g******r
发帖数: 29 | 1 如题
thx
比如说 先有 W_t
构造一个 B_t = W_t 0<=t<=1
B_t = B_1 - (W_t-W_1) t > 1
这样的B_t也是一个布朗运动吗 |
a*********r
发帖数: 139 | 2 Of course not. You can take two independent Brownian motions, then their
correlation is zero. |
t***l
发帖数: 3644 | 3 这问题问的就够可以了，想不到答的可以更离谱。
虽然结论对的，但这论据。。。难道zero不是常数吗？
【在 a*********r 的大作中提到】
: Of course not. You can take two independent Brownian motions, then their
: correlation is zero.
|
w**********y
发帖数: 1691 | 4 给两个例子:
1. A_t = W_t; B_t = W_t, corr = 1
2. A_t = W_t; B_t = \sqrt{a}*W_{t/a}, a>1, corr=1/sqrt{a} |
t***l
发帖数: 3644 | 5 例子2中的B_t都不是Martingale，怎么会是BM呢？
而且两个corr都差乘个t
我猜楼主说的是corr一定为常数乘以t吗？
我来给个非常数的例子:
假设W_t和B_t是两个独立的BM，又假设a(t)是个取值0到1之间的函数，让b(t)=sqrt{1-
a(t)^2}。那么 a(t)W_t + b(t)B_t就是一个BM，它和W_t之间的corr是a(t)t
【在 w**********y 的大作中提到】
: 给两个例子:
: 1. A_t = W_t; B_t = W_t, corr = 1
: 2. A_t = W_t; B_t = \sqrt{a}*W_{t/a}, a>1, corr=1/sqrt{a}
|
w**********y
发帖数: 1691 | 6 你确定么?-王小丫说.
1-
【在 t***l 的大作中提到】
: 例子2中的B_t都不是Martingale，怎么会是BM呢？
: 而且两个corr都差乘个t
: 我猜楼主说的是corr一定为常数乘以t吗？
: 我来给个非常数的例子:
: 假设W_t和B_t是两个独立的BM，又假设a(t)是个取值0到1之间的函数，让b(t)=sqrt{1-
: a(t)^2}。那么 a(t)W_t + b(t)B_t就是一个BM，它和W_t之间的corr是a(t)t
|
t***l
发帖数: 3644 | 7 我指出了你还看不出来？随机过程怎么学的？
【在 w**********y 的大作中提到】
: 你确定么?-王小丫说.
:
: 1-
|
w**********y
发帖数: 1691 | 8 俺笨.没学过随机过程..请大侠赐教...给具体算算....
【在 t***l 的大作中提到】
: 我指出了你还看不出来？随机过程怎么学的？
|
t***l
发帖数: 3644 | 9 我上面都演算给你看了，你哪里不明白？
【在 w**********y 的大作中提到】
: 俺笨.没学过随机过程..请大侠赐教...给具体算算....
|
w**********y
发帖数: 1691 | 10 不明白你咋这么自信...
这是做矿工必须的品质么?俺得学学...
【在 t***l 的大作中提到】
: 我上面都演算给你看了，你哪里不明白？
|
|
|
t***l
发帖数: 3644 | 11 我晕，看来你是真还没懂自己的例子都错了。。。
做矿工是要自信，但这个明摆着的错的例子，不需要什么自信就能指出的吧。
【在 w**********y 的大作中提到】
: 不明白你咋这么自信...
: 这是做矿工必须的品质么?俺得学学...
|
a*********r
发帖数: 139 | 12 The question is ill-posed, you have to guess the meaning. My guessing is
whether Cov(B_1(t), B_2(t)) must be a non-zero constant for two Brownian
motions B_1 and B_2.
【在 t***l 的大作中提到】
: 这问题问的就够可以了，想不到答的可以更离谱。
: 虽然结论对的，但这论据。。。难道zero不是常数吗？
|
t***l
发帖数: 3644 | 13 比较确信问的不是你猜的这个。
【在 a*********r 的大作中提到】
: The question is ill-posed, you have to guess the meaning. My guessing is
: whether Cov(B_1(t), B_2(t)) must be a non-zero constant for two Brownian
: motions B_1 and B_2.
|
a*********r
发帖数: 139 | 14 I didn't see this. In the second example, B_t is a Brownian by scaling
invariance property of Browian motion, so it must a martingale.
