t**8
发帖数: 4527 | 1 【 以下文字转载自 Military 讨论区 】
发信人: ts78 (ts), 信区: Military
标 题: Re: FDA grants ‘emergency use’ coronavirus test that can
发信站: BBS 未名空间站 (Sat Mar 21 15:39:40 2020, 美东)
Diagnostics company Cepheid on Saturday said it has received emergency
authorization from the U.S. Food and Drug Administration to use its rapid
molecular test for point-of-care patients that can detect the virus that
causes COVID-19 in 45 minutes.
This is the first coronavirus test that can be conducted entirely at the
point-of-care f... 阅读全帖 |
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a****y
发帖数: 1035 | 2 【 以下文字转载自 Biology 讨论区 】
【 原文由 aubrey 所发表 】
Published online May 1, 2003: http://www.sciencemag.org/feature/data/sars/
Characterization of a Novel Coronavirus Associated with Severe Acute
Respiratory Syndrome
In March 2003, a novel coronavirus (SARS-CoV) was discovered in association
with cases of severe acute respiratory syndrome (SARS). The sequence of the
complete genome of SARS-CoV was determined, and the initial characterization
of the viral genome is presented in this report. The genome of |
|
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d******e
发帖数: 2265 | 4 YDKN , Finance , Major Banks ==> 2014-11-20 00:00:00:JONES STEVEN W as
Officer, Purchase 2000 shares at $18.70, total value is 37,400
X , Basic Industries , Steel/Iron Ore ==> 2014-11-10 00:00:00:BURRITT DAVID
B as Officer, Purchase 26489 shares at $37.36 - $37.75, total value is 995,
000
X , Basic Industries , Steel/Iron Ore ==> 2014-11-11 00:00:00:FILHO MARIO
LONGHI as Officer, Purchase 28362 shares at $35.21 - $35.25, total value is
999,000
YUM , Consumer Services , Restaurants ==> 2014-12-16... 阅读全帖 |
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z********4
发帖数: 533 | 5 准备eb1a中，但是有好几个引用下不来，请大家帮帮忙，list在下面，请发到我的邮箱
z***********[email protected] 先谢谢各位了！
1.Antigenic diversity and seroprevalences of Torque teno viruses in children
and adults by ORF2-based immunoassays
J Gen Virol February 2013 vol. 94 no. Pt 2 409-417
2.Tunicamycin induces resistance to camptothecin and etoposide in human
hepatocellular carcinoma cells: role of cell-cycle arrest and GRP78
JL Hsu, PC Chiang, JH Guh - Naunyn-Schmiedeberg's archives of …, 2009 -
Springer
3.p53 antagonizes the unfolded prote... 阅读全帖 |
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m******n
发帖数: 194 | 6 个人觉得低价HDB的COV会略降，高价HDB的COV会降的比较多，condo（各种档次）价格
都会降
总体价格1年内降10%，如果加息，会降更多。
留着1年后再看，错了正常，蒙对了也没什么。 |
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m*****n
发帖数: 511 | 7 不知道是我看的area太火还是咋回事，cov降这种事情从来也砸不到我头上。。。
jurong east toh guan area，昨天刚看的，５房估价５１０，０００，cov还要４万。
。。 |
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l*****8
发帖数: 16949 | 8 那个QB感觉不错的，就是int多了点。不过才二年级。
DR虽然TD多，但不管1-cov 2-cov,打定主意就往人堆里扔。完全是靠天吃饭。也就是ND
的角卫太差，老被吃死。 |
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Y******e
发帖数: 20256 | 9 市长差点心脏病发。我遗憾的是hoke没有play前叫time out，这是他的弱点。对进攻不
