.v.s with symmetric p.d.f.]
if the # is even, then use the tree diagram to reduce to
products of the elements in the covariance matrix. |
v*******o
发帖数: 10 | 9 r.v.s? "random variables"?
can you explain more details? Thanks!!!
【在 p*****k 的大作中提到】
: jpmx, as already pointed out, the answer is 0.
: the rule of thumb is:
: if there are odd number of normal r.v.s, the expectation of
: their product (central moment) vanishes;
: [of course it also applies to r.v.s with symmetric p.d.f.]
: if the # is even, then use the tree diagram to reduce to
: products of the elements in the covariance matrix.
|
t*******y
发帖数: 637 | 10 for the odd number case, do we need every one of them has zero mean?
【在 p*****k 的大作中提到】
: jpmx, as already pointed out, the answer is 0.
: the rule of thumb is:
: if there are odd number of normal r.v.s, the expectation of
: their product (central moment) vanishes;
: [of course it also applies to r.v.s with symmetric p.d.f.]
: if the # is even, then use the tree diagram to reduce to
: products of the elements in the covariance matrix.
|
|
|
A**T
发帖数: 362 | 11 -E[B_t1 * B_t2 * B_t3] = E[(-B_t1)(-B_t2)(-B_t3)] = E[B'_t1 * B'_t2 * B'_t3]
(By letting B'_t = -B_t. Again B'_t is a BM) = E[B_t1 * B_t2 * B_t3].
【在 v*******o 的大作中提到】
: r.v.s? "random variables"?
: can you explain more details? Thanks!!!
|
b***k
发帖数: 2673 | 12 这个证明对吗？
用你的方法我可以证明E[B_t^2]=0.which is wrong of course.
t3]
【在 A**T 的大作中提到】
: -E[B_t1 * B_t2 * B_t3] = E[(-B_t1)(-B_t2)(-B_t3)] = E[B'_t1 * B'_t2 * B'_t3]
: (By letting B'_t = -B_t. Again B'_t is a BM) = E[B_t1 * B_t2 * B_t3].
|
A**T
发帖数: 362 | 13 does -E[B_t*B_t] = E[-B_t * -B_t] ?
【在 b***k 的大作中提到】
: 这个证明对吗？
: 用你的方法我可以证明E[B_t^2]=0.which is wrong of course.
:
: t3]
|
k*******d
发帖数: 1340 | 14 I think this is proof is correct and is the simplest way to solve this
problem.
t3]
【在 A**T 的大作中提到】
: -E[B_t1 * B_t2 * B_t3] = E[(-B_t1)(-B_t2)(-B_t3)] = E[B'_t1 * B'_t2 * B'_t3]
: (By letting B'_t = -B_t. Again B'_t is a BM) = E[B_t1 * B_t2 * B_t3].
|
z****g
发帖数: 1978 | 15 B_t is not independent from itself.
However, B_t1, B_t2, B_t3 can be decomposed into independent variables
B_t1, B_t2-B_t1, B_t3-B_t2, which will make the proof correct.
【在 A**T 的大作中提到】
: does -E[B_t*B_t] = E[-B_t * -B_t] ?
|
k*******d
发帖数: 1340 | 16 他没有用到B(t)的indenpence,他用的是-B(t)也是BM，这个只能用来证明等于0而不能
真正去算出结果。
我原先想的办法也是去分成3个独立的变量，但这样需要一定的计算。
【在 z****g 的大作中提到】
: B_t is not independent from itself.
: However, B_t1, B_t2, B_t3 can be decomposed into independent variables
: B_t1, B_t2-B_t1, B_t3-B_t2, which will make the proof correct.
|
z****g
发帖数: 1978 | 17 It has noting to do with calculation.
To boil down the reason, it is because B_t and B_t won't have a joint
distribution filling the 2 dimensional space, while (B_t1, B_t2, B_t3)
will have a joint distribution filling the whole space.
(B_t1, B_t2, B_t3) or (B_t1, B_t2-B_t1, B_t3-B_t2) is just a matter of
space rotation which won't hurt the symmetric property, while (B_t, B_t)
is not.
The idea is just using the symmetric property of the distribution, not
even involving any calculation. That's why
【在 k*******d 的大作中提到】
: 他没有用到B(t)的indenpence,他用的是-B(t)也是BM，这个只能用来证明等于0而不能
: 真正去算出结果。
: 我原先想的办法也是去分成3个独立的变量，但这样需要一定的计算。
|
b***k
发帖数: 2673 | 18 all right, I buy it.
I think this is proof is correct and is the simplest way to solve this
problem.
t3]
【在 k*******d 的大作中提到】
: I think this is proof is correct and is the simplest way to solve this
: problem.
:
: t3]
|
l******i
发帖数: 1404 | 19 针对这题以上有很简单的trick，但是对于一般的函数例如(exp(-t*Wt)+3t+Wt^2)也有
framework
标准做法对于任意(t,Wt)的函数求期望：
Suppose Wt is Wiener process, to get E(W(t)*W(t+h1)*W(t+h2))
for any t and
for any positive numbers h1 and h2 with h2>h1>0:
Wt*W(t+h1)*W(t+h2)
=Wt*(Wt+Z1)*(Wt+Z1+Z2)
=Wt^3+(2Z1+Z2)Wt^2+Z1(Z1+Z2)Wt
where Z1=W(t+h1)-Wt and Z2=W(t+h2)-W(t+h1)=W(t+h2)-Wt-Z1,
Z1 and Z2 are independently normal distributed,
Z1 and Z2 have nothing to do with t.
d(Wt*(Wt+Z1)*(Wt+Z2))
=d(Wt^3+(2Z1+Z2)Wt^2+Z1(Z1+Z2)Wt)
=3Wt^2d |