1-
【在 t***l 的大作中提到】
: 例子2中的B_t都不是Martingale，怎么会是BM呢？
: 而且两个corr都差乘个t
: 我猜楼主说的是corr一定为常数乘以t吗？
: 我来给个非常数的例子:
: 假设W_t和B_t是两个独立的BM，又假设a(t)是个取值0到1之间的函数，让b(t)=sqrt{1-
: a(t)^2}。那么 a(t)W_t + b(t)B_t就是一个BM，它和W_t之间的corr是a(t)t
|
a*********r
发帖数: 139 | 15 The ID "geometer" should clarify the question.
【在 t***l 的大作中提到】
: 比较确信问的不是你猜的这个。
|
w**********y
发帖数: 1691 | 16 自信之前也得double check一下吧..
你知道correlation的定义么?covariance呢? 任何一个random variable跟自身的
correlation都是1这是个常识吧,你觉得呢?
你再去check一下martingale的定义..然后你再check一下2*W_{t/4}是不是martingale;
假设你定义那个 X_t = tW_t + \sqrt{1-t^2}B_t..
你算算 E(X_{1/2} | F__{1/4})是什么..是X_{1/4}么?
【在 t***l 的大作中提到】
: 我晕，看来你是真还没懂自己的例子都错了。。。
: 做矿工是要自信，但这个明摆着的错的例子，不需要什么自信就能指出的吧。
|
a*********r
发帖数: 139 | 17 I agree with you.
First, the question is ill-posed and ambiguous. The question-poser should
step forward to clarify that. Otherwise, we gonna try to guess the meaning.
Second, your two examples are both Brownian motions. This follows
immediately from the three basic invariance properties of Brownian motion.
Third, the example of X_t = tW_t + \sqrt{1-t^2}B_t is not a Brownian because
it's obviously not a martingale. For example, we can take a(t)=t, 0\leq t\
leq 1.
martingale;
【在 w**********y 的大作中提到】
: 自信之前也得double check一下吧..
: 你知道correlation的定义么?covariance呢? 任何一个random variable跟自身的
: correlation都是1这是个常识吧,你觉得呢?
: 你再去check一下martingale的定义..然后你再check一下2*W_{t/4}是不是martingale;
: 假设你定义那个 X_t = tW_t + \sqrt{1-t^2}B_t..
: 你算算 E(X_{1/2} | F__{1/4})是什么..是X_{1/4}么?
|
t***l
发帖数: 3644 | 18 B_t is a martingale?? See below.
Since a>1, we can find s such that t/a < s < t. Therefore,
E[B_t | F_s] = E[sqrt(a) * W_{t/a} | F_s] = sqrt(a) * E[W_{t/a} | F_s]
note t/a < s, so the most right conditional expectation is just sqrt(a) * W_
{t/a} = B_t.
That is, E[B_t | F_s] = B_t, which is not B_s. So B_t is not a martingale
【在 a*********r 的大作中提到】
: I didn't see this. In the second example, B_t is a Brownian by scaling
: invariance property of Browian motion, so it must a martingale.
:
: 1-
|
t***l
发帖数: 3644 | 19 correlation确实我定义搞错了，我一直想的是covariance。
但是你说的那个 2*W_{t/4}是不是martingale呢？可以说是，也可以说不是，这得看你
取得F_t是什么了，具体你看我之前回avidswimmer的一个证明。
至于你说X_t是不是martingale，这还比较显然，两个martingale的linear
combination一定是martingale的。
martingale;
【在 w**********y 的大作中提到】
: 自信之前也得double check一下吧..
: 你知道correlation的定义么?covariance呢? 任何一个random variable跟自身的
: correlation都是1这是个常识吧,你觉得呢?
: 你再去check一下martingale的定义..然后你再check一下2*W_{t/4}是不是martingale;
: 假设你定义那个 X_t = tW_t + \sqrt{1-t^2}B_t..
: 你算算 E(X_{1/2} | F__{1/4})是什么..是X_{1/4}么?
|
a*********r
发帖数: 139 | 20 As I pointed, B_t in example 2 is a martingale because it is a Brownian
motion by scaling invariance property.
