f*******i
发帖数: 8492 | 1 不好意思，因为我现在做一个project，需要做出RDF的图形，现在的问题是我拿到了数
据，但是对于
RDF的定义比较模糊，希望懂这方面的人能帮我指点一下。 只考虑一维。
g(r)=density(dr)/average density 这个应该是g(r)的数学表述吧？
我觉得用数据表述起来比较简单
r dr r到dr范围间的颗粒数 r+dr整个范围内的颗粒数
0 0.2 0 0
1.0 0.2 2 2
1.2 0.2 3 5
1.4 0.2 5 10
1.6 0.2 7 17
请问，我如果想得到g(1.4)这一点的值应该怎么表述呢？
A g(1.4)=（5/0.2）/(10/1.4)
B g(1.4)=（5/0.2）/(17/1.6)
谢谢~~ |
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b*p
发帖数: 242 | 2 B
density at r relative to one ball at r=0 / average density = r.d.f |
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t**********m
发帖数: 205 | 3 The joy of viXra
https://telescoper.wordpress.com/2011/05/19/
Posted in The Universe and Stuff with tags arXiv, astronomy, Cosmology,
Physics, scientific publishing, viXra on May 19, 2011 by telescoper
From time to time on this blog I post rants about the state of scientific
publishing, open access, the importance of the arXiv for astronomy and
cosmology, and so on.
This morning, however, I discovered an “alternative” side to the whole
business of online science, a site by the name of viXra. Mos... 阅读全帖 |
|
t**********m
发帖数: 205 | 4 http://blog.vixra.org/about/
1
CV: Philip Gibbs
Nationality: British
Education Summary
Secondary school: Currie High School, Midlothian, August 1972 - July 1978
Scottish higher grades:
English B, Mathematics A, Physics A, Chemistry A, Engineering Science A
Scottish sixth year studies:
Physics A, Maths algebra A, Maths analysis A, Maths Statistics A
higher education:higher education:higher education:higher education:higher
education: higher education: higher education:higher education: higher
edu... 阅读全帖 |
|
i*****e
发帖数: 218 | 5 向大家请教一个问题：
统计力学中常提到的Ising Model 到底算是经典的， 还是量子的 ？
如果假设自旋只有向上， 向下两个状态。
如果Ising Model算是经典的， 可以给它定义density matrix吗 ？
还是density matrix只能适合量子力学体系 ？
多谢大家。 |
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t***o
发帖数: 24 | 6 有个粗浅的解释， chi square distribution在 df<2的时候，在0附近是有density的，
可以趋近， df>=2的时候，在0的density=0，无法趋近。 至于为什么要用generating
function, 不太明白, 可能是正正经经证明这是non-central chi-square?
：（ |
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a***r
发帖数: 594 | 7 is this joke? the form of brownian motion has strong physical meaning and is
definitely NOT a convenient way out due to laziness.
the sqrt(t) behavior in diffusion processes is a direct result of density
leaving spatial vicinity at a rate proportional to the density gradient.
df/dt = - c * df/dx.
this is studied and established by theoretical and experimental physicists
hundreds years ago.
more fundamentally, the df/dt ~ -df/dx relationship is a result of the
assumption that the underlying r |
|
j*****4
发帖数: 292 | 8 2.
The water level will drop w.r.t the boat.
If the object has average density greater than water density(sink to the bot
tom), water level will drop w.r.t the bank.Otherwise,it will keep the same.
to |
|
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s********t
发帖数: 31 | 9 Conclusion is correct but the logic is not correct.
If the object sinks, boat will rise NOT because of water level drops.
Reason: If object floats, boat still rises even though water level will be the same.
A extreme case: the object has very large density thus no volume. So throwing
it to water is as if the boat has smaller density. So boat will rise and water level drops, both wrt to land. |
|
B****n
发帖数: 11290 | 10 Let f(x) denote the density (or probability) of X.
The distribution of min(X,1) is the same as the distribution of the sum
P(X<=1)*G+P(X>1)*1, where the random variable G has conditional density f(x|
X<=1).
So the variance of X is equal to P(X<=1)^2*Var(G).
Var(X)>=E(Var(X|I{X<=1}))>=P(X<=1)*Var(X|X<=1), where I is an indicator
function.
