q********e
发帖数: 1255 | 1 conjugate part, convergent
what, r u doing calculus homework? |
|
L*********s
发帖数: 3063 | 2 a_n=1/ (n[(n+1)^(1/2) + n^(1/2)]) < 1/(n^1.5)
the series converges |
|
m*********w
发帖数: 408 | 3 Can we say convergence is a numerical validation? I think the numerical
solution depends on the model (model is a simplified system of the real
world due to its assumptions), so the numerical solution may not be real
solution. |
|
d*******e
发帖数: 1649 | 4 你说的是两个东西。
一般意义上convergence指的是数值解收敛到解析解，在给定了方程和边界条件的情况
下。一旦模型给定，这个是独立于真实世界的。
validation指的是模型的准确性，亦即用某个模型去模拟真实世界得出的解是否准确。
但是这个也取决于人们的定义。比如有人说，我有某某方法可以解一个方程，解析解可
以算出来，用这个方法可以数值上收敛到解析解，所以是numeric validation，这样也
可以，不能叫错。 |
|
n***p
发帖数: 7668 | 5 Depends on how you define convergence. There is always a gap between
numerical simulation and numerical analysis.
approaches |
|
jl
发帖数: 398 | 6 Does the absolute convergence
sum |a_n| = K imply lim n a_n = 0?
It seems this is not right if |a_n| is not a monotone sequence.
Is there any conditions on those a_n where |a_n| > 1/n ?
多谢! |
|
B********e
发帖数: 10014 | 7 good example, thanks
based on it, it is not hard to see that
\lim sup {n a_n} could be infinity, for instance
a_n=1/sqrt(n), if n is k^4, k=1,2,...
a_n=0, otherwise
more generally {n |a_n|}
could contain subsequence converging to any positive number. |
|
x********9
发帖数: 31 | 8 ..You can't apply the martingale convergence since B.M is not uniformly
bounded in L_1. |
|
A***d
发帖数: 25 | 9
The following is an example:
Suppose Xn follows exponential distribution with df n/(n+1)exp{-nx/(n+1)}, X
follows exponential
distribution with df exp(-x). Further assume that Xn and X are independent.
Obviously, Xn converges in law to
X. However, by using the convolution formula, we can compute that P(|Xn-X|>=
e)->exp(-e)>0 for any e>0. |
|
B***y
发帖数: 83 | 10
If | b | > 1 then the series will be divergent.
If | b | < 1 then the series will be convergent.
The interesting case happens when |b| = 1, hmm... |
|
B***y
发帖数: 83 | 11
Notice that it is a second order iteration equation, thus first write it as a
vector equation:
Y_(n+1) = A(n) * Y_(n) ,
where Y_n = (a(2n-1), a(2n))^t, where "t" means "transpose".
Easy to check that A(n) asymptotically behaves like (b 0
0 b )
Thus if |b| > 1, then the series Y_n will diverge, if |b|< 1 then the series
will converge. |
|
r********t
发帖数: 41 | 12 Is there any result talking about the convergence of linear regression (or
likelihood based) with diverning number of predictors?
Say n observations yi, and xi.
the dimension p of xi diverges as n, say p=p(n).
however the response may depend only on the first 10 predictors.
Many thanks! |
|
S******y
发帖数: 1123 | 13 Hi. guys,
I am trying to write my own Stochastic Gradient Ascent for logistic
regression in R. But it seems that I am having convergence problem.
Am I doing anything wrong, or just the data is off?
Here is my code in R -
lbw <-
read.table("http://www.biostat.jhsph.edu/~ririzarr/Teaching/754/lbw.dat"
, header=TRUE)
attach(lbw)
lbw[1:2,]
low age lwt race smoke ptl ht ui ftv bwt
1 0 19 182 2 0 0 0 1 0 2523
2 0 33 155 3 0 0 0 0 3 2551
#-----R implementation of l |
|
s*****i
发帖数: 62 | 14 You mean almost sure converge? I can not type formula, so you check the
formula and you will know that Roughly speaking,
The set for A.S is that exist a fixed set that it has probability one.
for In Probability, the set can be changing and the limit of its probability
is one.
examples like walking on (0,1] by 1/2, 1/3 and go on, use indicator function
you get conv in P, but not A.S |
|
f*********e
发帖数: 45 | 15 Usually, when fitting mixed model , if we run across the un converge problem
,
what we should do ? what I can think about is
1.try different type of variance structure like CS ...
