j*******e
发帖数: 2168 | 1 right, right, still more or less a guess, but I somehow think you have
better chances.
1) I tend to place HA in LHO because of the opening lead;
2) In case RHO cashes HQ, you are certain now;
3) If you play HT, the unlucky case is HQ shows up in RHO; If you play HK,
the unlucky case is HA shows up in RHO, the latter has better indication of
card placement.
Anyway, it's just my thought. I am here to ask for advice:) |
|
b***y
发帖数: 2804 | 2 What do you mean "still any way"? The contract is virtually cold now. You
can over-ruff and play a club, hoping that LHO shows out, in which case you can make an overtrick. If LHO follows, you ruff in dummy (club should be 3-3), finesse spades, cash SA, then club (dummy pitches heart). RHO can ruff this, but has to give you DK. Make only 2.
Actually if LHO follows club, you can even try a simpler line: pitch H on the club, then just SA and SQ. RHO gets SK, but that's it.
The point is, once you p... 阅读全帖 |
|
b***y
发帖数: 2804 | 3 Best play will go down, wrong play (with lucky guess) will make.
LHO cashes 3 spades (keep last spade un-cashed) and exit with a heart. RHO
pitches 1 diamond and 1 heart, with 5 cards left: HQJ, DKQT. Now double-
dummy, declarer can just play hearts and score H10. But declarer has no idea
that LHO started with 4 hearts. He more likely will play LHO for 6313 shape
, that would leave RHO with HQJx and DKQ. So he wins HA in hand, plays DA
and another D, trying to establish his DJ. Lights out!
Still... 阅读全帖 |
|
b***y
发帖数: 2804 | 4 The line also wins if RHO has Kxx. If RHO has Jxx, it doesn't matter anyway:
even if you guess to duck second spade to LHO SK, RHO will get a ruff in
club. |
|
b***y
发帖数: 2804 | 5 These two lines have very similar odds.
Line 1 wins when RHO has KJx (3 cases), Jx (3 cases)
Line 2 wins when RHO has Jxx (3 cases), xx (3 cases)
Technically speaking, line 1 is slightly superior since it also wins when
RHO has stiff SJ. But this is very insignificant.
How was the bidding go? Something like 1D-1S-4S? |
|
i****e
发帖数: 642 | 6 The real hand:
AJ
KQ83
AJT84
KT
Qxxx
xxx
-
A98643
KT8
AJT
9763
Q72
Either way, you will be down. But you are not alone, since I would play RHO
for SQ too. At table, both declarers made 10 tricks. They are stars, I have
to admit :)
Questions:
1. Given the bidding, what will you lead, S or C?
2. If C is led, does it do any good to play CK at first trick?
3. continue with Q#2, after CK holds, is there any reason not to safely play
RHO for SQ?
The actual play:... 阅读全帖 |
|
i****e
发帖数: 642 | 7 The trick seems to prevent RHO from gaining lead by possible D Qxxx or Qx.
Cash 3 hearts, discarding a C. DAK, and throw in LHO by trump.
If RHO can get in with D Qxx, we still have C guess. I will play RHO for CQ,
but I think I won't need that guess. |
|
j*******e
发帖数: 2168 | 8 I didn't think I would have any problem establishing a H, since RHO would
likely return H; but maybe you are right, playing H first may increase the
chances of endplaying RHO.
RHO |
|
a****s
发帖数: 524 | 9 S AJ95432
H 6
D AQ8
C 98
MP , Both
Your RHO opens a 3rd seat 1D, you overcall 1S, doubled by left, RHO rebid 2D
. You try 2S, doubled again by LHO, explained as show cards and don't like
sell 2S. RHO fumbled and bid 2NT.
Distrusting the whole affair, you have a strong feeling these guys are
stealing from you. taking a deep breath you bid 3S, passed out quickly.
LHO leads D6, dummy comes down:
S ---
H AQ943
D 1092
C QT732
S AJ95432
H 6
D AQ8
C 98
Seems these guys are at least as twice crazy as yo... 阅读全帖 |
|
b***y
发帖数: 2804 | 10 The more failure cases, the worse the line, right? Case 3 fails on: RHO
stiff K, RHO Kx, RHO K8. How does that make it better? |
|
b***y
发帖数: 2804 | 11 Playing diamond to Q is inferior. Compared with playing to 8, you lose the
case where RHO has KJxx.