So Var(X)>=P(X<=1)^2*Var(X|X<=1)=Var(min(X,1)) |
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B*********s
发帖数: 306 | 11
I don't think this is correct. The *expected arrival time* of the slow car
is 3 minutes. It doesn't mean that you will see the slow car within 3
minutes for sure. So, even if you condition on the fast car arriving in
more than 3 minutes, you are not guaranteed to see the slow car first. The
question is asked in the setting of modeling the arrival process as
Poisson, where the inter-arrival times are exponentially distributed. With
independent arrival, the joint density is the product of the two... 阅读全帖 |
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l****o
发帖数: 2909 | 12 Exactly. by the way, how do you think about the link between risk neutral
density and real world density? especially regarding the volatility
estimation in real world and risk neutral world?
is |
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l****o
发帖数: 2909 | 13 Exactly. by the way, how do you think about the link between risk neutral
density and real world density? especially regarding the volatility
estimation in real world and risk neutral world?
is |
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p******y
发帖数: 5 | 14 Not really, the question asks for density for X+Y, not the joint density of
X and Y. |
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z****g
发帖数: 1978 | 15 essentially you try to find unconditional density function of the event with
conditional density function given by models. |
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m******2
发帖数: 564 | 16 无drift时候那个
Hitting Time Density = 2 pdf(M,W) M=W
这看起来很奇怪，但是它是事实。
Shreve用偏微分推导了半天的Joint Density for Maximum of Brownian Motion and
its terminal value 在w=m那点的密度恰为在t时刻打到m的概率密度的一半。
有没有高手给理解一下啊？
这可是考验功力的问题啊！ |
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y***s
发帖数: 23 | 17 f(x) is the density of N(0,1)
f(x/sqrt(t)) is the density of N(0,t)=W_t in distribution, which is
even. |
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f*********5
发帖数: 367 | 18 Risk neutral measure?
——面试小白
我在想是
1. 指用BS模型，\mu 取银行利率，sigma取implied volatility吗？如果是的话这个
implied volatility是如何算的？利用at the money options的价格估计:
Option Price ~= 0.4*sigma*sqrt{tau}*S 来反推的吗？tau是time to maturity. 反
推就是
Implied volatility=OPrice/(0.4*sqrt{tau}*S) 这样算吗？
2. 还是直接对比BS公式给出的比如European call的价格公式c=S*N(d1)-K*exp(-r*tau
)*N(d2)和市场上option的价格，把那d1和d2里面的参数sigma当成未知数，用牛顿法之
类的找这个非线性方程的数值解？
3. 又或是希望我给出一个risk-neutral probability density function of stock
price at time T? 然后我要先假定have the European call p... 阅读全帖 |
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s**********y
发帖数: 353 | 19 我觉得这道面试题对初学者能答出从option prices 得到 Risk neutral density
就可以了。具体操作直接用 market prices 是得不到可以直接用的RND，有的地方
density甚至可能是负的。前面有人说过了，大概两种做法。 用parametric approach,
比较简单的方法可以假设RND是几条lognormal curve 叠加，fit the parameters
according to market prices; non-parametric approach, 你可以先得到IV，smooth
IV,
然后 再用 price 求导。 |
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h******5
发帖数: 58 | 20 Thank you all for sharing the info. I almostly solved the issue by using the
GSL library. The last problem I got is that the optimizer is not strong
enough when the initial parameters guess are not close to what they should
be. I used 13 sample data to fit a normal distribution density function (x =
returns, y=densities ). Based on the sample data, mu shoulbe be close to 0
and sigma should be close to 0.02. When I put 0 as inital mu and 0.15 as
initial sigma, the result is good. But when I try 0... 阅读全帖 |
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F****t
发帖数: 4 | 21 Expansion本质上是对terminal density做的。你去做calibration实际上是fit
terminal density。这样对于path dependentoption比如knock out，你如何认为你的
path上的any时刻的distribution是吻合市场的呢？
这个问题heston本身也是有的。local vol会好很多，因为local vol calibration是
fit whole surface not one smile |
|
L*******t
发帖数: 2385 | 22 明白你的意思了，可以看看SP Zhu 2006，expand american options的，这个似乎不像
是对Terminal Density做的啊。还可以看Fouque, Sircar的一系列文章，他们的
diffusion也不是对terminal density做的。
你说的expansion是那种？ |
|
L*******t
发帖数: 2385 | 23 我觉得取决于业界怎么用模型。。。
和我聊天的那个教授，已经在develop一些不用recalibrate的一些非参数时间序列模型
了，过段时间和我做连续时间的。
recalibration我总觉得不好。违背了模型的精神。我的density很好用，算起来非常方
便，而且calibrate就是求解一个线性方程组，求个矩阵的逆就好了。我觉得是这个模
型的优势。而且我的density是动态非静态的。
随时间有个dynamics。
现在急需知道业界的一些做法，实习又被拒了，还是决定向板上的大牛前辈们讨教，不
去实习了，apply起来太心酸。 |
|
L*******t
发帖数: 2385 | 24 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 |
|
s**e
发帖数: 103 | 25
Don't understand. The bigger density will definitely over close the
universe, it's actually put an upper bound on density of any unknown
matter spicies. |
|
z***e
发帖数: 5600 | 26
what does the LHS mean?