When we talk about a martingale, we must specify the filtration. If the
filtration is not specified, the convention is the filtration is taken to be
the one generated by the stochastic process in question.
In your argument, you used the original filtration generated by a Brownian W
_t and a scaled Brownian motion B_t, then you ended up with a plausible, but
wrong conclusion.
W_
【在 t***l 的大作中提到】
: B_t is a martingale?? See below.
: Since a>1, we can find s such that t/a < s < t. Therefore,
: E[B_t | F_s] = E[sqrt(a) * W_{t/a} | F_s] = sqrt(a) * E[W_{t/a} | F_s]
: note t/a < s, so the most right conditional expectation is just sqrt(a) * W_
: {t/a} = B_t.
: That is, E[B_t | F_s] = B_t, which is not B_s. So B_t is not a martingale
|
|
|
a*********r
发帖数: 139 | 21 Yes. Linear combination of two martingale is a martingale. But unfortunately
, neither tB(t) nor \sqrt{1-t^2}B_t is a martingale! Also, the combination
is a martingale.
【在 t***l 的大作中提到】
: correlation确实我定义搞错了，我一直想的是covariance。
: 但是你说的那个 2*W_{t/4}是不是martingale呢？可以说是，也可以说不是，这得看你
: 取得F_t是什么了，具体你看我之前回avidswimmer的一个证明。
: 至于你说X_t是不是martingale，这还比较显然，两个martingale的linear
: combination一定是martingale的。
:
: martingale;
|
t***l
发帖数: 3644 | 22 当然和filtration有关，我同意如果用本身做filtration，那个是martingale。
但是具体到例子2，难道两个随机过程用不同的filtration来举例子？
be
W
but
【在 a*********r 的大作中提到】
: As I pointed, B_t in example 2 is a martingale because it is a Brownian
: motion by scaling invariance property.
: When we talk about a martingale, we must specify the filtration. If the
: filtration is not specified, the convention is the filtration is taken to be
: the one generated by the stochastic process in question.
: In your argument, you used the original filtration generated by a Brownian W
: _t and a scaled Brownian motion B_t, then you ended up with a plausible, but
: wrong conclusion.
:
: W_
|
t***l
发帖数: 3644 | 23 tB_t和sqrt{1-t^2}B_t不是Martingale？？？
首先B_t是martingale吧，难道乘上个非随机的函数就变成不是martingale了？？？
unfortunately
【在 a*********r 的大作中提到】
: Yes. Linear combination of two martingale is a martingale. But unfortunately
: , neither tB(t) nor \sqrt{1-t^2}B_t is a martingale! Also, the combination
: is a martingale.
|
u******s
发帖数: 157 | 24 From the definition, if we have B_t = c * W_t/c^2,
then B_t has independent increment,
and E[B_t - B_s] = 0, V[B_t - B_s] = t - s.
So I think B_t is a Brownian Motion.
W_
【在 t***l 的大作中提到】
: B_t is a martingale?? See below.
: Since a>1, we can find s such that t/a < s < t. Therefore,
: E[B_t | F_s] = E[sqrt(a) * W_{t/a} | F_s] = sqrt(a) * E[W_{t/a} | F_s]
: note t/a < s, so the most right conditional expectation is just sqrt(a) * W_
: {t/a} = B_t.
: That is, E[B_t | F_s] = B_t, which is not B_s. So B_t is not a martingale
|
u******s
发帖数: 157 | 25 I think, E[t * B_t | f(s)] = t * E[B_t | f(s)] = t * B_s
【在 t***l 的大作中提到】
: tB_t和sqrt{1-t^2}B_t不是Martingale？？？
: 首先B_t是martingale吧，难道乘上个非随机的函数就变成不是martingale了？？？
:
: unfortunately
|
t***l
发帖数: 3644 | 26 哦，那是我想错了。
【在 u******s 的大作中提到】
: I think, E[t * B_t | f(s)] = t * E[B_t | f(s)] = t * B_s
|
u******s
发帖数: 157 | 27 Re, agleed.
I guess, if we use the filtration generated by itself, f(t) = sigma(c*W_t/c^
2), then it is a MTG.
be
W
but
【在 a*********r 的大作中提到】
: As I pointed, B_t in example 2 is a martingale because it is a Brownian
: motion by scaling invariance property.