Diamond to 8 will win any time LHO has DJ, or RHO has both K and J.
You can play HK first and maybe set up 4th heart. But in order to cash it,
you still need to guess diamonds (and you won't be able to guess right if
RHO has both K and J). Against strong opponents, you won't have a real clue
on who has what, so in that sense, the outcome is also already determined,
even if double-dummy you could h... 阅读全帖 |
|
i****e
发帖数: 642 | 12 LHO shows out on first H. You win HA, ruff another S, and play H to RHO's HK
. As expected, RHO returns DT. You win DA, and clear last trump from RHO.
Any change to your plan? |
|
i****e
发帖数: 642 | 13 这也是我的感觉。
HA
C to T and K
H to K
C to J and A
H to Q, RHO follows
S to Q, RHO with 9
Sx J K and A
S to T, RHO pitches H
现在的形势很好：已经有8个赢墩，只输了CAK和SA。 |
|
i****e
发帖数: 642 | 14 大吼一声，100% 的路线找到了！
The difficulty for us is, when RHO has 7 minors (4-3 or 5-2), we may need to
guess his major holding, which can be either 1-5 or 4-2.
But if we end up with this 4-card ending,
KT9
-
T
-
A43
K
-
-
RHO must have DJ, plus 3 major cards. And he already discards a H. Now we
play HK, and we can tell he has 2 or more hearts originally! The trick is to
delay the DT discard from dummy, and force RHO to discard a H first.
The interesting thing is that this works the same way if LHO has l... 阅读全帖 |
|
a****s
发帖数: 524 | 15 Match point
S Deals E-W Vul
East:
♠ AK4
♥ K43
♦ K95
♣ K1042
RHO 3rd seat open 1S, you 1NT, Passed to RHO, who repeat 2S.
This is passed to partner, who doubles
(more or less a takeout), RHO passes, You? |
|
a****s
发帖数: 524 | 16 Match point
S Deals E-W Vul
East:
♠ AK4
♥ K43
♦ K95
♣ K1042
RHO 3rd seat open 1S, you 1NT, Passed to RHO, who repeat 2S.
This is passed to partner, who doubles
(more or less a takeout), RHO passes, You? |
|
x***u
发帖数: 6421 | 17 Ft~mv
m~rho*vSt
F~rho*v^2S
rho:空气密度
v:物体速度，排出空气的近似速度
m:t时间内排出的空气质量
S：物体截面积 |
|
j****u
发帖数: 1413 | 18 中国科大咨询业校友调查
中国科大供职管理咨询公司（外资）校友调查
阅读全文及图表请点击：http://www.ustcif.org/default.php/content/2445/
立即捐赠低调的牛校，为中国科大打赏！
——您的公益捐赠可获美国与中国个税减免!欢迎指定项目，捐赠方式附页尾
如您认为本调查对您有启发，敬请捐赠母校。
咨询业[1]被认为是工作内容最具挑战的行业之一。有数百位中国科大校友供职咨询业
。中国科大新创校友基金会（简称：新创基金会）发布科大咨询业（外资公司）校友初
步调查。本文附上超过30位咨询业的校友简介，或许其职业经历对年轻校友有更形象的
启发。
新创基金会调查表明：多数外资咨询业校友有海外留学背景；多人有Ph.D.、MBA或J.D.
