h*****u
发帖数: 204 | 1 对固定的T, \int_0^T W_tdt 是normal random variable吗？ 如果是， 怎么证明？
我只算出来期望是0， 方差是 T^3/3, 有没有办法证明它是normal random variable？ |
l******i
发帖数: 1404 | 2 我觉得不是normal.....
最初的想法：
Set S_t=\int_0^t W_s ds;
Then: E[S_t]=0; E[S_t^2]=t^3/3.
注意：N(0,t^3/3)对应的moment generating function是exp{(u^2)*(t^3)/6}；
如果S_t是normal，那么moment generating function必须一致，
也就是说E[exp{u*S_t}]=exp{(u^2)*(t^3)/6}成立；
i.e. E[exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}]=1;
Let P_t=exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}
If S_t is normal, usually P_t is a martingale,
then it will easy to prove E[exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}]=1.
(Such technique has been used by Shreve many times in his book.)
Apply Ito's on P_t:
dP_t=P_t*{u*W_t-(u^2)*(t^2)/2}dt;
However the drift part cannot be gone! So probably it is not normal.
Then I am considering about using
the moment identification to find the contradiction point.
Here is 严格证明：
Set S_t=\int_0^t W_s ds. Then: E[S_t]=0; E[S_t^2]=(t^3)/3.
如果S_t是normal的话，distribution就是N(0,(t^3)/3)，mean=0 and var=(t^3)/3.
注意：N(0,t^3/3)对应的fourth moment是3*var^2=(t^6)/3；
通过计算可知E[S_t^4]=(t^6)/5，故矛盾！ |
l******n
发帖数: 9344 | 3 T*W_T-int_0^T t dW_t = \int_0^T W_tdt
Left side is normal
【在 l******i 的大作中提到】
: 我觉得不是normal.....
: 最初的想法：
: Set S_t=\int_0^t W_s ds;
: Then: E[S_t]=0; E[S_t^2]=t^3/3.
: 注意：N(0,t^3/3)对应的moment generating function是exp{(u^2)*(t^3)/6}；
: 如果S_t是normal，那么moment generating function必须一致，
: 也就是说E[exp{u*S_t}]=exp{(u^2)*(t^3)/6}成立；
: i.e. E[exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}]=1;
: Let P_t=exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}
: If S_t is normal, usually P_t is a martingale,
|
h*****u
发帖数: 204 | 4 We know that T*W_T is normal
and int_0^T t dW_t is normal,(integrand is non-random)
If T*W_T and int_0^T t dW_t are independent.
then it is normal.
But T*W_T and int_0^T t dW_t are NOT independent.
Then I don't know.
and for this one I think T*W_T and int_0^T t dW_t are NOT independent.
So could you show me how do you get that? Thanks
【在 l******n 的大作中提到】
: T*W_T-int_0^T t dW_t = \int_0^T W_tdt
: Left side is normal
|
h*****u
发帖数: 204 | 5 感觉应该是的, 从积分的定义
\int_0^T W_tdt=sum W_t(j)(t(j+1)-t(j)),
可以认为是无穷个正太分布的和，所以感觉上我觉得是的，但是不知道怎么证明。
【在 l******i 的大作中提到】
: 我觉得不是normal.....
: 最初的想法：
: Set S_t=\int_0^t W_s ds;
: Then: E[S_t]=0; E[S_t^2]=t^3/3.
: 注意：N(0,t^3/3)对应的moment generating function是exp{(u^2)*(t^3)/6}；
: 如果S_t是normal，那么moment generating function必须一致，
: 也就是说E[exp{u*S_t}]=exp{(u^2)*(t^3)/6}成立；
: i.e. E[exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}]=1;
: Let P_t=exp{u*\int_0^t W_s ds-(u^2)*(t^3)/6}
: If S_t is normal, usually P_t is a martingale,
|
p********6
发帖数: 1802 | 6 The summation of two normal is normal no matter they are independent or not.
【在 h*****u 的大作中提到】
: We know that T*W_T is normal
: and int_0^T t dW_t is normal,(integrand is non-random)
: If T*W_T and int_0^T t dW_t are independent.
: then it is normal.
: But T*W_T and int_0^T t dW_t are NOT independent.
: Then I don't know.
: and for this one I think T*W_T and int_0^T t dW_t are NOT independent.
: So could you show me how do you get that? Thanks
|
h*****u
发帖数: 204 | 7 Let A is a normal random variable
let B=-A is a normal random variable.
But C=B+A=0
C=0 is not a normal random variable.
not.
