j****j
发帖数: 270 | 1 Hi all,
I am looking for the Radon-Nikodym derivative when both Brownian Motion and
Poisson Process present? For example, when the Bt and Nt are independent, we
have a Radon-Nikodym derivative which is the product of the RN derivative
of the Brownian Motion part and the RN derivative of the Poisson Process
Part. But what if they are not independent? |
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k*******l
发帖数: 69 | 2 whatever
if u knew what Radon-Nikodym derivative means, u would understand why i was
asking that. and the book contains change of measure for Levy processes
not |
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m*****z
发帖数: 357 | 3 再问一个
已知BM Bt，对应probability measure P
和一个与Bt有关的Bt*， 对应Probability measure P*
问 Radon-Nikodym derivative dP*/dP 是多少？
我就是想知道求的过程是怎么样的
再次谢谢 |
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发帖数: 1 | 4 什么Radon nikodym derivative 一般人理解的了吗？ |
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s**********e
发帖数: 33562 | 5 那你应该说数学分析啊。黎曼积分就足够弄广义积分了，哪里需要用测度论和勒贝革积
分啊？
step function 通过 RC network 又哪里需要实分析的知识来求解啊？难不成你要用
Radon–Nikodym的微分定义再在RC network上定义啥测度之类的？懂点微分方程和
傅立叶变换就够了。电路原理和信号与系统里的知识都足够解决这些问题，哪里要用近
代数学的知识啊？ |
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s**********e
发帖数: 33562 | 6 那你应该说数学分析啊。黎曼积分就足够弄广义积分了，哪里需要用测度论和勒贝革积
分啊？
step function 通过 RC network 又哪里需要实分析的知识来求解啊？难不成你要用
Radon–Nikodym的微分定义再在RC network上定义啥测度之类的？懂点微分方程和
傅立叶变换就够了。电路原理和信号与系统里的知识都足够解决这些问题，哪里要用近
代数学的知识啊？ |
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g****t
发帖数: 31659 | 7 你本科不是数学系的吧? 我可以肯定,在我本科的年代,北大/复旦/北大新讲/...的数学分
析课本,不够解决带step function的微分方程.事实上这几套教材,按我多年前的记忆,
连提都没提带step function输入的微分方程.
美国这边的数学分析,或者国内新的数学分析教材,讲的是什么,我就不太清楚了.
那你应该说数学分析啊。黎曼积分就足够弄广义积分了，哪里需要用测度论和勒贝革积
分啊？
step function 通过 RC network 又哪里需要实分析的知识来求解啊？难不成你要用
Radon–Nikodym的微分定义再在RC network上定义啥测度之类的？懂点微分方程和
傅立叶变换就够了。电路原理和信号与系统里的知识都足够解决这些问题，哪里要用近
代数学的知识啊？ |
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a**a
发帖数: 416 | 8 发信人: ukim (X人), 信区: Science
标 题: Heroes in My Heart (41)
发信站: BBS 水木清华站 (Thu May 16 09:59:50 2002)
开始说说波兰的数学家，从Banach开始, 最最伟大的波兰数学家。
Banach在数学界的登场是一段美丽的传说// :"-))
1916年的一个夏夜，Steinhaus在一个公园里散步，突然听到了一阵阵的谈话声，更确
切的是有几个词让他感到十分的惊讶，当听到“Lebesgue积分”这个词的时候，他就毫
不犹豫的走向了谈话者的长椅，原来是Banach和Nikodym在讨论数学。Steinhuas就这样
子发现了Banach,并把他带到了学术界。他说：“Banach是我一生最美的发现。”
波兰学派的人似乎喜欢在咖啡馆里讨论数学，Kuratowski和Steinhaus是有钱人，他们一
般在高档的罗马咖啡馆里谈论数学；Banach,Ulam和Mazur穷一些，整天呆在一个苏格兰
咖啡馆里，那里的老板挺不错，即使过了营业时间，也不会赶他们。这样子很多年轻的
数学家都来到这里，每次有什么 |
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k*******l
发帖数: 69 | 9 don't know what you want to get
the RN derivative of which measure wrt which measure???
possibly u could look at Oksendal's Applied Stochastic Control of Jump
Diffusions, Ch. 1
and
we |
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j****j
发帖数: 270 | 10 say by changing of measure, you can get rid of the drift of the brown motion
, and you can change the intensity of the poisson process, ... |
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k*******l
发帖数: 69 | 11 that's what i was asking, what measure are you changing into? risk-neutral?
forward-measure? or other numeraire-denominated measures? or if u r not in
quant finance, then u r probably asking for other things.
i don't know what to say. check out the book (chapter 1) i recommended in my
previous post.
motion |
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j****j
发帖数: 270 | 12 It really doesn't matter. RN is a very useful notion that applies in many
fields. I am looking for very general treatment about change-of-measure, not
specific to finance. Thanks
?
my |
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j****j
发帖数: 270 | 13 You sure it talks about change-of-measure when BM and Poisson are not
independent? Thanks! |
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Q***5
发帖数: 994 | 15 For (1.3), you can pull a big gun: Radon-Nikodym Thm.
We know that m(A)< B p(A) for any open set. So for any Borel measuable set D
, if p(D) = 0, then D can be covered by a open set, which is of arbitrarily
small measure. As a result, m(D) is arbitarily small, there for m(D) = 0.
Hence m<=0, and h is Borel
measurable, L_1 w.r.t p.