熟悉。我看到双方的阵势，就知道这个球不太行。看了球很久了，有时候感觉就出来，
不知道为什么。但是，一看两边阵势，我就可以看出这个球pass run都不行。这种2pt
cov的play太过简单了，连一点motion都没有，就靠QB扔球，成功率当然低，这就是OC
水平，看别人鸭子，2pt cov变化无穷。 |
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g****g
发帖数: 1828 | 10 标准差（Standard Deviation），在概率统计中最常使用作为统计分布程度（
statistical dispersion）上的测量。标准差定义为方差的算术平方根，反映组内个体
间的离散程度。测量到分布程度的结果，原则上具有两种性质：
1. 为非负数值，
2. 与测量资料具有相同单位。
一个总量的标准差或一个随机变量的标准差，及一个子集合样品数的标准差之间，有所
差别。其公式如下所列。
标准差的观念是由卡尔·皮尔逊 (Karl Pearson)引入到统计中。
目录
[隐藏]
* 1 阐述及应用
* 2 标准差的定义及简易计算公式
o 2.1 标准计算公式
o 2.2 简化计算公式
o 2.3 随机变量的标准差计算公式
o 2.4 样本标准差
o 2.5 连续随机变量的标准差计算公式
o 2.6 标准差的性质
* 3 范例
* 4 正态分布的规则
* 5 标准差与平均值之间的关系
* 6 几何学解释
... 阅读全帖 |
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e*******c
发帖数: 1479 | 11 安徽大学生命科学学院2011级硕士研究生胡亚伟，在中科院微生物所高福教授指导下，
一直从事人高致病性病毒的研究。2013年7月7日，以她为并列第一作者的论文《高致病
性新型人冠状病毒MERS-CoV与其细胞受体CD26结合的分子基础》，发表在世界自然科学
最高级别的学术期刊Nature上。该文章阐述了新型人冠状病毒MERS-CoV与其细胞受体
CD26结合的分子基础，为人类了解冠状病毒的分子结构，找到防治该种病毒感染的手段
做出了重要贡献。 |
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l*******3
发帖数: 1074 | 12 我做生物信息，CPU不重要，内存和硬盘是瓶颈，尤其是内存！
如果你用velvet做de novo assembly，请计算一下你需要多少内存，比如：
http://denovoutils.appspot.com/velvetmem
我的机器16gb，昨天用20 cov的mate-pair reads做hg_chr14(107m bp)的assembly，内
存就已经不够了（略大于16gb），无奈只能用15 cov的做。
现在单条8gb的太贵，建议用x58的板子，上24gb内存，至于cpu，上一代i7就已经可以
了，比如920/960等等。
ssd当然可以考虑，不过不是必须。 |
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n******7
发帖数: 12463 | 13 这个图似乎有些问题
1.microarray data 为啥也有RPKM值？
2.很明显microarray data 的range很小，不是很可比
3. CoV是SD除以mean，mean值小的时候就容易特别大了。一个低表达gene，两次测到1
个reads，两次测到4个reads，reads的CoV是0.69，so what？ （RPKM同理） |
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t******n
发帖数: 6242 | 14 matlab进行矩阵运算很快，尽量避免loop循环。可是下面这个问题有什么好办法解决呢？
对很多位置坐标进行下面的运算
sigma = [x y] * cov^-1 * [x y]'
cov是2x2矩阵
是不是只能用for语句每次进行一个坐标的运算？有没有办法对所有的坐标同时进行以
提高速度？谢谢。 |
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G****n
发帖数: 145 | 15 100+ citation articles of Qian Yingyi
Regional decentralization and fiscal incentives: Federalism, Chinese
stylestanford.edu 中的 [PDF]
Carnegie Mellon FulltextH Jin, Y Qian… - Journal of Public Economics, 2005
- Elsevier
Aligning the interests of local governments with market development is an
important issue for
developing and transition economies. Using a panel data set from China, we
investigate the
relationship between provincial government's fiscal incentives and
provincial market ...
被引用次数：... 阅读全帖 |
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D******n
发帖数: 2965 | 16 解个线性方程就性了。
1. Let X = (W, X_{sub}). 因为正态分布，对于每个W_j, 都可以写成一个关于X_sub
的线性方程加上一个独立的正态分布变量， i.e.,
W = b+ B * X_{sub} + V
where b is (n-m)*1 vector, B is (n-m)*m matrix, V independent of X_sub and
be joint normal with
zero mean, variance sigma_V -- (n-m)*(n-m) matrix.
use variance and covariance info of X to solve B, and sigma_V: covariance
between W and X_sub for B, and variance of W for sigma_V (the latter you
might not need given 2.)