: When we talk about a martingale, we must specify the filtration. If the
: filtration is not specified, the convention is the filtration is taken to be
: the one generated by the stochastic process in question.
: In your argument, you used the original filtration generated by a Brownian W
: _t and a scaled Brownian motion B_t, then you ended up with a plausible, but
: wrong conclusion.
:
: W_
|
t***l
发帖数: 3644 | 28 原来我的那些例子才是错的哈。
不过非常数的例子还是有的：
还是取a(t),b(t)两个关于时间t的函数，使得a(t)^2 + b(t)^2 = 1。
随后W_t和B_t是两个独立的BM。
让X_t是a(t)dW_t + b(t)dB_t的Ito积分，也就是
dX_t = a(t)dW_t + b(t)dB_t
这样X_t就是一个BM了，而且W_t和X_t的Corr是
( \int_{0}^{t} a(s) ds ) / t，这是一个关于t的函数，不是常数，比如取a(t) =
cos(t)，那么Corr就是 sin(t) / t，这次应该是对了。
【在 u******s 的大作中提到】
: I think, E[t * B_t | f(s)] = t * E[B_t | f(s)] = t * B_s
|
g******r
发帖数: 29 | 29 嗯 就这个意思
我开始就是问 两个布朗运动的 quadratic variation是不是常数xt
也就是correlation的意思
我只是不确定我后面的例子是不是布朗运动
你判断是 dX_t = a(t)dW_t + b(t)dB_t 是布朗运动是用levy判定吧
其实我那个例子就是 a(t) = 1_{t<=1} - 1_{t>1}
那应该就没问题了
【在 t***l 的大作中提到】
: 原来我的那些例子才是错的哈。
: 不过非常数的例子还是有的：
: 还是取a(t),b(t)两个关于时间t的函数，使得a(t)^2 + b(t)^2 = 1。
: 随后W_t和B_t是两个独立的BM。
: 让X_t是a(t)dW_t + b(t)dB_t的Ito积分，也就是
: dX_t = a(t)dW_t + b(t)dB_t
: 这样X_t就是一个BM了，而且W_t和X_t的Corr是
: ( \int_{0}^{t} a(s) ds ) / t，这是一个关于t的函数，不是常数，比如取a(t) =
: cos(t)，那么Corr就是 sin(t) / t，这次应该是对了。
|
a*********r
发帖数: 139 | 30 When the filtration is not explicitly specified, it is understood to be the
one generated by the process. That's the convention!
【在 t***l 的大作中提到】
: 当然和filtration有关，我同意如果用本身做filtration，那个是martingale。
: 但是具体到例子2，难道两个随机过程用不同的filtration来举例子？
:
: be
: W
: but
|
|
|
a*********r
发帖数: 139 | 31 How did you get this conclusion. What if that function is a function in t???
You should think carefully before jumping to any conclusion.
【在 t***l 的大作中提到】
: tB_t和sqrt{1-t^2}B_t不是Martingale？？？
: 首先B_t是martingale吧，难道乘上个非随机的函数就变成不是martingale了？？？
:
: unfortunately
|
a*********r
发帖数: 139 | 32 Yes. You basically proved the scaling invariance property again.
【在 u******s 的大作中提到】
: From the definition, if we have B_t = c * W_t/c^2,
: then B_t has independent increment,
: and E[B_t - B_s] = 0, V[B_t - B_s] = t - s.
: So I think B_t is a Brownian Motion.
:
: W_
|
t***l
发帖数: 3644 | 33 convention吧，你也得看看这个例子啊。老抱着convention一成不变有意思吗？
难道说例子中给的两个随机过程在不同的filtration里，随后来算他们的correlation？如果这也是你的convention我没话说。
还有，楼主的问题哪里ill-posed了？我本把corr想成cov了，所以觉得问题提的不对。现在看看提的没问题。
the
【在 a*********r 的大作中提到】
: When the filtration is not explicitly specified, it is understood to be the
: one generated by the process. That's the convention!
|
a*********r
发帖数: 139 | 34 First, how can you define the correlation between two stochastic process?
