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s*****V
发帖数: 21731 | 2 标准模型简史
- 作者：Steven Weinberg 译者：卢昌海 -
译者序： 2003 年， 物理学家们相聚在欧洲核子中心 (CERN) 纪念中性流发现三十周
年及 W 与 Z 粒子发现二十周年。 著名理论物理学家 Steven Weinberg 在纪念会上作
了题为 "The Making of the Standard Model" 的演讲。 这一演讲经整理后发表于
Eur. Phys. J. C34 5-13, 2004， 本文便是据此而译。 Weinberg 是电弱统一理论的
提出者之一， 亲身参与了标准模型诞生过程中一系列激动人心的进展， 因此他的这篇
文章具有很大的参考价值。 在翻译本文的过程中恰逢今年的 Nobel 物理学奖颁给了美
国物理学家 D. J. Gross, H. D. Politzer 和 F. Wilczek， 以表彰他们对 “发现强
相互作用理论中的渐进自由” 所做出的贡献。 这是物理学家因标准模型领域中的工作
又一次获奖。 标准模型虽已不再 fancy， 却枝繁叶茂、 沉稳如昔。 最后提醒读者一
下， 原文所附的参考文献实在太多， 为了节... 阅读全帖 |
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e***n
发帖数: 10 | 3 假設無窮遠處一個人把線向上提一小段距離dr。。。 |
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z****i
发帖数: 406 | 5 还挺严谨的吧。
最后一步用的是Wald's equation.
a Markov approach:
Let the desired expectation be E, then
E = p*1 + (1-p)*(2E+1)
of
movement |
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b******e
发帖数: 118 | 6 如果用Wald's equation， 为什么E[Sn]＝1呢？我的理解是如果最终head和tail的数目
一样，E[Sn]不应该是0吗？但现在有已知条件，第一次flip是tail，这个怎么考虑进去？ |
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z****i
发帖数: 406 | 7 x_i = 1 with probability p;
= -1 with probability 1-p.
so E[x_i] = 2p-1 for each i, i.e, each flip.
The process stops when \sum x_i = 1.
S_n = \sum_i^n x_i
where n is the stopping time, i.e., # of flips when S_n = 1.
We know at stopping time n, S_n = 1.
apply Wald's equation,
1=S_n = E(S_n) = E(n) *E(x_i)
we have E(n) = 1/(2p-1).
one |
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p******e
发帖数: 756 | 9 可以解释一下怎么用wald identity么
我怎么也绕不开S_\tau的分布或者期望.thx |
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f*******d
发帖数: 339 | 11
suggest:
R. Wald: general relativity |
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o******6
发帖数: 538 | 12 Ha should be "at least one coefficent is not equal to 0"
You can use wald test or likelihood ratio test; |
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z********n
发帖数: 710 | 13 no one knows?
Please help out leh! |
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q**j
发帖数: 10612 | 14 from my distant memory the test depends upon the null hypothesis and the
functional form, ie, the variance could vary if the null varies, that is the
reason that many people do not like it much. not sure if it is ture for
your case. |
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s*****h
发帖数: 232 | 15 could be small sample size.
if you have other variables in the model and this variable is correlated
with another variable. it also could cause the smaller standard error and
wide confidence intervals. |
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z********n
发帖数: 710 | 16 To qqzi,
Thank you for your reply. I think what you talked about reminds me of
something similar. I really appreciate that.
Thanks again. |
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t**i
发帖数: 688 | 17 Why smaller stderr would lead to wider conf. int.? I think it should be the
other way around. |
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s*****h
发帖数: 232 | 18 sorry, should be big standard error.