(1) B* Sigma_{X_sub} = Cov (W, X_sub)
(2) B*Sigma_{X_sub} * B... 阅读全帖 |
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ET
发帖数: 10701 | 17 第一，这是定性描述, 还是没法着手。到底有啥计算办法没？
比如，我给了signal frequency, 那么sample/hold的这个RC的bw = 1/RC
R = Ron
显然如果signal frequency 高，RC 都必须小，或者2者中一个小。R小就要W/L 大； 大
的W/L就会带来precision的问题，
(1)clock feedthrough , related to Cov=w*Lov*Cov
delta v = Vck WCov/(WCov+Ch)
(2)charge injection, delta V = wl*cox(vdd-vin-vth)/2*C,
这里就给了w & c的limitation.
(3) sqrt(kt/c)
写到这，我想起来了点了。
hold cap 应该由给定的dynamic range spec.来确定 - 这样得出一个C.
由C在来计算Ron, 然后确定switch size.
剩下的就是优化。。 |
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a***n
发帖数: 328 | 18 【 以下文字转载自 Statistics 讨论区 】
发信人: akhan (neopho), 信区: Statistics
标 题: 请教variance
发信站: BBS 未名空间站 (Thu Mar 1 17:37:29 2007)
有两组随机数X,Y
在什么情况下var(X+Y)=var(X-Y)?
是cov(X,Y)=0就可以么？如果是的话，怎样才能使cov(X,Y)=0？
非常感谢！ |
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h***i
发帖数: 3844 | 19 What is true is that if the random variable pair (X,Y) follows the bivariate
normal distribution, and Cov(X,Y) = 0, then X and Y must be independent.
But what is not true is that if each of X and Y is normally distributed, and
Cov(X,Y) = 0, then X and Y must be independent.
two random variables are each normally distributed, that does not
necessarily mean that they jointly follow a bivariate normal distribution
bell |
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s******h
发帖数: 539 | 20 Give any two INDEPENDENT r.v's X, Z such that
X~N(0,1), P(Z=1)=P(Z=-1)=1/2
Let Y=Z*X
Then
X,Y~N(0,1)
Cov(X,Y)=0
Because
a) P(Y<=y)=P(-X<=y|Z=-1)*P(Z=-1)+P(X<=y|Z=1)*P(Z=1)
=1/2*(P(X>=-y)+P(X<=y))
=P(X<=y)
b)
E(XY)=E[E(XY|Z)]=E[ZE(X^2|Z)]=E[ZE(X^2)]=EZ*E(X^2)=0
EX=0=EY=0
So Cov(X,Y)=0.
But
1) |X|=|Y|=> X and Y are not independent.
2) Let U=X+Y, then U is not Normally distributed.
P(U<=u)=P(U<=u|Z=-1)*P(Z=-1)+P(U<=u|Z=1)P(Z=1)
=1/2*I{u>=0}+1/2*P(X<=u/2) |
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g******n
发帖数: 339 | 21 use chlosky decomposition s.t. cov=d’d, then generate two independent x1~N(
0,1), x2~N(0,1), y=d'(x1,x1)~BVN(0,cov). |
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a********l
发帖数: 55 | 22 1. X and Y are uncorrelated because Cov(X,Y)=0.
2. If X is uniformly distributed on [0,1], then Cov(X,Y)>0 and hence X and Y
are not uncorrelated. |
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q******1
发帖数: 220 | 23 不好意思,漏打了个条件,the correlation coefficient of Y, Z is 0.8.
刚刚又考虑的一下，想出来了。。。
从correlation coefficient的公式说起r(X,Y)=cov(X,Y)/sqrt(Var(X),Var(Y)),
where cov(X,Y)=E(XY)-EXEY and X, Y are r.v. on (S, P). So r(X,Y)=cosA, A is
the angle between the vector X-EX and Y-EY in L^2(P).