2, specifiy different estiamation method. (mle - reml)
3. reduce covariates.
What do you think about it ?
Thank you so much! |
|
s*r
发帖数: 2757 | 16 i did see complex models cause convergence problems. so 1 and 3 |
|
q**j
发帖数: 10612 | 17 the inability to converge is often a result of inherent model inconsistency,
which is more severe for complicated models. so the first thing to do is to
make sure that the system of equations you estimate is consistent within
itself. |
|
n******0
发帖数: 298 | 18 When there is a dummy explanatory variable X such that all observations
having one of the values of X (0 or 1) occur in censored observations, the
Cox regression model does not converge. How to deal with it? |
|
h******3
发帖数: 190 | 19 Is there any theory that says the algorithm will eventually converge to the
right distribution regardless of the starting values?
thanks~ |
|
h******3
发帖数: 190 | 20 谢谢你. 我其实只想知道是不是最终都可以converge到true distribution regardless
of starting values. |
|
j*****e
发帖数: 333 | 21 查资料说这indicates degeneracy of the Nelder--Mead simplex. 但不懂
degeneracy是什么意思，这个代表什么意思呢，为什么会这样的结果，有什么办法得到
convergence=0呢？谢谢大虾指教！ |
|
x********i
发帖数: 54 | 22 在数据满足什么条件下，K mean 永远不会converge |
|
x********i
发帖数: 54 | 23 如果均匀分布的话，最后converge的情况就是所有的cluster重合在一起。这也算是收
敛。我想问会不会出现oscillate的情况。 |
|
发帖数: 1 | 24 请教懂Bayesian的大虾们，我用Gibbs sampling （MCMC）但是无论sample多久都没法
达到convergence， model好像也不存在不能identify的情况，有什么其他的
alternative的sampling方法吗？
上次在ENAR的时候听有个同行提起过，有一种替代的Gibbs sampling的方法，但是我忘
了名字了。：（
有知道的大虾求赐教。 |
|
g******2
发帖数: 234 | 25 what do you mean by converge? do you mean stationary? Can you show a trace
plot of your samples? |
|
c********g
发帖数: 1106 | 26 Metropolis-Hasting?
Model code有没有错误？
Data足够支持model?
不同链的初始值差太多了？试试相近的初始值？还是不converge简化模型试试。 |
|
f**d
发帖数: 768 | 27 这是一本计算神经科学的优秀著作，全文拷贝这里（图和公式缺），有兴趣的同学可以
阅读
如需要，我可以分享PDF文件（－－仅供个人学习，无商业用途）
From Computer to Brain
William W. Lytton
From Computer to Brain
Foundations of Computational Neuroscience
Springer
William W. Lytton, M.D.
Associate Professor, State University of New York, Downstato, Brooklyn, NY
Visiting Associate Professor, University of Wisconsin, Madison
Visiting Associate Professor, Polytechnic University, Brooklyn, NY
Staff Neurologist., Kings County Hospital, Brooklyn, NY
In From Computer to Brain: ... 阅读全帖 |
|
l*s
发帖数: 783 | 28 ☆─────────────────────────────────────☆
runPython (凸-.-) 于 (Mon Oct 15 00:29:15 2012, 美东) 提到:
还在犹豫：
语言上C#强于Java；
框架上MVC的ASP更易用，
但是JAVA的基于开源的生态系统已经很强大，
大公司都在用，找工作前景很不错。
感觉上java和j2EE的是中大公司搞，
C#和ASP是中小公司搞或者非IT的大公司。
可以看出来薪水还是有点差别的，平均来说。
JAVA略高
☆─────────────────────────────────────☆
a9 (嗯) 于 (Mon Oct 15 08:33:29 2012, 美东) 提到:
搞电子商务的很多大公司都在用.net
☆─────────────────────────────────────☆
NeverLearn (24K golden bear) 于 (Mon Oct 15 11:06:35 2012, 美东) 提到:
Java is paid high simply b/c it's c... 阅读全帖 |
|
s*********l
发帖数: 103 | 29 > Q9：为什么有时候K means算法不能converge？
http://en.wikipedia.org/wiki/K-means_clustering
The algorithm has converged when the assignments no longer change. Since
both steps optimize the WCSS objective, and there only exists a finite
number of such partitionings, the algorithm MUST converge to a (local)
optimum. There is NO guarantee that the global optimum is found using this
algorithm.