【在 p********6 的大作中提到】
: The summation of two normal is normal no matter they are independent or not.
|
p********6
发帖数: 1802 | 8 Why not? It is a normal with zero variance
【在 h*****u 的大作中提到】
: Let A is a normal random variable
: let B=-A is a normal random variable.
: But C=B+A=0
: C=0 is not a normal random variable.
:
: not.
|
h*****u
发帖数: 204 | 9 Maybe my example is not so good, I will try to figure out a better one,haha
【在 p********6 的大作中提到】
: Why not? It is a normal with zero variance
|
h*****u
发帖数: 204 | 10 Let X have a standard normal distribution.
Let Z be independent of X, with Z equally likely to be +1 or -1 (i.e., Pr[Z=
+1] = Pr[Z=-1] = 1/2).
Let Y = X Z (the product of X and Z).
[ShreveII example 2.2.10 page 62-63] we know Y is standard normal
but Prob(X+Y=0)=1/2 so X+Y is not normal.
not.
【在 p********6 的大作中提到】
: The summation of two normal is normal no matter they are independent or not.
|
|
|
p********6
发帖数: 1802 | 11 OK, independence is required
Z=
【在 h*****u 的大作中提到】
: Let X have a standard normal distribution.
: Let Z be independent of X, with Z equally likely to be +1 or -1 (i.e., Pr[Z=
: +1] = Pr[Z=-1] = 1/2).
: Let Y = X Z (the product of X and Z).
: [ShreveII example 2.2.10 page 62-63] we know Y is standard normal
: but Prob(X+Y=0)=1/2 so X+Y is not normal.
:
: not.
|
l******i
发帖数: 1404 | 12 当然要independence。
按你的说法：
任意多个normal variable的joint distribution都是multivariate normal了。
明显不可能啊，
那Shreve在4.7节证明几个Gaussian process例子的时候那么费劲干吗。。。。。。。
not.
【在 p********6 的大作中提到】
: The summation of two normal is normal no matter they are independent or not.
|
l******i
发帖数: 1404 | 13 Here is 严格证明：
Set S_t=\int_0^t W_s ds. Then: E[S_t]=0; E[S_t^2]=(t^3)/3.
如果S_t是normal的话，distribution就是N(0,(t^3)/3)，mean=0 and var=(t^3)/3.
注意：N(0,t^3/3)对应的fourth moment是3*var^2=(t^6)/3；
通过计算可知E[S_t^4]=(t^6)/5，故矛盾！
【在 h*****u 的大作中提到】
: 感觉应该是的, 从积分的定义
: \int_0^T W_tdt=sum W_t(j)(t(j+1)-t(j)),
: 可以认为是无穷个正太分布的和，所以感觉上我觉得是的，但是不知道怎么证明。
|
x******a
发帖数: 6336 | 14 consider the property of Characteristic function
【在 h*****u 的大作中提到】
: 感觉应该是的, 从积分的定义
: \int_0^T W_tdt=sum W_t(j)(t(j+1)-t(j)),
: 可以认为是无穷个正太分布的和，所以感觉上我觉得是的，但是不知道怎么证明。
|
x******a
发帖数: 6336 | 15 it seems to me only joint normal is required.
【在 p********6 的大作中提到】
: OK, independence is required
:
: Z=
|
p********6
发帖数: 1802 | 16 But if the correlation of X Y is const, then X Y can be represented by two
independent normal, then the summation seems to be normal...
【在 l******i 的大作中提到】
: 当然要independence。
: 按你的说法：
: 任意多个normal variable的joint distribution都是multivariate normal了。
: 明显不可能啊，
: 那Shreve在4.7节证明几个Gaussian process例子的时候那么费劲干吗。。。。。。。
:
: not.
|
p********6
发帖数: 1802 | 17 Is the joint normal equivalent with the X Y has const correlation?
【在 x******a 的大作中提到】
: it seems to me only joint normal is required.
|
l******i
发帖数: 1404 | 18 no
xiaojiya说的是(X,Y)如果jointly normal，
那么任意linear combination of X and Y must be normal，
自然而然summation就是normal的，
这是exactly jointly normal的定义，和你说的correlation一点关系也没有。
【在 p********6 的大作中提到】
: Is the joint normal equivalent with the X Y has const correlation?
|
l******i
发帖数: 1404 | 19 Since correlation is always const given any two normal variables,
I really don't know what you are talking about here............