学位；也有职场校友“转行”进入咨询业。这可能是中国科大与相关校友机构首次发布
的咨询业校友调查，期望帮助年轻的中国科大人了解职业抉择的一种选择。
咨询业校友在各公司分布
综合基金会数据中心与LinkedIn数据，新创基金会掌握的供职外资咨询公司的科大校友
约105人，分布在24个公司。新创基金会估计[2]，至少有160名科大校友... 阅读全帖 |
|
s******c
发帖数: 331 | 19 工作当然很好，不过receptor和arrestin的mutations很多人都用过，fusion protein
也有不是新点子。XFEL做GPCR结构的也有一些先例，结构解析确实需要一些技巧，不是
可以靠软件自动完成的。
Rho是个比较特例的GPCR，很多机理并不能推广，比如Rho和G protein complex很多年
前大家都可以拿到，但是拿到稳定的非Rho的GPCR和G protein complex就很难。我个人
觉得做GPCR和arrestin complex的结构，未来的方向应该是电镜，因为GPCR和arrestin
的结合非常的不稳定，否则arrestin signaling也不会那么晚才发现，而且需要很多
post-translational修饰，或者arrestin的结合需要很特殊的lipid的环境，这些对晶
体都是很大的局限。
GPCR- |
|
F****I
发帖数: 270 | 20 首先，定义就是定义，只要是自洽的，只有接受不接受，是不是行业普遍接受的。
#9 算符的local与否，这里的local按Parr&Yang (2.1.24)的定义，和L(S)DA/GGA无关
。LDA的定义是
E_{xc}^{LDA}[\rho]=\int \rho(r) \varepsilon(\rho) dr (Parr&Yang的7.4.1)
包括后续的GGA,(8.7.1)等，在数学上很容易验证他们对应的算符都是local的，因为
LDA/GGA的泛函形式不存在HF里交换算符的情况。
然后，是Kohn-Sham方案的定义:
按Parr&Yang p146, v_eff(r) is a local operator in the sense of (2.1.24).
这个问题来自于，如果一个人从头到尾完整推导Kohn-Sham scheme（暂时不考虑non-
interacting v-representability), 对应的potential operator只能是local的。
我没有引进对KS scheme推广和修正的定义。我做的判断是,B3LYP等hydrid fu... 阅读全帖 |
|
b********e
发帖数: 58 | 21 \begin{equation}
\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty}
e^{-y^2}dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
e^{-(x^2+y^2)}dxdy =\int_{0}^{2\pi}\int_{0}^{\infty} e^{-\rho^2}
\rho d\rho d\theta = 2\pi /2 = \pi
\end{equation}
Notice:
\begin{equation}
\int_{-\infty}^{\infty} e^{-x^2}dx =\int_{-\infty}^{\infty}
e^{-y^2}dy
\end{equation}
we prove the conclusion that
\begin{equation}
\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}
\end{equation} |
|
l*3
发帖数: 2279 | 22 考虑R^3中的一个对称陀螺
(所谓R3^中的对称陀螺, 就是指在两个主轴方向上的转动惯量相等, 特殊的来说, 如果
在三维柱坐标中用 rho(r,theta,z)表示陀螺的密度分布函数 (非陀螺主体部分rho=0),
可以选到一个柱坐标, 使得rho只与r和z有关, 则就一定是对称陀螺. 更特殊的情形也
可以想象成平面上的一个区域绕三维空间中某一轴旋转扫过的点形成的均匀质量的三维
图形 (旋转体), 我们小时候玩的大都是对称陀螺)
其底端有一个支点.
考虑两种运动情形:
1. 将此支点固定.
2. 让此支点能够在水平支撑面上滑动, 支撑平面完全光滑.
现在, 让这两个陀螺有同样的较大初始角动量, 角动量方向沿唯一的非主轴方向, 也即
刚说的柱坐标中的z轴方向, 然后使该轴向微微倾斜, 与竖直方向有一个小夹角. 按如
上两种情形使得陀螺在重力场和支点约束 (两种不同的约束) 条件下运动.
试问, 哪一种运动更稳定?