...
As it is a heuristic algorithm, there is no guarantee that it will converge
to the global optimum, and the result m... 阅读全帖 |
|
z****g
发帖数: 1978 | 30 Ok, forget 1) and 2), you don't need them here.
For 3), clearly you see it converges by measure. So if you wanna
almost everywhere convergence you should add some conditions.
I can't remember all the results, but there are definitely some
conditions you should meet to make the result a.e converge. One that I
can remember
seems to be: the random variable is defined on bounded interval on real
axis, which does not apply in your case. You probably need to check real
analysis or measure theory. I ch... 阅读全帖 |
|
t***f
发帖数: 113 | 31 看看哪位能帮我。谢了。
【 以下文字转载自 EE 讨论区 】
发信人: tsrif (tsrif), 信区: EE
标 题: 问个学术问题
发信站: BBS 未名空间站 (Thu Feb 16 16:12:49 2012, 美东)
怎么证明下面这个公式，converge或者不converge。如果converge，怎么加快converge
。如果不converge，怎么让它converge。a和b都是random number，都>=0，Xn+1和Xn也
都>=0。
Xn+1=Xn+a-b |
|
t***f
发帖数: 113 | 32 怎么证明下面这个公式，converge或者不converge。如果converge，怎么加快converge
。如果不converge，怎么让它converge。a和b都是random number，都>=0，Xn+1和Xn也
都>=0。
Xn+1=Xn+a-b |
|
a****5
发帖数: 2 | 33 Reference:
Here are the books I used for preparation. All the commons are just my
opinion.
1. "A Practical Guide to Quantitative Finance Interviews": You should be
fine with most of the standard
BT/Probability/Math/Option pricing questions if you truly UNDERSTAND every
part of the book.
2. "C++ Primer": It is too lengthy and is full of details. But it is very
comprehensive and you can find
everything there (before or after you are asked the questions :) )
3. "Effective C++": The first 10 ~ 20 it... 阅读全帖 |
|
发帖数: 1 | 34 12月 谈谈间谍问题
我一直感觉Reuters 有日本的间谍，他们会通过报道来传递针对华人屠杀的计划，和发
布不利于华人的报道，制造舆论。最重要他们后边的势力和犯罪集团有关。有必要好好
调查这个组织。
How top U.S. colleges hooked up with controversial Chinese companies
By Steve Stecklow and Alexandra Harney
Filed Dec. 2, 2016, noon GMT
New Oriental, China’s biggest private educator, has been accused of
academic fraud. Thanks to two enterprising Americans, it has also gained
access to leading U.S. college admissions officers.
SHANGHAI/SHELTER ISLAND, New York - Thomas Benson once ran a small... 阅读全帖 |
|
c*****n
发帖数: 33 | 35 We only need to prove that the sum of a_n(f(z^n)-f(0)) converges to an
analytic function. To show this one only need to show that the sum uniformly
converges on any compact subset of the unit disc. Note that on any compact subset of
the unit disc. Note that we have |f(z^n)-f(0)|\leq M|z|^n. Using this and that
$|a_n|$ is uniformly bounded since it is convergent, one can show the sum is
uniformly convergent on any compact subset of the unit disc. The trick is also used
in proving convergence of |
|
r*****t
发帖数: 286 | 36 ☆─────────────────────────────────────☆
maglion (da木头) 于 (Sun Mar 4 18:49:11 2007) 提到:
Two single linked list A and B
They may or may not converge; i.e. there may be a pair i and j so that A[i].
next = B[j].next
What's your best algorithm to detect whether A and B converge, and the
converging point?
☆─────────────────────────────────────☆
cocojumbo (Nick) 于 (Sun Mar 4 21:39:28 2007) 提到:
there is a fast algorithm to tell if they converge. but as
for where to converge, no ideas.
].
☆─ |
|
t******l
发帖数: 32 | 37 这个问题问得好。
SLLN and WLLN deal with iid random variables, i.e. how \bar{X} converge to \
mu.