We can only talk about the correlation between two random variables B_1(t)
and B_2(t) for a fixed t given that B_1 and B_2 are two stochastic process.
That's why I say it's ill-posed.
You miss the point! When we talk about martingales, we must specify the
filtration. However, when we talk about correlation between B_1(t) and B_2(t
) (note t is fixed), it has nothing to do with filtration!
I don't want to be harsh. But you have lots of wrong concepts. I will not
reply to your posts anymore if they don't make any sense like this one.
correlation？如果这也是你的convention我没话说。
。现在看看提的没问题。
【在 t***l 的大作中提到】
: convention吧，你也得看看这个例子啊。老抱着convention一成不变有意思吗？
: 难道说例子中给的两个随机过程在不同的filtration里，随后来算他们的correlation？如果这也是你的convention我没话说。
: 还有，楼主的问题哪里ill-posed了？我本把corr想成cov了，所以觉得问题提的不对。现在看看提的没问题。
:
: the
|
t***l
发帖数: 3644 | 35 你觉得你说的这些我是不懂？太把自己当回事了吧。
这里又不是发paper，难道得一步一步写清楚。楼主说的显然是correlation process啊。
我当然知道correlation不关filtration什么事情。。。我的意思是你讨论两个BM，却
不用同一个filtration，你见过吗？举个例子看看？
你想想一个多维的BM，每个分量上的BM都用的同一个filtration吧，为啥？
.
(t
【在 a*********r 的大作中提到】
: First, how can you define the correlation between two stochastic process?
: We can only talk about the correlation between two random variables B_1(t)
: and B_2(t) for a fixed t given that B_1 and B_2 are two stochastic process.
: That's why I say it's ill-posed.
: You miss the point! When we talk about martingales, we must specify the
: filtration. However, when we talk about correlation between B_1(t) and B_2(t
: ) (note t is fixed), it has nothing to do with filtration!
: I don't want to be harsh. But you have lots of wrong concepts. I will not
: reply to your posts anymore if they don't make any sense like this one.
:
|
g******r
发帖数: 29 | 36 well, given 2 random processes X Y
i think we can define the correlation of X Y
as the correl(X_t, Y_t),
if dX_t = a_t dB_t
dY_t = b_t dW_t where B W two brownian motions
the quadratic variation of X_t and Y_t
E[\int_0^t a*b ] = E[\int_0^t a dB * \int_0^t b dW] = cov(X_t, Y_t)
is the quadratic variation of B and W
.
(t
【在 a*********r 的大作中提到】
: First, how can you define the correlation between two stochastic process?
: We can only talk about the correlation between two random variables B_1(t)
: and B_2(t) for a fixed t given that B_1 and B_2 are two stochastic process.
: That's why I say it's ill-posed.
: You miss the point! When we talk about martingales, we must specify the
: filtration. However, when we talk about correlation between B_1(t) and B_2(t
: ) (note t is fixed), it has nothing to do with filtration!
: I don't want to be harsh. But you have lots of wrong concepts. I will not
: reply to your posts anymore if they don't make any sense like this one.
:
|
a*********r
发帖数: 139 | 37
That's you! From the beginning, you assume a very arrogant position towards
all others. Even if I told you {tB(t)}_t is not a martinagle. You still
resisted thinking about what others say until somebody did this for you. You
should rein in your arrogance and anger and learn the stuff seriously. This
may hurt, but frankly, you know very very little about stochastic calculus
(even the simplest Brownian motion version).
啊。
Exactly the opposite. When you write a paper, you don't write every single
step! You actually skip most steps. However, no matter what you do, you need
to articulate what you are saying.
Correlation process? Never heard about this. Your new concept? Define it
please!
Hey, remember, we were talking about two independent one-dimensional BM.
Someone give two examples of one-dimensional BM. You claimed the second is
not a BM because it is not a martingale! Then you made a wrong argument. I
had to point out the filtration is so important to make it clear why your
argument is wrong! It's subtle. By the way, when we talk about BMs (no
matter how), there is no need to refer to the filtratioin generated by it
until we touch on the martingale property!