the |
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S*x
发帖数: 705 | 19 use wald chi-square
factor |
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D******n
发帖数: 2836 | 20 no matter how u test, they are different! |
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q******n
发帖数: 272 | 21 You are abusing Statistics. Statistics is based on random sample. |
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g**r
发帖数: 425 | 22 您就5个数，只能谈分布，就看看variance有多大就行了，或者把这个variance除
以4开方，看看这个STD有多小就行了
您要是有5 组数，那个TEST才用的上。 |
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g**r
发帖数: 425 | 24 你可以换个角度这样看：
你把假设换成：每一组的均值都和总均值相等。
这样在套你GOOGLE来的公式的时候，那个NULL就是0了，而你的那个HAT，就是和均值的
DIFFERENCE。
你WIKI出来的那个FORMULA，只对比较两组数据管用，你再GOOGLE出来个对多组数据的
（矩阵化了的），比较四组（如果四组成了，第五组不用比较，因为你的均值和四组数
据就包含了所有信息）就行了。
这个最普通的REGRESSION的软件做ANOVA的时候应该包含这个结果的，没人用手算的。 |
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m********g
发帖数: 46 | 27 You are really helpful, thanks! |
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t******l
发帖数: 32 | 28 Yes, that is Wald test.
你也可以代入想测的的值，那样就是score test.
Under the null, the two tests are asymptotically equivalent. |
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q**j
发帖数: 10612 | 29 that is why wald test is not very well respected. |
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h******e
发帖数: 1791 | 30 用stepwise的方法选择变量，得到两个A和B,都是continuous的，两者之间的
interaction被排除了，如下：
Parameter DF Estimate SE Wald Pr>ChiSq
A 1 0.1297 0.0435 8.9076 0.0028
B 1 -0.3586 0.1732 4.2843 0.0385
可见两者的作用是相反的。
但如果单独用A或B做logistic model，得到：
A的estimate是0.0565，B的estimate是0.1665，两者的作用是同向的。如果单独对二者
做linear regression的话，发现存在正相关性。
请问该如何解释logistic regression的结果。谢谢。 |
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h******e
发帖数: 1791 | 31 前辈，有这么一个logistic regression的问题，请帮帮忙：
In this logistic regression (real clinical data) with 22 severe diseased and
25 less diseased, I used stepwise method to select covariates with SAS
procedure logistic. The objective is to examine variables which could cause
disease progression. I finally got two: A and B. Both are continuous. Their
interaction A*B was ruled out by the SAS procedure.
Parameter DF Estimate SE Wald Pr>ChiSq
A 1 0.1297 0.0435 |
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j*****e
发帖数: 182 | 32 I forgot to mention that some of the p-values associated with the intercept
estimates in your SAS output are useful, too. These tests are Wald tests.
If you want to perform likelihood ratio test, you have to combine categories
, refit the model to the data, and pull out the -2log likelihood values by
yourself. |
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f*******r
发帖数: 383 | 33 Fisher, Pearson, Neyman
Tukey, Rao, Wald, Le Cam, Stein
Efron, Huber, Donoho, ... |
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n******0
发帖数: 298 | 34 The outcome is Cancer vs. Normal
The explanatory varibles are four protomics peaks
The following is the results:
Analysis of Maximum Likelihood Estimates
Standard Wald
Parameter DF Estimate Error Chi-Square Pr >ChiSq
Intercept 1 -1.7465 1.4500 1.4509 0.2284
c14 1 -2.4761 0.8573 8.3430 0.0039
c25 1 7.7777 2.4119 10.3987 0.0013
c46 |
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r*******e
发帖数: 510 | 35 算Relative risk (RR)
不用normal approximation, 算WALD CI （这个应该是BASED ON CHI-SQUARE的吧）。
哪位同学解释下， 或给点参考资料。
谢谢 |
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a*****n
发帖数: 8 | 36 我怎麽覺得Wald interval 不是 BASED ON CHI-SQUARE的吧 ？ |
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j*****e
发帖数: 182 | 37 First, the square of a standard normal r.v. is chi-squared with df=1.
Second, in SAS, CI of relative risk is constructed based on the asymptotic
normality of log(estimated RR).
It is impossible to have a Wald CI without some sort of normality assumption
. This is a concept issue. |
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g******n
发帖数: 339 | 38 Support this.