Therefore, original question becomes a simple question in linear algebra. |
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m******t
发帖数: 4077 | 24 no, it is linear estimator then. For Gaussian R.V., the best estimator is
the linear one. The formula is
E[X|Y] = E[X] + Cov(X, Y)Cov(Y, Y)^{-1}(Y -E[Y]). |
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f*********1
发帖数: 117 | 25 Agree.
no, it is linear estimator then. For Gaussian R.V., the best estimator is
the linear one. The formula is
E[X|Y] = E[X] + Cov(X, Y)Cov(Y, Y)^{-1}(Y -E[Y]). |
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m*****e
发帖数: 692 | 26 It's not listed in ecnometrics book but it's quite similar to the univariate
regression.
y = aX1 + bX2 + error
cov(y,X1) = aVar(X1) + bcov(X1,X2)
cov(y,X2) = acov(X1,X2) + bVar(X2) |
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L*****e
发帖数: 169 | 27 P=aS1+bS2,
a和b是数量，是未知量，S1和S2是股价,是随机变量。
Var(P)=a^2 Var(S1)+b^2 Var(S2) +2ab Cov (S1,S2)
R=Cov(S1,S2)/sqrt(Var(S1)Var(S2)),
Var(S1),Var(S2)和R已知，剩下就是求Var(P)最小值问题。
a E(S1)+ b E(S2)= E (P)是约束条件，带进去画一个curve求最小值。
不知道题中收益率相等啥意思，我是业余级的。 |
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i**w
发帖数: 71 | 28 thanks for sharing!!
w1^2*var(x1) + w2^2*var(x2) + 2*w1*w2*cov(x1,x2)
if x1 and x2 independent, cov=0
degenerate roots
conbination of e^(-x) and x*e^(-x)
exp(mu - .5*sigma^2), where mu-mean, sigma-standard deviation
//x follows log-normal
of
transformation of random variable. since pdf is given, x is continuous: f_Y=
f_X(f_inv(y))/f'
where f_X is the pdf of X, f_inv = inverse of f,f' = derivative of f
//otherwise for discrete
P(Y=y)=P(X=f_inv(y))
这个板上见过很多次
有没有什么比较decent的回答??抛砖引玉一下 |
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l******8
发帖数: 32 | 29 I Concur. I use E[XY] = Cov(X,Y) and Cov(W_s,W_t) = min(s,t). |
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o****e
发帖数: 80 | 30 周的书第100页 correlation of max and min
X1 and X2 be IID R.V.
Y=min(X1,X2), Z=max(X1,X2)
我的问题是，
Y,Z就是X1, X2中各取一个， 为什么 cov(Y,Z) != cov(X1,X2),前者是1/36，后者是0 |
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d***y
发帖数: 65 | 31 cov(Y,Z)=E(YZ)-E(Y)E(Z)
=E(X1X2)-E(Y)E(Z)
=E(X1)E(X2)-E(Y)E(Z)
We have E(X1)=E(X2)=1/2
E(Y)=1/3
E(Z)=2/3
So, cov(Y,Z)=1/36 |
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w**********y
发帖数: 1691 | 32 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 |
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t*******y
发帖数: 637 | 33 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 |
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z****g
发帖数: 1978 | 34 shrinkage method: linear combination of PCA/Factor based cov and sample
based cov |
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s******s
发帖数: 63 | 35 现在算的cov(Y1^2,Y2^2),不是cov(Y1,Y2) |
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R**T
发帖数: 784 | 36 多谢分享！
Var(A+B) = Var(A) + Var(B) + 2*Cov(A,B)
Cov(A,B) = correlation*sqrt(Var(A)*Var(B)) |
|
H****E
发帖数: 254 | 37
基本上就是用sum of expectation=expectation of sum展开，再代入。。。
但是我卡在一堆cov(x1,x2), cov(y1,y2)。。。 |
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H****E
发帖数: 254 | 38
基本上就是用sum of expectation=expectation of sum展开，再代入。。。
但是我卡在一堆cov(x1,x2), cov(y1,y2)。。。 |
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j****p
发帖数: 86 | 39 mission impossible
如果是求cov的范围，还make sense，只需用var cov matrix positive definite |
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w*********o
发帖数: 434 | 40 拍屁股想想也知道不可能
如果两个随即变量的分布都不知道，只知道mean,std，
cov 可以是任何东西
比如随即变量完全独立，或者完全线性相关
这个啥狗屁面试题啊，如果能知道mean,std推得cov，统计学家吃饱了撑的在搞个新概
念出来啊 |
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d********t
发帖数: 9628 | 41 你抹平return算cov根本就是错误的。我的例子就说明两个本来高度相关的strategy用
你的方法就变得不相关了。cov只是衡量相关度的条件之一。 |
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w**********y
发帖数: 1691 | 42 Without losing generalization, assuming std of x1, x2, y are all 1.