Where are these theorems from?
two
【在 p********6 的大作中提到】
: But if the correlation of X Y is const, then X Y can be represented by two
: independent normal, then the summation seems to be normal...
|
p********6
发帖数: 1802 | 20 Shreve 习题4.13？
【在 l******i 的大作中提到】
: Since correlation is always const given any two normal variables,
: I really don't know what you are talking about here............
: Where are these theorems from?
:
: two
|
|
|
l******n
发帖数: 9344 | 21 Yes, they are not independent, correlation is T^3/2
【在 h*****u 的大作中提到】
: We know that T*W_T is normal
: and int_0^T t dW_t is normal,(integrand is non-random)
: If T*W_T and int_0^T t dW_t are independent.
: then it is normal.
: But T*W_T and int_0^T t dW_t are NOT independent.
: Then I don't know.
: and for this one I think T*W_T and int_0^T t dW_t are NOT independent.
: So could you show me how do you get that? Thanks
|
h*********7
发帖数: 23 | 22 非独立的正态分布之和还是正太分布....绿皮书里面貌似有这道题目.. 因为 Ho Lee
model算 Bond Price的时候要用到这个
【在 h*****u 的大作中提到】
: We know that T*W_T is normal
: and int_0^T t dW_t is normal,(integrand is non-random)
: If T*W_T and int_0^T t dW_t are independent.
: then it is normal.
: But T*W_T and int_0^T t dW_t are NOT independent.
: Then I don't know.
: and for this one I think T*W_T and int_0^T t dW_t are NOT independent.
: So could you show me how do you get that? Thanks
|
k*****y
发帖数: 744 | 23 please see if there is anything wrong with this argument?
【在 h*****u 的大作中提到】
: 对固定的T, \int_0^T W_tdt 是normal random variable吗？ 如果是， 怎么证明？
: 我只算出来期望是0， 方差是 T^3/3, 有没有办法证明它是normal random variable？
|
k*****y
发帖数: 744 | 24 能不能详细说一下4th moment怎么算？谢谢~
【在 l******i 的大作中提到】
: Since correlation is always const given any two normal variables,
: I really don't know what you are talking about here............
: Where are these theorems from?
:
: two
|
l******i
发帖数: 1404 | 25 Set S_t=\int_0^t W_s ds.
把这些式子写成两边积分的形式，然后从下往上推：
d(S_t^4)=4*S_t^3*W_t*dt
d(S_t^3*W_t)=3*S_t^2*W_t^2*dt+S_t^3*dW_t
d(S_t^2*W_t^2)=2*S_t*W_t^3*dt+2*S_t^2*W_t*dW_t+(1/2)*2*S_t^2*dt
E[S_t^2]=T^3/3
d(S_t*W_t^3)=W_t^4*dt+3*S_t*W_t^2*dW_t+(1/2)*6*S_t*W_t*dt
E[W_t^4]=3*t^2
d(S_t*W_t)=W_t^2*dt+S_t*dW_t
E[W_t^2]=t
【在 k*****y 的大作中提到】
: 能不能详细说一下4th moment怎么算？谢谢~
|
k*****y
发帖数: 744 | 26 第3，4行是不是少了对W的二阶导数？
【在 l******i 的大作中提到】
: Set S_t=\int_0^t W_s ds.
: 把这些式子写成两边积分的形式，然后从下往上推：
: d(S_t^4)=4*S_t^3*W_t*dt
: d(S_t^3*W_t)=3*S_t^2*W_t^2*dt+S_t^3*dW_t
: d(S_t^2*W_t^2)=2*S_t*W_t^3*dt+2*S_t^2*W_t*dW_t+(1/2)*2*S_t^2*dt
: E[S_t^2]=T^3/3
: d(S_t*W_t^3)=W_t^4*dt+3*S_t*W_t^2*dW_t+(1/2)*6*S_t*W_t*dt
: E[W_t^4]=3*t^2
: d(S_t*W_t)=W_t^2*dt+S_t*dW_t
: E[W_t^2]=t
|
x******a
发帖数: 6336 | 27 it is a normal distribution and you know mean and variance.
【在 k*****y 的大作中提到】
: 能不能详细说一下4th moment怎么算？谢谢~
|
l******i
发帖数: 1404 | 28 Moment generating function does not work for this problem.
And your proof is correct.
【在 k*****y 的大作中提到】
: please see if there is anything wrong with this argument?
|
h*****u
发帖数: 204 | 29 Thanks, I think you are right
【在 k*****y 的大作中提到】
: please see if there is anything wrong with this argument?
|
R**T
发帖数: 784 | 30 这个是对的
heard from street那本上好像有讲
【在 k*****y 的大作中提到】
: please see if there is anything wrong with this argument?
|