You can prove that h is bounded by B a.e. w.r.t p: otherwise, there exists
delta>0, such that p(E)>0, where E = (h>B+delta), w |
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b**********t
发帖数: 6 | 16 问题背景：
随机过程和系统的仿真，尤其碰到稀有事件时，常常采用一个新的概率测度替代原有的
自然概率测度仿真，以增大稀有事件的发生概率（change of measure,重要抽样）。比
如我们想通过仿真（Monte Carlo）估计样本空间中的某个集合A的概率（很小，比如10
的-8次方级），如果采用自然概率测度，则平均要仿真一亿次才能得到一个样本落在这
个集合里面。这样需要仿真多个亿次才能得到有效的估计。为克服这个问题，我们可以
采用一个新的概率测度，在这个新的测度下，让集合A出现的概率变大，甚至大到1，也
就是说仿真的每一个样本点都落在集合A里面。当然，在结果统计时，一个样本点不能
算1个，而必须乘以likelihood ratio来纠偏。进一步，假设集合A的一个划分：子集A1
, A2, A3. 其中A3的自然概率比A1, A2小很多（比如A1, A2在10的-8次方级，而A3在10
的-12次方级）. 这样估计集合A的概率时我们可以忽略A3。于是我们可以选择一个新的
概率测度，让全部样本点都落在A1,A2里面。
问题描述：
对于这样一种做法，有一些事件（A3）在原有的自然概率测 |
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n******t
发帖数: 189 | 17 class A content(two semester):
Set theory/fundamentals. Axiom of choice, measures, measure spaces, Borel/
Lebesgue measure, integration, fundamental convergence theorems, Riesz
representation.
Radon-Nikodym, Fubini theorems. C(X). Lp spaces (introduction to metric,
Banach, Hilbert spaces). Stone-Weierstrass theorem. Basic Fourier analysis.
Theory of differentiation.
class B content(two semester):
Probability spaces. Distributions/expectations of random variables. Basic
theorems of Lebesque theor... 阅读全帖 |
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g******8
发帖数: 542 | 18 Thank you very much! I will check out the book you recommended. Thanks! The
articles I browsed (coz I did not fully understand) was somehow they
specify a state price density (pricing kernel) and also use something like
Ito's lemma, Radon-Nikodym derivative (change of measure) etc. but I could
not fully understand how they did it.
in |
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w******g
发帖数: 271 | 19 Hi Big Brothers, hot sisters,
I'm still confused about the change of numeraire technic,
say P and Q are equivalent measures,
then dP/dQ= D , I also checked the Radon-Nikodym theory, could anyone
explain more in detail about this formula?
I just know V(s)/M(s)= E[V(T)/M(T) |Fs], V and M are numeraires, this is
clear for me. But how does the dP/dQ come from? I don't get the point of the
derivative here.
Sorry, can't type Chinese, can anyone answer in Chinese? |
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L**********u
发帖数: 194 | 20 在calculus中，求积分的时候，你会做变量替换，变量替换后，在新的坐标系下，被积
函数是原来的函数乘上Jacobian。这里的Radon-Nikodym derivative D就是 ”
Jacobian“。 |
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r**m
发帖数: 7 | 21 Let w_t be the brownian motion under the original risk-neutral measure Q
_0, and \tilde(w)_t be the brownian motion under Q_s(with s(t) as
numeraire). Change the probability measure from Q_0 to Q_s using Girsanov
theorem with exp(-1/2\sigma^2t+\sigma w_t) as the Radon-Nikodym
derivative.
We have \tilde(w_t) = w_t -\sigma t. Therefore, the original drift rt
becomes (r+sigma^2)t. |
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m*****z
发帖数: 357 | 23 求出来一个具麻烦无比的东西
exp{-BtXt - 0.5Xt^2}
其中，Xt = intergral from 0 to t（f（Bs）ds）
f(x)是个优点麻烦的函数
整个Xt不能用Ito's fumula 替代掉
化不掉。。。我要死了。。。 |
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L*******t
发帖数: 2385 | 24 首先，SP500用drifted BM来model会被毙掉啊。。。。
其次，一个东西的prob，就是这个事件indicator function的expectation，你要换测
度，需要乘一个Radon Nikodym derivative，然而，你不知道这个事件和RN
derivative的correlation，所以你说的方法可行，不过需要聪明一点的处理方式。。
。。
Q_ |
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L*******t
发帖数: 2385 | 25 Define event A = {sup_{t in [0,T]} (W^P(t) + mu*t) >= b}
= {sup_{t in [0,T]} (W^Q(t) ) >= b}
Define xi(t) = Radon Nikodym derivative from Q to P
= exp(-0.5*mu^2*t+mu*W^Q(t))
Then, P(A) = E^P[Indicator_function(A)] = E^Q[xi(T)*Indicator_function(A)]
Note that, there is an exp(W^Q) in xi(T), and there is a sup_{t in [0,T]} (W
^Q(t)} in event A.
So, first, you have the joint density of max(W^Q) and W^Q, which can be
found in a lot of lecture notes, then, compute ... 阅读全帖 |
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j******n
发帖数: 91 | 26 我能理解Radon Nikodym derivative，在上述题目中（drift=a, vol=1）是 exp(B_t-0
.5*t)。然后怎么继续推导？ |
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s*****n
发帖数: 2174 | 27 说说我用过的书:
周民强的 <<实变函数论>>, 本科时候的教材, 很不错.
测度里面最经典的是Halmos 的 Measure Theory, GTM系列, 这本书很严谨, 不过Radon-Nikodym theorem的那一块, 老师说思想有点老, 用讲义代替.
也看过Ash的Probability & Measure Theory, 这本书基本就针对概率讲测度论, 相对通俗易懂, 数学感较差. 不过这本书里面有不少错, 有时让人摸不着头脑. |
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