N(x) where N is the standard normal distribution? Then no such c
exists because if x-> -infinity, LHS=1 and RHS=0
Again, what do x(1),..,x(n) mean? The joint density function of
n independant RVs X1...Xn is, by definition,
g(x1,x2,..,xn)=f1(x1)...fn(xn)
where fi are the density of xi. In your case all fi=f, there should be
no n! factors here.
-Z. |
|
c**********g
发帖数: 222 | 27
>>> it is proved that the freely expanding (adiabatic expansion)
photon field
will remain the balckbody spectrum and in thermal equilibrium as long
as it is
in equilibrium at the begining. You can try to prove it. it is fun.
>>>> yes, the photon energy is not conserved. The energy density drops
at the
rate of V^(-4/3) or (1+z)^4 or a^(-4). V is the considered volume. a
is the scale
factor. for the ideal fluid with p=w \rho. (w is a constant. \rho is
density and
p is pressure), then \rho drop |
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i*******n
发帖数: 166 | 28
it is energy density per unit frequency
I would like to say:
\pho(\nu) d\nu is the energy density of the photon with energy
at the range h\nu to h(\nu+d\nu)
To my point of view, some other equations in physics books also
has such kind of inaccurate description which make students uncomfortable
and sometimes it may lead to wrong understanding.
It is still a unresolved problem to find a suitable energy-momentum
Tensor in GR. Currently, energy conservation is regard |
|
i*******n
发帖数: 166 | 29
Brown dwarfs are those stars with mass lower than, say,
0.08 M_sun so that they cannot ignite Hydrogen in their
lifetime. Dark matter cannot be all contributed by those
stuffs. In fact, if BBN is right, the observation has
already limited the baryonic density (in unit of critical
density) to be less than 0.1. And most of the baryonic
matter cannot be seen. |
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h****o
发帖数: 49 | 30 I found that heat capacity=density*specific heat, then since
the density
has the unit of kg/m3, specific heat is J/kgk, then heat
capacity should
have the unit of J/m3k, right? (Holman, heat transfer, the
last
paragragh of page 4)
thermal expansion coefficient has the unit of ppm-k;
ppm=parts per million, who can find some explantion about
ppm? I can not understand it.
Thx. |
|
J**Y
发帖数: 34 | 31 It is easy to write down the density function of max{xi} based on
f(x). See the chapter about order statistics in any probability text
book. The ersulting density is still the function of the parameters in
f(x). Then use MLE to get the estimate of these parameters, that is,
solving the parameters according to first-order conditions. |
|
t****x
发帖数: 1650 | 32 想把两个histogram重合在一张图上，并在上面加上相应的density curve/smooth contour.
知道在s-plus里作两个分开的，各自有density curve，不知道怎么重叠起来，
在exel里可以重叠起来，但是没有kennel curve之类的加上去 |
|
l*******l
发帖数: 204 | 33 in R try:
hist(x,freq=F)
lines(density(x))
hist(y,freq=F,add=T,border=2)
lines(density(y),col=2) |
|
l*******l
发帖数: 204 | 34 hist(x,freq=F,main="Histogram of X and Y",xlab="")
lines(density(x))
hist(y,freq=F,add=T,border=2,lty=3)
lines(density(y),col=2)
No familiar with S+. Sorry |
|
s******h
发帖数: 539 | 35 I'll give you an example that says x, y, z are marginally normal and
pairwise independent but jointly can have infinitely many possible
situations.