WALD/SCORE/likelhood ratio tests are asymptotically equivalent, all are
based on asymptotically normality of the test statistics.
assumption |
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f***a
发帖数: 329 | 39 我觉得有两个概念很容易弄混，一个是model的goodness of fit，一个是model
selection。
前面一个是判断某一个model和data是不是fit很好，后一个判断一些candidate models
中哪一个和data最吻合。用model selection方法的时候，当你从candidate models找
出最吻合的那个model时，不能直接就判断那个model的goodness of fit就是很好的。
有可能所有candidate models都差，矮子中拔高个的结果而已。
有很多常用的方法，不过最好具体model、具体数据类型具体分析，找到最合适的方法
。譬如AIC就有很多变种，对应不同类型的数据结构。我列举一些希望对楼主有用。
General model selection methods:
1) common tests: wald Z, chi-square, t, F etc
2) likelihood ratio test and its variants
3) information criteria methods: AIC,BIC, a |
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h******3
发帖数: 190 | 40 I am confused about the two tests.
Does the degree of freedom of both the tests follow the following?
degree of freedom = number of unknown parameters in H1- number of unknown
parameters in H0 |
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d********t
发帖数: 837 | 41 It's assuming asymptotic normal distribution, why do you need degree of
freedom? |
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h******3
发帖数: 190 | 42 It can be more than 1 dimension. The statistic is chi-square. |
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d********t
发帖数: 837 | 43 It's still related to the multivariate normal, so the d.f. equals the
dimension of your multivariate normal. |
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h******3
发帖数: 190 | 45
And the degree of freedom is?
You need to determine the dimension of multivariate normal.
I don't get your logic. |
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d******g
发帖数: 130 | 46 I think it should be the number of parameters in your model. |
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y*****n
发帖数: 5016 | 47 if you have eminer, then you can use the "interactive grouping node". it can
not only bin each variable into woe, but also calculate the information
value and Gini for each variable. you can prescreen variables there. some
may argue that variables with low IV may still be picked up in the stepwise
regression. However, your bosses may want to see the "stand alone"
relationship between each model attribute and the target. therefore, you
want to make sure that each candidate variables has high IV b... 阅读全帖 |
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n******r
发帖数: 1247 | 48 同意，所以没有说具体方法，因为方法很多，即使同一个公司，方法也不断的在改进。
目前做project我们用treenet从近千个选到top50，
然后做empirical analysis，transformation， regression选变量，加interaction，
这几步反复。
以前有先对变量cluster，主成份，快速选取的。最复杂最慢但效果最好的有在
logistic regression里，对somer's D做敏感度分析一个一个选的。treenet的好处
是data preparation方便，linear nonlinear关系都能找到，top 50变量，加上
interaction一般模型可以达到很好的效果。
再具体也不说了，这个东西是靠日常在实际运用中积累经验的，不是广靠听就能会的。
对于面试被问到如何从上千个变量里选，我个人的观点是，除非你有过实际操作的经验
，否则也只能说听来的皮毛，不要太指望能和有经验的面试者做深入讨论。
对于楼主被问到的transformation的问题，这个对即使没有大型数据处理经验的fresh
graduate也是fair quest... 阅读全帖 |
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S********a
发帖数: 359 | 49 > unadj.lrm
结果如下
Coef S.E. Wald Z P
Intercept -3.60925 0.132424 -27.26 0.0000
pmavg 0.02205 0.009483 2.33 0.0201
我想取pmavg 的standard error （i.e.,0.009483）
怎么样写R code呢？
> unadj.lrm$SE 好像不对呀 |
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S********a
发帖数: 359 | 50 完整的结果是这样的， 如何取SE呢？谢谢大家
> unadj.lrm
Logistic Regression Model
lrm(formula = combpreecl ~ pmavg, data = mydata2)
Frequencies of Responses
0 1
118691 4360
Obs Max Deriv Model L.R. d.f. P C 123051
7e-10 5.41 1 0.02 0.504
Dxy Gamma Tau-a R2 Brier
0.008 0.015 0.001 0 0.034
Coef S.E. Wald Z P
Intercept -3.60925 0.132424 -27.26 0.0000
pmavg 0.02205 0.009483 ... 阅读全帖 |
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