p********a
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TNEGIETNI (lovewisdom) 于 (Tue Oct 11 20:01:18 2011, 美东) 提到:
朋友告诉我说他要来，还有Harvard的Carl Morris等，$200的Registration fee对我来
说不是一笔小数目。我其实并不欣赏他的bootstrap法，曾公开批评过这个方法的逻辑
错误。如果去，我跟他讲什么呢？版上的恶朋好友可否给点建议？谢谢。
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angelsun (安吉笋) 于 (Tue Oct 11 20:38:46 2011, 美东) 提到:
在哪里？什么时候？讲座是什么topic的？
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statcompute (statcompute) 于 (Tue Oct 11 23:51:23 2011, 美东) 提到:
$200 for a lecture by Efron? it i... 阅读全帖 |
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h***i
发帖数: 3844 | 2 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 |
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B****n
发帖数: 11290 | 3 This is the application of second borel cantelli lemma.
Suppose the number of letters in a book is n, the number of letters in a typew
ritter is m, then the probability of typing a book within n letters is
1/m^n. Let this event be A1. A2 is typing a book in next n letters. ....
sum(Ai)=infinity Ai are independent P(Ai io.)=infinity |
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h***n
发帖数: 276 | 4 多谢回答。
不过当我使用中心极限定理时，在贝努利实验下去估计真实概率p，得到的所需要样本数
目n是形如下面的式子
n>=p*(1-p)*f(tolerance to real p)
好像结论和你的相反，真实概率离1/2近的反而要多些？
还有既然大家收敛速度有快慢，如何理解Glivenko-Cantelli theorem 给出的uniform
convergence的结论呢？ |
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c*****a
发帖数: 49 | 5 我在wiki上看到了这个不等式，但是老板说要引用参考书。不知道那本经典的教科书上
有讲。望大虾们指点一本。
多谢！ |
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A*******r
发帖数: 768 | 6 查题目叫 inequalities 的书至少有两本 |
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a***s
发帖数: 616 | 7 This seems impossible except some degenerated cases.
Let $Y_k := K(x^*, X_k)$. Since $\{X_k\}_{k \geq 1}$ are i.i.d., $\{Y_k\}_{k
\geq 1}$ are also i.i.d. Since the value of the kernal is between 0 and 1,
$Y_k$'s are nonnegative. Then Borel-Cantelli lemma says $\sum_{k=1}^{\infty}
Y_k = \infty$ with probability one unless $Y_k = 0$ with probability one.
Therefore for the infinite sequence $K$ to be in $l^1$ with positive probabi
lity (one, actually), we must have $Y_k = 0$ with probability one. ... 阅读全帖 |
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z****e
发帖数: 702 | 8 用borel-cantelli引理，似乎证明的是事件概率的和吧，
此处是随机变量的和。
{k
,
infty}
probabi
mean |
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M****i
发帖数: 58 | 9 You are welcome.
Generally speaking, the martingale convergence theorem can not be applied
here because I(t) is only a semimartingale, not a martingale:
I(t)=g(t)*M(t),
where
g(t)=exp(-\int_0^t c(r)dr)
M(t)=\int_0^t c(s)*exp(\int_0^s c(r)dr) dB(s).
Note that g(t) is of finite variation and only M(t) could be a martingale.
This can also be verified by my previous example
c(t)=1/(1+t). In this case I(t)=B(t)/(1+t), which is not a martingale, but
it converges to 0 a.s. by iterated logarithm.
For yo... 阅读全帖 |
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w*********i
发帖数: 77 | 10
Borel-Cantelli should work |
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c******r
发帖数: 300 | 11 1. with probability one, you have Xi \neq 0 for infinitely many times, check
second Borel-Cantelli lemma for more details.
1 |
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m****r
发帖数: 237 | 12 By using Cantelli's inequality, assuming the mean is mu, standard deviation
is sigma, we have:
P(X-mu>=k*sigma)<=1/(1+k^2)
Take k=1, we have,
P(X-mu>=sigma)<=1/2, which is
P(X>=mu+sigma)<=1/2
Similarly, from P(mu-X>=k*sigma)<=1/(1+k^2),
we get P(mu-X>=sigma)<=1/2, which is
P(X<=mu-sigma)<=1/2
Therefore, the median of X shoule lie between mu-sigma and mu+sigma. (median
(X)-E(X))^2<=var(X) follows. |
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d******e
发帖数: 7844 | 13 样本的empirical distribution当然不是true distribution。
可惜你不明白Glivenko–Cantelli定理，唉。 |
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d******e
发帖数: 7844 | 14 Glivenko–Cantelli可以看成是中心极限定理的推广，不过大师应该也不明白中心极限
订立成立。
但是大师要知道，你天天在那里嚷嚷说t-test NB，这玩意也是建立在中心极限极限之
上的。
如果没有中心极限定理做收敛性的保障，t-test就是坨屎了。
统计工具之所以有效，大样本收敛性分析是理论保障之一，连sample mean这种最简单
的计算也是有大数率来保障的。
可惜大师只会脑补，什么都已一拍脑门YY，就“想”出来了。 |
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a****e
发帖数: 150 | 15 小小纠正一下,Glivenko–Cantelli 是 law of large number 的推广, CLT的推广是
Donsker. |
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