Then you can think in this way,
y = b1*x1 + b2*x2 + sqrt(1 - b1^2 - b2^2)*x3,
where x1, x2, x3 are all i.i.d
cov(y,yhat) = b1^2 + b2^2
var(y) = 1
var(yhat) = b1^2 + b2^2
Thus, cor(y,yat) = cov(y,yhat)/sqrt(var(yhat)) = sqrt(b1^2+b2^2) |
|
c*****d
发帖数: 186 | 43 So cov(Wu, Xu) = a Cov(Xu, Xu)=aVar(Xu), get a.
Thank you very much!
linear |
|
c*****d
发帖数: 186 | 44 So cov(Wu, Xu) = a Cov(Xu, Xu)=aVar(Xu), get a.
Thank you very much!
linear |
|
s******h
发帖数: 539 | 45 直接算，然后利用independence和对称性。
Cov((X2 - X1)^2, (X2 - X3)^2)
= Cov(X2^2 - 2X2X1, X2^2 - 2X2X3)
= Var(X2^2) - 2Cov(X2^2, X2X3) - 2Cov(X2^2, X1X2) + 4Cov(X1X2, X3X2)
= Var(X2^2)
|
|
l*********s
发帖数: 5409 | 46 you don't have to read code to understand the mechanism. For instance, simple
linear regression formula is beta=cov(x,y)/cov(x,x). Both the numerator and
denominator are accumulative in nature and can be computed progressively, just like "
mean".
To
update glm, you can throw away old data and only keep those running totals of sum of squares.
That is why you can chop data into pieces small enough to fit into the RAM.
This |
|
k*******a
发帖数: 772 | 47 inthat=ybar-xbar*slopehat
so cov(inthat,slopehat)=cov(ybar,slopehat)-xbar*var(slopehat)
可以证明第一项=0
因为 ybar=sigma(yi)/n
slopehat=sigma(ki*yi)
ki=(xi-xbar)/Sx^2
所以第一项~sigma((xi-xbar)*var(yi))
因为var(yi)都一样，而且sigma(xi-xbar)=0
所以第一项=0 |
|
b*****n
发帖数: 685 | 48 我数学不是很好，老是靠经验，试了试：
eigen(cov(matrix(rnorm(10), nrow = 2)))
eigen(cov(matrix(rnorm(10), ncol = 2)))
您数学牛，给解释一二？ |
|
z**********i
发帖数: 12276 | 49 我只保留了一个COVARIATE IN THE MODEL, 如果加上其他的,花的时间更长.
加RANDOM 之前,只要2分钟,但如果加了RANDOM,花几个小时或更久.
CODE:
proc nlmixed data=ami1_04Q1 tech=newrap qpoints=50;
parms b0=-2.53 b24=-0.055 rho=0.007 varRin=0.90 cov=-0.011
varRslp=0.0028 ;
eta=(b0+ (b24+Rslp)*nqt+Rin);
expeta = exp(eta);
p = expeta/(1+expeta);
N=totdenom; r=R_totnum;
A=p*(1-rho)/rho;
B=(1-p)*(1-rho)/rho;
loglike=(lgamma... 阅读全帖 |
|
z**********i
发帖数: 12276 | 50 这个就是BETA BINOMIAL加RANDOM EFFECTS,该怎么变一下呢? 多谢!
proc nlmixed data=ami1_04Q1 tech=newrap ;
parms b0=-2.53 b24=-0.055 rho=0.007 varRin=0.90 cov=-0.011
varRslp=0.0028 ;
eta=(b0+ (b24+Rslp)*nqt+Rin);
expeta = exp(eta);
p = expeta/(1+expeta);
N=totdenom; r=R_totnum;
A=p*(1-rho)/rho;
B=(1-p)*(1-rho)/rho;
loglike=(lgamma(n + 1)-lgamma(r + 1)-lgamma(n-r+1))
+lg... 阅读全帖 |
|