(所谓谁更稳定, 是指如下概念: 如果初始角动量都一样, 那么当倾角越来越大时, 陀
螺必然晃得越来越厉害, 到达某个角度后陀螺就倒了, 不倒情况下对应的最大倾角越大
, 则... 阅读全帖 |
|
m********g
发帖数: 46 | 23 Consider the transformation:
(x,y) -> (\rho, \theta) -> (r,q)
r=\rho*cos(2\theta)
q=\rho*sin(2\theta)
Then Jacobi(r,q)/(x,y)=2. It means that r,q are independent with the same
joint distribution as x,y. |
|
x****x
发帖数: 87 | 24 假定市场上有两个不同index，比如是两个stock index futures（F1, F2）的报价；可
以得到高频的历史数据，比如10分钟。
现在想要预测未来10－30分钟 F1-F2的spread的走势；开始假设F1,F2都是lognormal，
用历史数据去算各自的mean/vol；此外假设
dw1×dw2 ＝ rho dt； 也是用过去的50个历书数据去算这个rho。 有了mean，vol和rh
o，用monte carlo去simulate F1-F2 的动态画出图像，
感觉不太管用：一方面是rho变化很迅速，二是monte carlo 模拟path不同，F1-F2就不
同了。55555
请问直接把F1,F2的历史数据对减得到 Spread ＝F1-F2,然后用利率的各种short rate
model去算这个spread的expected value如何？理论上很离谱啊。但是这些short rate
model都有解析解，不需要simulate了，calibrate to last historial spread可以了吧
，可能一个spread不够去calibrate。 |
|
m***s
发帖数: 605 | 25 我提供一个硬算的笨办法。
E(y2)=E(E(y2|x,y))=E(rho*x+(1-rho)*y))=1/2
var(y2)=var(E(y2|x,y))+E(var(y2|x,y))
note r=rho
let u=r*x+(1-r)*y, then x-u=(1-r)*(x-y), y-u=r*(y-x)
=var(u)+E(r*[x-u]^2+(1-r)*(y-u)^2)
=[r^2+(1-r)^2]*1/12 +r*E[(x-u)^2]+(1-r)*E[(y-u)^2]
=[r^2+(1-r)^2]*1/12 + r* E[(1-r)^2*(x-y)^2]+(1-r)*E[r^2*(x-y)^2]
=[r^2+(1-r)^2]*1/12 + r*(1-r)*E[(x-y)^2]
=[r^2+(1-r)^2]*1/12 + r*(1-r)*1/6
=1/12 |
|
c******r
发帖数: 300 | 26 En, I was treating the covariance to be 1/2 instead of 1/sqrt(2), but the
idea applies similarly, let
X = 1/\sqrt{2}(U-V)
Y = U
P(X>0|Y<0)=P(U>V,U<0)/P(U<0)= (1/8)/(1/2)=1/4
I think in general for rho > 0, the result will be
\frac{1}{pi}arctan(rho/\sqrt(1-rho^2)) |
|
o*p
发帖数: 77 | 27 problem 3
1/4 ?
suppose x,y are N(0,1)
then Cholesky decomposition x=rho*y+sqrt(1-rho^2)*z
y, z are independent
rho is correlation between x and y =1/sqrt(2)
draw graph in y ,z plane
then p(x>0|y<0)
=p(x>0 intersect with y<0）/p(y<0)=(1/8)/(1/2)=1/4 |
|
w**********y
发帖数: 1691 | 28 多谢分享.大概做了做..欢迎补充和指正.
- sqrt(i)=?
e^{\pi/4 i} or - e^{\pi/4 i}
- You and me roll a dice,first one gets a six wins. You roll first. what
is the probability of you winning?
P(I win) = P(Y !win and I win) = 6/11
- A stair of n steps. Each time you step up 1 or 2 steps. How many
different ways are there to reach the top? what is the asymptotic limit?
Fibonacci sequence ..limF(n)/F(n-1)==x for n>2, solve x, and F(n) ~ x^{n-1}
- Moment generating function of standard model.
statistic book…
- Write a si... 阅读全帖 |
|
t*******y
发帖数: 637 | 29 第二题应该是6/11吧
能讲讲这个吗? - X1 and X2 are independent random variable with pdf f and g.
what is what is the pdf of X=X1+X2
Jacobian matrix for X1+X2 and X1-X2..
多谢分享.大概做了做..欢迎补充和指正.
- sqrt(i)=?
e^{\pi/4 i} or - e^{\pi/4 i}
- You and me roll a dice,first one gets a six wins. You roll first. what
is the probability of you winning?
P(I win) = P(Y !win and I win) = 5/6*1/6
- A stair of n steps. Each time you step up 1 or 2 steps. How many
different ways are there to reach the top? what is the asymptotic... 阅读全帖 |
|
A**u
发帖数: 2458 | 30 X,Y 可以这么产生 若rho > 0 的话
U,V,W独立[0,1].