【在 t***l 的大作中提到】
: 你觉得你说的这些我是不懂？太把自己当回事了吧。
: 这里又不是发paper，难道得一步一步写清楚。楼主说的显然是correlation process啊。
: 我当然知道correlation不关filtration什么事情。。。我的意思是你讨论两个BM，却
: 不用同一个filtration，你见过吗？举个例子看看？
: 你想想一个多维的BM，每个分量上的BM都用的同一个filtration吧，为啥？
:
: .
: (t
|
t***l
发帖数: 3644 | 38 你读书读太多了，如果就这种思维方式，做quant是不适合的。为了你要的严谨：这里
我假设了你想做quant才来这个版的。
correlation process从字面意思就一目了然了吧。为了你我定义一下吧：给定两个随
机过程A_t和B_t，我们可以定义随机过程C_t := corr(A_t,B_t)，明白了？
没什么好争的，就为了那两个不知所云的例子。。。反正楼主出的问题解决了。
towards
You
This
calculus
need
【在 a*********r 的大作中提到】
:
: That's you! From the beginning, you assume a very arrogant position towards
: all others. Even if I told you {tB(t)}_t is not a martinagle. You still
: resisted thinking about what others say until somebody did this for you. You
: should rein in your arrogance and anger and learn the stuff seriously. This
: may hurt, but frankly, you know very very little about stochastic calculus
: (even the simplest Brownian motion version).
: 啊。
: Exactly the opposite. When you write a paper, you don't write every single
: step! You actually skip most steps. However, no matter what you do, you need
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a*********r
发帖数: 139 | 39
Quant is a technical job anyway. Basic rigorous is necessary in any
technical job. Actually I train quants. Which program you were from? If you
were our quant program. you would have already failed.
Another piece of advice, don't just focus on technical ability, improve your
personality.
When something is well-know, it's not obvious. People may think it's cross-
variation, may think it's covariance function, may think...etc. Your
ignorance makes this obvious to you, but not to others.
Confidence is based on solid knowledge and competence not ignorance and
arrogance.
【在 t***l 的大作中提到】
: 你读书读太多了，如果就这种思维方式，做quant是不适合的。为了你要的严谨：这里
: 我假设了你想做quant才来这个版的。
: correlation process从字面意思就一目了然了吧。为了你我定义一下吧：给定两个随
: 机过程A_t和B_t，我们可以定义随机过程C_t := corr(A_t,B_t)，明白了？
: 没什么好争的，就为了那两个不知所云的例子。。。反正楼主出的问题解决了。
:
: towards
: You
: This
: calculus
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t***l
发帖数: 3644 | 40 不好意思，从来没读过什么quant的program，个人觉得没这个必要。
就这点数学，自己看看足够了。
you
your
【在 a*********r 的大作中提到】
:
: Quant is a technical job anyway. Basic rigorous is necessary in any
: technical job. Actually I train quants. Which program you were from? If you
: were our quant program. you would have already failed.
: Another piece of advice, don't just focus on technical ability, improve your
: personality.
: When something is well-know, it's not obvious. People may think it's cross-
: variation, may think it's covariance function, may think...etc. Your
: ignorance makes this obvious to you, but not to others.
: Confidence is based on solid knowledge and competence not ignorance and
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|
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w**********y
发帖数: 1691 | 41 如果非要讨论到这个filtration啊和制造一个'correlation process'啊,这玩意就更麻
烦了,大家做矿工,又不是要研究纯理论,都是听说些名词就开始瞎掰活.差不多别说错就
行了..
correlation在统计中,就是基于两个random variable的定义,不是关于随机过程.
你可以定义A_t和B_t的correlation,当然也可以定义A_n和B_m的correlation,与所谓的
什么filtration没有任何关系.
filtration的数学含义是: a collection of nested sigma-algebra.
常识的情况下,我们是assume你要算的是A_t和B_t的correlation.那么,这一切的定义都
是基于概率空间的:
这个概率空间是(\Omega, F_t, P),粗略的说,这个\Omega 是 [0,t] \cross R 的函数
空间上一个集合(就去理解成所有可能的从0到t走出来的曲线的集合吧),然后F_t就是定
义在\Omega上的类似于Borel集(空交并补)的sigma algebra.你这个A_t和B_t完全可以
定义在这同一个\Omega的.
那么如果你让t变化,这一堆F_t在一起统称filtration.不存在啥玩意A_t有一个
filtration啊,B_t有一个filtration啊之类的...