Uniform convergence deals with with random functions derived from iid random
samples, say emprical distribution F_n(x)=\sum_{i=1}^nI(X_n
converges to F(x) uniformly, ie the convergence does not depend on x. Of
course, pointwise we have F_n(x)->F(x), by SLLN or WLLN, but uniform
convergence gives us more. |
|
p********2
发帖数: 9939 | 38 我想run一个regression allowing for error correlation within certain clusters.
比如说，year 和 firm。
proc genmod的一个选项是repeated subject。看了看好像这就是用来specify一个
cluster where errors are correlated within this cluster.但是我要specify两个
clusters。它要我写成year*firm。这是什么意思呢？为什么有*。表示interaction？
if yes,怎么个interaction法呵？如果有三个cluster呢？
还有一个问题，我得model不能converge
WARNING: The negative of the Hessian is not positive definite. The
convergence is questionable.
WARNING: The procedure is continuing but the validity of the model fit i... 阅读全帖 |
|
T*******x
发帖数: 8565 | 39 Gamma函数的积分定义可以计算z的实部大于0处的函数值，再加上递推公式Gamma(z+1)=
z Gamma(z)，可以推导任意z处函数值。
黎曼Zeta函数积分定义为Alpha(z)/Gamma(z)。其中Alpha函数为另一个积分，和Gamma
非常相像。但是只能定义在z实部大于1的地方，因此黎曼Zeta函数的积分定义也只能定
义在实部大于1的地方。黎曼Zeta函数的其他定义，比如级数定义，乘积定义，都是如
此。这也不奇怪，因为黎曼Zeta函数在z=1点趋于无穷，定义不过去。
黎曼Zeta函数也有functional equation，
Zeta(z)=f(z)Zeta(1-z)，
根据z实部大于1处的函数值可以退出z实部小于0处函数值，但是z实部在01之间的Zeta
函数值推不出来，要另想办法。
最有效的应该还是一种functional equation。这种functional equation是非常有用的
，黎曼零点的计算肯定也依靠它。
理论上也应该考虑一下，泰勒展开也就是解析函数的解析展开，的radius of
convergence。比如考虑Alpha函数，也就是黎曼Z... 阅读全帖 |
|
m******x
发帖数: 35 | 40 有些是lz自己面的有些是各处收集来的 红/绿皮书的题就不贴了 可能有些时间的原因
难免可能记错一些 请大家
包含!~
待lz想起来会不定期更新
---
explain EM algorithm, use EM algorithm to find SVD of a given matrix
---
Assume if you write an online training model, estimate how many obs the
parameters start to converge
Whats the convergence rate of online training algorithm (e.g. stochastic
gradient descent)
Convergence rate of gradient descent?
How many points are needed to converge give dim of feature space ?
Derive gradient descent formula
---
reservoir sampling... 阅读全帖 |
|
j*****h
发帖数: 3292 | 41 RIM 有讨论B股， h股， ADR，price converge的问题， 其实不单stock price
converge, living std 也是converge的， 从亚洲4小龙， 到中国，(非洲不能算， 太
特别， 人太搞了)。以后可能是印度。给人的感觉是资本主义没路了， 其实只是落后
地区在趋同。
关键是这个converge要花多长时间来完成的问题。
我看不出回国目前有什么意义，趋同并不代表赶上啊，除非有资源，回去能鱼肉，而不
是被鱼肉 |
|
|
|
t******l
发帖数: 10908 | 44 我不做这块 machine learning，我不知道这个具体案例。
但我知道很多 global optimizer 乱走，如果修 bug 的要求是 Out-From-Box，
push-button solution，也就是完全机器自主，没有人工干预，很多不是简单的
说写两个 if，fix 几个choice 就能解决的。因为 global optimizer 是 converge
到那个解的，不管是 problem solving 阶段 converge，还是 machine learning
里 training 阶段的 converge。你如果多加 constraints 针对对这种边际情况，
改善特定这种情况一般不难，但常常会让其他边际情况甚至主流情况 converge 得变差。
或者只能改善零星几种情况，但代价是花不必要的计算时间，这样一个一个情况修到
机器自主不出错的要求，多半最后代码行数和运行时间撑爆掉。要修到机器完全自主不犯
错，很多不那么容易。
当然这不是说不能改进，只是说老板要开支票，不开支票只能给个 patch 或 magic
number 对付这一种特定情况，同... 阅读全帖 |
|
h***i
发帖数: 3844 | 45 不知道对否
1. by law of large number, empirical risk will converge to theoretic risk
this is why we like to minimize empirical risk. called ERM
2. but inf of empirical risk will not always converge to inf of theoretic
risk. so
1 is not good enough.