X = U
Y = U * 1_{W < rho} + V * 1_{ W > rho } |
|
l******i
发帖数: 1404 | 31 我只会最笨的解法，
Step1: Given Y1, Y2 are iid from N(0,1), you can write the pdf of jointly
distribution of (Y1, Y2), denoted by f(y1, y2).
Step2: Change of variable:
X1 = Y1;
X2 = \rho*Y1+((1-\rho^2)^(1/2))*Y2;
Then X1, X2 are two random variable ~ N(0,1) with correlation \rho.
Now the pdf of (X1, X2) should be
f(y1(x1,x2), y2(x1,x2))/abs(Jacobi determinant of the change of variable)
Step3:
cdf of (X1, X2) can be calculate from step2, denoted by F(x1, x2).
Thus F(m,m) is the cdf of max(X1,X2).
Thus F'(... 阅读全帖 |
|
i********c
发帖数: 7033 | 32 恩，明白了。我看了下hagan的文章,是说alpha, beta, rho,nu四个参数，其中beta
是提前定好的，rho, nu相对稳定，不需要经常update。而alpha needs to be updated
frequently，因此可以每天用atm vol 直接update alpha，再定期(say monthly)
update rho, nu就行了。
那么，像我前面所说，atm价格不直接available的话，calibrate的时候就直接用4
个参数一起update了？ 还是说，先用interpolation得到atm vol，再用atm vol
update其它参数呢？
我会用两种方法都试一下，更想知道业内通常的做法，在ATM价格不一定available
的时候，选择哪种方法update vol surface（直接求4个参数，还是通过interpolation
求得atm vol然后calibrate）? 我现在的underlying是上证ETF。谢谢指教了:)
ATM |
|
t********t
发帖数: 1264 | 33 你还是找篇文章先读读吧，Obloj在Risk Magazine发的那片
有两种方法，一般是beta->rho,mu calibrated from all available points->alpha
implied from rho, mu and ATM vol
另一种是beta->rho,mu,alpha calibrated all together from all available points
你说的情况，大概只能用第二种。第一种更稳定，而且不需要频繁calibration，因为
alpha is implied by some closed form |
|
f*******d
发帖数: 339 | 34
What are you talking about? "The more mass, expand more"? For your
information,
the Friedman equation is
d^2 R /dt^2 = -4pi G (rho+3p)/3
note the negative sign in the RHS.
R is the cosmological scale factor, rho the density, p the pressure. This
is exactly the opposite of what you say, i.e. in general relativity not only
mass deaccelerate the expansion, the pressure does so too. The only
exception
are those things with negative pressure such as cosmological constant.
So, as long as rho+3p >0, t |
|
c**********g
发帖数: 222 | 35
The SIGN of the cosmological constant Einstein introduced in order to offest
the expansion of the universe is OPPOSITE to the SIGN of the current
favorite cosmological constant.
In the cosmological dynamical equation: H^2+k/a^2=\rho. here, a is
the scale fator of the universe, H is hubble constant. k=-1,0,1.
\rho=\rho_matter+\rho_lambda is the cosmological density. static universe
requires H=0. when k=0, \rho=0. a negative cosmologicl constant (but pressure
is positive) can satisfy this |
|
c**********g
发帖数: 222 | 36 in fact, thing is subtle here.
The two equations are:
(1) H^2+k/a^2=\rho
(2) \ddot{a}/a=-(\rho+3P).
static universe requires that a=constant and H=0. so, \rho=constant.
so \rho_matter=0 and \rho_lambda=k/a^2;
but then, (2) \ddot{a}/a=-2 \rho_lambda.
so the only static solution is \rho_matter=0 and \rho_lambda=0, namely,
an empty universe.
so, I am confused that how can Einstein build a static universe.
在 fisherdad (渔父) 的大作中提到: 】 |
|
c**********g
发帖数: 222 | 37
>>> 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 |
|
r*f
发帖数: 731 | 38
RIght, right, right.
My problem is just how to express -div(\rho*u) in terms of
\rho=\rho(a,t) and x=x(a,t) where a is the initial spatial
coordinate. |
|
r*f
发帖数: 731 | 39 The problem is how to get grad(\rho) in term of \rho=\rho(a,t), and
x=x(a,t). :-) |
|
c*w
发帖数: 4736 | 40 What I meant by m=0 is that the rho of the gas inside the
bubble is 0.