Consider (X, Y, Z) with the following density function (don't mess those '{'
,'}'):
f(x, y, z;theta)
= 1/sqrt{2*pi}^3*exp{-(x^2 + y^2 + z^2)/2}*
{1 + theta*xyz*exp{(x^2 + y^2 + z^2)/2}*I{-1<= x, y, z <= 1}},
theta can be any real number such that theta < exp{-3/2}.
In this case, f(x,y), f(x, z), f(y,z) are all normal densities with x, y, z
pairwise i |
|
s******h
发帖数: 539 | 36 From the example I gave, you can not express P(x, y, z), or equivalently f(x
, y, z) in terms of f(x, y), f(y, z) even though they are bivariate normal
densities and x, z are independent. This is because, in the above example, f
(x, y), f(y, z) are independent product of N(0, 1) densities, so no matter
what you do, you are not able to make f(x, y, z), which contains parameter
theta, in terms of f(x, y), f(y, z).
However, if you change the condition 'x and z are independent' into 'x and z
are ind |
|
c********8
发帖数: 5 | 37 有一个关于 RMA background correction 的公式：
Model S as the sum of a signal X and a background Y, S=X+Y, where we assume
X is exponential (α) and Y is Normal (µ,σ2), X, Y independent
random variables.
Background adjusted values are then E(X|S=s), which is
a + b[f(a/b) - f((s-a)/b)]/[F(a/b) - F((s-a)/b) - 1],
where a = s - µ- ασ2, b = σ, and f and F are the normal
density and cumulative density, respectively.
请提供一些推导此公式的线索。谢谢！ |
|
g*****k
发帖数: 623 | 38 这个很赞，根据你的提示，这道题可以这么解
Ｕ1和Ｕ2是 iid U[0,1]
X1 = min(U1, U2) and X2 = max(U1, U2)
density function of X_1 和 X_2 是
2-2x_1 和 2x_2
只不过我在求joint density 时还是卡住了。
不知道如何求 p(X1, X2) = ...
但是非常感谢你，因为你的解释证明了X_1不是uniformly distributed on [0,1]
我已经非常开心了。多谢。 |
|
l*********s
发帖数: 5409 | 39 uniform distribution of bivariate(u1,u2) with support of a unit square,
their difference is the intercept of the familiy of diagonal lines,with support(-1,1),which
corresponds to the corner, so density is 0, the mode is 0, with a density of
the 1/sqrt(2). |
|
a*****3
发帖数: 601 | 40 看了三遍没看懂 - 贱妾有时间帮忙看看这个‘小鸟’定理说的是什么。
support(-1,1),which : corresponds to the corner, so density is 0, the mode
is 0, with a density of the 1/sqrt(2). |
|
P******V
发帖数: 83 | 41 已知 x0 is a sequence of (0,1), 通过计算得到density f(x0) (continuous). 现在
我想根据这个已经得到的density f(x0) , 来在（0，1）之间随机取得100个samples，
请问在R里面应该如何操作啊？谢谢 |
|
T*******I
发帖数: 5138 | 42 一个有用的SAS code，用于在同一空间里画两个分布曲线。
data test;
drop i;
do group='A','B';
do i=1 to 10;
x=20+ 8*rannor(12345);
output;
end;
end;
run;
proc print; run;
proc transpose data=test out=test2;
by group;
var x;
run;
proc transpose data=test2 out=test3(drop=_name_);
id group;
run;
proc print; run;
ods listing close;
ods html;
proc sgplot data=test3;
density A / lineattrs=(color=red pattern=1) legendlabel='A';
density B / lineattrs=(color=blue pattern=2) legendlabel='B';
xaxis label='Normal Curves';
run;
ods h... 阅读全帖 |
|
z****g
发帖数: 1978 | 43 你MLE还没入门
MLE的本质是最大熵，大样本时log-likelihood function是逼近分布的熵的，这个也是
真实世界里的终极规律。所以虽然一般来说log-likelihood function直接用的是density
function，但是这个是连续变量的情况。一般情况下应该是distribution function的微分，所以
你不能直接把变换过以后的数值带到standard normal distribution的density里
你那个残差的概念，只不过是入门。
统计两大估计方法：MLE类和Momentum类，第一类是直接基于最大熵的，第二类是基于
分布的Momentum Generating function的Taylor逼近。
N( |
|
d******e
发帖数: 7844 | 44 Covariance Matrix Rank deficient.