3. under some constrain for example, uniform converge condition, inf of
empirical risk will converge to inf of theoretic risk.
4. Glivenko-Cantelli theorem.
5. VC theorem(a generalization of GC theorem), the famous inequality,
theoretic risk <=empirical ri |
|
o**a
发帖数: 76 | 46 Ahlfors said that the convergence of \sum a_n is neither sufficient nor
necessary condition for the convergence of \prod (1+a_n).
(a_n are complex numbers)
I've worked out the example that \prod (1+a_n) converges but \sum a_n doesn't.
For example, a_n=-1/(n+1), \sum a_n -> -\infty, but \prod (1+a_n) -> 0
Actually, as long as for all a_n, -1 < a_n < 0, then \prod (1+a_n) is positive
and decreasing, so converges, but obviously \sum a_n may diverge.
My question is, can you think of a series {a_n} s |
|
M*****d
发帖数: 100 | 47 Still no.
Suppose X_i's are iid with mean 0 and std 1. Let Y_n=\sqrt n \bar X.
If Y_n converge to a r.v. in probability, Y_n must be Cauchy convergent in
prob, i.e.,
\lim_{n\rightarrow\infty}\sup_{m>n}P(|Y_m-Y_n|>\epsilon)=0
The LHS > \lim_{n\rightarrow\infty}P(|Y_{2n}-Y_n|>\epsilon) and
Y_{2n}-Y_n=(X_1+...+X_n)/(\sqrt {2n} - \sqrt n) + (X_{n+1}+...+X_{2n})/\sqrt
{2n}.
(X_1+...+X_n)/(\sqrt {2n} - \sqrt n) converges to -(1-1/\sqrt 2)Z_1 and (X_{
n+1}+...+X_{2n})/\sqrt {2n} converges to 1/\sqrt 2 |
|
B****n
发帖数: 11290 | 48 It can be proved using one of the equivalent definitions for convergence in
distribution.
Xk->Y is distribution if and only if ET(Xk)-ET(Y)->0 for any bounded
continuous function (E means expectation).
I think it is called something like Helly's Theorem.
|ET(Xk)-ET(Y)|<=|ET(Yk)-EY|+|ET(Yk)-ET(Xk)|
The first term on the right side converges to 0 due to the fact that Yk-Y in
distribution.
The second term on the right side converges to 0 because Xk-Yk converges in
probability and T is a continuous... 阅读全帖 |
|
z**********g
发帖数: 16 | 49 继续发表有关斥力子的文章，请指正。
“斥力子理论”对物理学的影响1
庄一龙
近年来，由中国学者创立的“斥力子理论”，经过中国科学院《科技促进发展》《
科技成果管理与研究》、《科技中国》、《中国科技财富》、《中国科技成果》等主流
科技杂志报导后，引起了学术界广泛重视。那么，这一理论究竟有那些观点震动了学术
界，值得人们如此关注呢？
“斥力子理论” 是一种新物质作用理论，实际上也是一种物理哲学理论。它的代
表性文章是《论“斥力子”的存在及其意义》，另外还包括《相对性效应和牛顿定律的
本质》、《物理学中精细结构常数的理论推导与解释》等几十篇论文。同牛顿的苹果从
树上掉下来的万有引力理论相对应，这是一个可称作“苹果如何转移到树上去的‘万有
斥力’理论”。该理论认为，地面的物体一旦动起来，不管方向如何，都具有一种挣脱
地球引力的能力，就好象物体吸收了一种对抗万有引力的物质那样。“斥力子理论”把
普朗克能量子看作是一种对抗引力的实物粒子，称它为“斥力子”，并认为物体运动状
态的改变就是由于吸收或释放斥力子造成的。
在思维方法上，它有两个根本突破：其一,是把抽象的能量概念实物化，使能量有
一个实物载... 阅读全帖 |
|