For example, the rho of water is 1000kg/m^3, and rho of
H2 is about 0.1kg/m^3, then the acceleration will be:
1000*V*g/(.1*V)=10000g~100000(m/s^2)
That is absolutely rediculous. |
|
h****a
发帖数: 234 | 41 use "retarded potential" ya...
phi(r,t)=1/(4pi) \int(d3r' \rho(r',t-|r-r'|/c) / |r-r'|)
A(r,t)=1/(4pi) \int (d3r' \j(r',t-|r-r'|/c)/|r-r'|)
where phi(r,t) and A(r,t) are the scalar and vector
potentials
(in Lorentz gauge), and
\rho(r,t) and j(r,t) are the charge and current densities,
for point particle
\rho(r,t)=q delta(r-R(t))
j(r,t)= q v(t)delta(r-R(t))
where R(t) is the trajectory of the electron.
Readily derivable from the above. |
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b****e
发帖数: 460 | 42 我不知道算没算错，但是结果很惊人
首先
[/dP=\rho (r)g(r)dr/]
($\rho (r)$是密度，P是压强，g(r)是重力加速度，r是距地心的距离)这是Pascal定理
的变形，原因是重力加速度变了，同时空气密度也变了。
重力加速度很好算
[/g(r)=\frac{GMr}{R^2}=g_0\frac{r}{R}/]
其中$g_0$是地球表面的重力加速度，$R$是地球半径。
现在假设空气是理想气体，4/5是氮气14，1/5是氧气16，所以估算出空气的摩尔密度是
14.4g/mol。
根据理想气体状态方程$PV=NkT$(V是气体体积，k是Boltzmann常数，$k=1.38\times 10
^{-23}J/K$，T是温度)得知
[/\rho=\frac{m_0P}{N_AkT}/]
(其中$m_0$是摩尔质量，$N_A$是啊佛家的罗常数)
所以
[/dP=\frac{m_0g_0rP}{N_AkTR}dr/]
于是
[/\frac{dP}{P}=\frac{m_0g_0}{N_AkTR}rdr/]
做积分，左边从$P |
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g******s
发帖数: 733 | 43 在网上查了一下，Stokes' Law是用来求微小球体在连续粘滞流体中的阻力用的，公式为F=6pi mu R V，其中mu为粘度viscosity 1.8e-5(Newton second meter^-2)，R为球体半径，V为球体速度。但是看到一篇Science 1966年vol. 161 pp.1322的一篇文章，用Stokes' Law来求悬臂梁在连续粘滞流体中振动的阻力。
文章是这么说的，Using Stokes’ law for the damping force per unit area, we obtain … 求出来的转动阻尼系数(单位Newton*second*meter/rad) 为c=1.03*24 mu M/rho/d/w，其中mu为粘度viscosity 1.8e-5，M为质量惯性矩形 kg m^2，rho为悬臂梁材料密度，d和w分别为悬臂梁厚度和宽度。对悬臂梁，因为M=rho*d*w*L^3/12，c可简化为c=1.03*2 mu L^3。(或c=1.03*2 mu d L^2)。
哪位前辈能帮忙说说是怎么从Stokes’ Law推导出阻尼系数公式的？
先 |
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a*****n
发帖数: 5158 | 44 有A, B两个数，每个都有error(是one sigma normal distribution的）
并且有一个error coefficient (rho)
我想random 取出一组数(即A and B)符合上述要求的，该怎么处理?
如果rho=0的话，好处理，如果rho不是0的话，该怎么办?
thanks |
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m********9
发帖数: 208 | 45 You can perform the hypothesis test. Let rho be the population correlation
coefficient.You are interested in testing
H0 rho = 0 vs Ha rho neq 0
then check the absolute value of the sample correlation coefficient (for
short, |r|). If |r| > r_critical, then reject the null, which implies the
linear relation exists between x and y. Here the r_critical can be found via
http://www.gifted.uconn.edu/siegle/research/correlation/corrchrt.htm.