举个简单的例子，x服从N(0,1), y = x，那么(x,y)服从正态分布。但是这个协方差矩
阵rank只有1。无法求逆，自然也无法算density。其实Rank deficient的情况很多，常
用的PCA的假设就是covariance是low rank的。
正态分布的定义是
A random vector is said to be multivariate normally distributed if every
linear combination of its components has a univariate normal distribution.
并不是由density function来定义的。 |
|
o****o
发帖数: 8077 | 45 is this what u r looking for?
x<-rnorm(20000)
idx<-replicate(10, sample(x, size=1000), simplify=T)
hist2<-function(x){
return(hist(x, plot=F))
}
hist<-apply(idx, 2, hist2)
plot(hist[[1]]$density~hist[[i]]$mids, type='s', lty=1, col=1)
for (i in (2:10)){
lines(hist[[i]]$density~hist[[i]]$mids, type='s', lty=i, col=i)
}
I felt that using SAS in this case will be a smooth solution, too |
|
t****r
发帖数: 702 | 46 sampling distribution 不是 population density的一个empirical estimator。。。
你说的事data distribution.
Sampling distribution 是statistics的distribution。比如你的population可以是任
意的density， 但是sample mean 的sampling distribution总是接近normal的。
standard |
|
t*****w
发帖数: 254 | 47 When I had my job interview, they always tested my SAS skill.However I use R
all the time. To help your preparation, read my R codes to see how much you
can understand it.
%in%
?keyword
a<-matrix(0,nrow=3,ncol=3,byrow=T)
a1 <- a1/(t(a1)%*%spooled%*%a1)^.5 #standadization in discrim
a1<- a>=2; a[a1]
abline(h = -1:5, v = -2:3, col = "lightgray", lty=3)
abline(h=0, v=0, col = "gray60")
abs(r2[i])>r0
aggregate(iris[,1:4], list(iris$Species), mean)
AND: &; OR: |; NOT: !
anova(lm(data1[,3]~data1[,1... 阅读全帖 |
|
a*****i
发帖数: 1045 | 48 第一年，完全没什么概念呢。已经开始选择论文题目了。现在有几个option，不知道有
没有好心人可以帮忙看看哪些方向比较好，或者对将来工作会有帮助。
自己之前定了跟statistical process control 有关的方向。但不怎么样。
1. Title : A comparison of different statistical process monitoring schemes
on real life data （这个是关于statistical process control的）
It is often desirable to provide online monitoring for industrial or
agricultural processes. Faced with the task of performing such monitoring,
practitioner will often choose to apply a control chart. However, there many
options available, and it i... 阅读全帖 |
|
p***o
发帖数: 44 | 49 上面的回答不完全对。drburnie 回答的是large p small n 导致的问题。这不是传统
意义上的curse of dimensionality。这个词是专在non parametric estimation 里才
用到的，近几年却因为high dim 的火热被人张冠李戴了很多。
直观解释的确是需要的数据随着dim 增加而迅速增长。但最早是专指kernel density
estimation 中 收敛速度会变慢。估计density 时，把数据按照小窗口来分，一个一个
小窗口来估计。单位面积内分割的小窗口的个数是维度的指数，如果每个小方格里需要
一个点，在三维下就已经需要1000个数据了。这个困难扩展到kernel smoothing 和其
他的non parametric regression。
如果把curse of dim 理解成“估计的精确度随着维数增加而下降”， 那就作为一个现
象永远存在。无论有多少样本，无论维数是多少。哪怕样本数是10000，或者更多，只
要维数增加了，哪怕只是从2加到3，它还是存在。 |
|
D***0
发帖数: 5214 | 50 原文:
For the first time since the Institute began comparing driver death rates
among vehicles in the 1980s (see Status Report, Nov. 25, 1989; on the web at
iihs.org), researchers adjusted for a variety of factors that affect crash
rates, including driver age and gender, calendar year, vehicle age, and
vehicle density at the garaging location. Previously, researchers had
adjusted only for driver age and gender.
但是
The adjusted driver death rates do a better job of teasing out differences
among veh... 阅读全帖 |
|