如果你非要定义一个'correlation process',看起来是很容易了,不就是一个关于t的函
数么..首先它不一定是个process,可能就是一个关于t的函数而已..然后,应该可能定义
成一个process(关于t的random process),但是这时候你要重新严格定义一个新的
extension space了,不是那么trivial的.
~~~~~~~~~~~~~
其实这些玩意很无聊..对工作没啥用..就还是老老实实按照MFE
program 常用的解释: filtration就是information啊,F_t就代表你知道时间t和它之前
的所有信息啊就行了..随机过程呢就是跟时间变化的随机变量..给定一个个时间t,S_t
就是个随机变量...其它所有的计算还都是统计概率里面基于随机变量的计算..别越讨
论越糊涂了... |
w**********y
发帖数: 1691 | 42 回到你这个ill-posed(?)题目啊:
如果按照你给的这个W_t, t \in [0,1] 和一个 B_t, t \in (1,\infty),然后要定义
correlation..那么这个correlation 只能定义成 corr (W_t1, B_t2), t分别在各自的
范围里.
所以你的correlation当然取决于你选的时间点.而不是什么随机过程的correlation..
俺吃饭去了.草草写了点看法,难免有错.
说实话,被timol说俺的例子错了之后,我还真是认认真真仔仔细细的想了一下才敢确保
自己没错的,然后才挖了个坑给你...
玩poker的时候,俺最喜欢也是最需要的就是9个人中有一个开始觉得一切显而易见,然后
开始冲动的时候..去年Xmas的时候,俺就是这么5分钟从200块翻成了1000块的..共勉共
勉..
【在 g******r 的大作中提到】
: 如题
: thx
: 比如说 先有 W_t
: 构造一个 B_t = W_t 0<=t<=1
: B_t = B_1 - (W_t-W_1) t > 1
: 这样的B_t也是一个布朗运动吗
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n*******e
发帖数: 107 | 43 灰常的深奥
【在 g******r 的大作中提到】
: 如题
: thx
: 比如说 先有 W_t
: 构造一个 B_t = W_t 0<=t<=1
: B_t = B_1 - (W_t-W_1) t > 1
: 这样的B_t也是一个布朗运动吗
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Q***5
发帖数: 994 | 44 I think both W_t and B_t in his example are defined on [0 \infty], another
way to write his B_t is:
B_s = \int_0^s a(t) dW_t
where a(t)=1 on [0 1] and a(t)=-1 on [1 \infty], i.e., B_t is a reflection
of W_t from t=1.
corr(W_t,B_t)= 1 on [0 1],
corr(W_t,B_t) converges to -1 as t goes to \infty. so his counter example is
a valid one
【在 w**********y 的大作中提到】
: 回到你这个ill-posed(?)题目啊:
: 如果按照你给的这个W_t, t \in [0,1] 和一个 B_t, t \in (1,\infty),然后要定义
: correlation..那么这个correlation 只能定义成 corr (W_t1, B_t2), t分别在各自的
: 范围里.
: 所以你的correlation当然取决于你选的时间点.而不是什么随机过程的correlation..
: 俺吃饭去了.草草写了点看法,难免有错.
: 说实话,被timol说俺的例子错了之后,我还真是认认真真仔仔细细的想了一下才敢确保
: 自己没错的,然后才挖了个坑给你...
: 玩poker的时候,俺最喜欢也是最需要的就是9个人中有一个开始觉得一切显而易见,然后
: 开始冲动的时候..去年Xmas的时候,俺就是这么5分钟从200块翻成了1000块的..共勉共
|
w**********y
发帖数: 1691 | 45 哦.没仔细看原题.那也是个不错的例子.
谢谢指正
is
【在 Q***5 的大作中提到】
: I think both W_t and B_t in his example are defined on [0 \infty], another
: way to write his B_t is:
: B_s = \int_0^s a(t) dW_t
: where a(t)=1 on [0 1] and a(t)=-1 on [1 \infty], i.e., B_t is a reflection
: of W_t from t=1.
: corr(W_t,B_t)= 1 on [0 1],
: corr(W_t,B_t) converges to -1 as t goes to \infty. so his counter example is
: a valid one
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