If your sample size is really large, then you can use the norm... 阅读全帖 |
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k*******a
发帖数: 772 | 46 因为conditional on xt
所以都写成 xt和 e 的项， 把xt 当constnt就可以
比如第二个 (rho*x_(t) + rho*e_(ti+1) + e_(t+2))所以variance就是 (rho^2 +1)*
sigma |
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z******n
发帖数: 397 | 47 我想我的看法不大受重视。所以构造了一个数值例子。为了使得结果能够重复，我固定
了随机数种子
set.seed(2)
library("mvtnorm")
n<-100
rho<-.9
bet<-c(.1,.1)
sigma<-matrix(c(1, rho, rho, 1), ncol=2)
x<-rmvnorm(n, sigma=sigma)
e<-rnorm(n,sd=.8)
y<-x%*%bet+e
data<-data.frame(y, x)
colnames(data)<-c("y", "x1", "x2")
mdl0<-lm(y~1, data=data)
mdl1<-lm(y~x1,data=data)
mdl2<-lm(y~x2,data=data)
mdl<-lm(y~x1+x2, data=data)
> anova(mdl0, mdl1, test="Chisq")[2, "Pr(>Chi)"]
[1] 0.03725746
> anova(mdl0, mdl2, test="Chisq")[2, "Pr(>Chi)"]
[1] 0.03311402
> anov... 阅读全帖 |
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z******n
发帖数: 397 | 48 这种情况我在4楼有讨论，结论仍然是各种情况都可能发生。下面是对应于你提到的情
况的一个数值例子
set.seed(29)
library("mvtnorm")
n<-100
rho<--.9
bet<-c(.1,.1)
sigma<-matrix(c(1, rho, rho, 1), ncol=2)
x<-rmvnorm(n, sigma=sigma)
e<-rnorm(n,sd=.5)
y<-x%*%bet+e
data<-data.frame(y, x)
colnames(data)<-c("y", "x1", "x2")
pv.x1<-summary(mdl)$coefficients["x1", "Pr(>|t|)"]
pv.x2<-summary(mdl)$coefficients["x2", "Pr(>|t|)"]
pv.jnt<-anova(mdl0, mdl, test="Chisq")[2, "Pr(>Chi)"]
> c(pv.x1, pv.x2, pv.jnt)
[1] 0.03195767 ... 阅读全帖 |
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c*******e
发帖数: 150 | 49 想请教一下版上的各位大牛们，如果
Linear Regression中Noise Term是一个AR(1) process，通常都有什么成熟的算法做
MLE 或者其它方法 fit ？
具体的说，模型可以表示为 Y(t) = X(t) \dot \beta + E(t),
X(t) 和 \beta 都是 K-维的向量，其它都是标量。
t = 1, 2, 3, ..., T 是手头的 sample，
但是和经典的 Linear Regression 不同，E(t) 不是 i.i.d. 的高斯白噪音，可以假定
E(t) 服从一下 model:
E(t) = \rho * E(t-1) + \sigma * Z(t)
\rho 和 \sigma 是 unknown parameter，Z(t) 可以认为是高斯白噪音。
所以全部的 parameters 包括 向量 \beta 和标量 \sigma, \rho
最好还是 maximum-likelihood 的方法，这样我可以保留后面做 log-Likelihood
Ratio
Test 的可行性，以便于 做 model comparison/s... 阅读全帖 |
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c*******e
发帖数: 150 | 50 DataSciences 版有一位朋友指出在 rho 给定情况下优化 beta 很简单（分别滤波X(t)
和Y(t) 序列，对应的噪音项就是白化了的\sigma * Z(t) 序列，i.i.d.了）；而在
beta 给定的情况下优化 rho 也很简单，所以我们可以 iteratively 地数值解出一个
fixed point [rho_star, beta_star]。（在 rho 和 beta 都给定的情况下解出
小sigma 的 MLE很容易 ）
从操作性上来看这样的算法非常地理想，实际编程试验后发现收敛也很快很稳定。最后
还是有点小好奇，有没有什么理论的结论保证这样的 fixed-point
[rho_star, beta_star] 也是globally maximizes the likelihood function 呢？不好
意思对于这个“MLE”的 optimality 比较在乎，因为下一步需要基于 MLE 的性质做
LLR test 等等，所以如果全局最优性有理论的保证基础可以很坚实一些，觉觉睡得更香
一些 *^_^*
有找 |
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