Q***5
发帖数: 994 | 1 For any stopping time \tau in formula one (the one with inner "max"), let
\tau2 = \tau (on S(\tau)>=K) and = \infty (on S(\tau)
You can prove that: \tau2 is a stopping time and when apply \tau_2 to
formula two, you get the same value as when you apply \tau to formula one. |
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p*****k
发帖数: 318 | 2 since there are total of 2^N-1 sums for an N-element set, so 2^(i-1)
optimizes the total sum.
OP's problem (or rather the discussion so far) is related to Erdos's
conjecture:
find a set with such property (i.e., distinctive subset sums) with the
smallest largest element.
(his conjecture is the asymptotic form with N->infty)
the solution was the so-called Conway-Guy sequence, which gives the upper
bound, and they verified up to N=8 that their solution is optimal.
(but the asymptotic form had been... 阅读全帖 |
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l******i
发帖数: 1404 | 3 到达3的概率是1，期望到达的时间是\infty。
我也这么想的，这不是基本结论吗？难道是我没有理解题意？ |
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x******a
发帖数: 6336 | 4 let A= \{ |B_T|<1 \} and B= \{ \sup_{0
P(B) \leq P(A) = P( B_1 \leq 1/\sqrt{T}) \to 0 as T\to \infty. |
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k*****y
发帖数: 744 | 5 如果x小于1，那么X_t是bounded below by x，所以\alpha只能取<0的数。这时前面修
正的指数那项也是<0，于是还是能用bounded convergence theorem，得到倒数第二个
等式。但是这时只能让\alpha -> 0^-，于是得到P(\tau < \infty) = 1。 |
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p*****k
发帖数: 318 | 6 note the symmetry, this is just twice of int from 0 to infty.
the trick then is to set x = (a/c)^(1/4) * exp(t) and use hyperbolic cosine
and sine. up to a constant factor, it's just a Gaussian integral again.
double check if the answer is exp[-2*sqrt(a*c)] * sqrt(pi/a)
more details can be found at:
http://www.wilmott.com/messageview.cfm?catid=26&threadid=72024 |
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C***m
发帖数: 120 | 7 Thank you for your comments. I was trying to see why var(max(X_i,1))
For the var, I was thinking E(W_t^4-3t^2)=0, so maybe E(W_\tau^4-3\tau^2)=0
as well. It seems doesn't work cause this method may lead a negative var... |
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c**********e
发帖数: 2007 | 8
Faint. max(X_i,1)<=1. Of course its var
0
I tried. This one might not have a nice close form solution unless special
case such as a=b or a=2b. |
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C***m
发帖数: 120 | 9 Thank you for your comments. I was trying to see why var(max(X_i,1))
For the var, I was thinking E(W_t^4-3t^2)=0, so maybe E(W_\tau^4-3\tau^2)=0
as well. It seems doesn't work cause this method may lead a negative var... |
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c**********e
发帖数: 2007 | 10
Faint. max(X_i,1)<=1. Of course its var
0
I tried. This one might not have a nice close form solution unless special
case such as a=b or a=2b. |
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l*****y
发帖数: 56 | 11 我是这样理解的
E(total payoff)=\sum_k=1^\infty E(total payoff| rolls=k)Pr(rolls=k)
然后前面算过conditional payoff 总是1/2, 那么最后的payoff应该也是1/2. |
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l*****y
发帖数: 56 | 12 我是这样理解的
E(total payoff)=\sum_k=1^\infty E(total payoff| rolls=k)Pr(rolls=k)
然后前面算过conditional payoff 总是1/2, 那么最后的payoff应该也是1/2. |
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k*****y
发帖数: 744 | 13 谢谢包子~
我不是很懂，不清楚E[ Y - a/(1-b) ]是什么意思。
但是我觉得能说的是 E[ Y(t) - a/(1-b) ] -> 0 当 b < 1 and t -> infty，因为
E[ Y(t) ] = a + b E[ Y(t-1) ]
=> E[ Y(n) - a/(1-b) ] = b^n E[ Y(0) - a/(1-b) ]
0 |
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k*****y
发帖数: 744 | 14 Apply the power series of sin to A.
Here is my 2cents.
Note that sin(x)^2 + cos(x)^2 = 1 is an identity as formal power series.
When you restrict to the n-th order polynomial, it gives
[sin(x)^2 + cos(x)^2](n) = 1
up to a residual of higher order terms.
In the case of sin(A) and cos(A), the residual about the n-th order tailor
polynomial consists of summands like A^i/i!, which makes the residual go to
0 as n goes to infty. So after taking the limit
sin(A)^2 + cos(A)^2 = 1. |
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n*******t
发帖数: 67 | 15 谢谢，不过前面那个 Hence... 是为什么呢？那一步本身为什么不需要用到 E(\tau)<\
infty？ |
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k*****y
发帖数: 744 | 16 There it only assumes P(tau < infty) = 1.
The equality follows directly from the property of martingale.
<\ |
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n*******t
发帖数: 67 | 17 谢谢，不过前面那个 Hence... 是为什么呢？那一步本身为什么不需要用到 E(\tau)<\
infty？ |
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k*****y
发帖数: 744 | 18 There it only assumes P(tau < infty) = 1.
The equality follows directly from the property of martingale.
<\ |
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M****i
发帖数: 58 | 19 一个简单算法是在平面上随便找一点x_0作为初始迭代点，然后找到数据点中到x_0距离
最远的那个点y_0（这个点不唯一时随便取一个就行），然后从x_0出发沿着连接x_0和y
_0那条线段走一段距离t_0到达下一个迭代点x_1，然后重复以上步骤即可。可以证明，
当每次走的步长(t_k)_k所成的级数\sum_{k=0}^{\infty}t_k发散但是平方收敛时（比
方说可以取t_k=1/(k+1)）该算法最后收敛于数据点的minimax center。这个算法在黎
曼流形上也是成立的，只是步长与流形的曲率上界和下界有关。有兴趣的话可以看看下
面的文章：
http://hal.archives-ouvertes.fr/index.php?halsid=1ka58mlobsd187 |
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r********2
发帖数: 19 | 20 I agree it is a martingale, but the stopping time is not integrable.
My thought is,
P(tau<\infty)=1
therefore, to make the game fair you should pay 10 as premium. |
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l******i
发帖数: 1404 | 21 你要用optional stopping theroem的话，
要先证明E[tau]< \infty，才能用M_0=E[M_{tau}]吧。
S |
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m******9
发帖数: 74 | 22 This probability is 0.
S_n^2 - n E X^2 is a martingale.
This shows that the expected time of leaving the region [-B, B]
is B^2/ E X^2 < \infty.
So 在－5和10之间不停晃动下去的概率是0 |
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w**********y
发帖数: 1691 | 23 [0, +infty]用一个s曲线 f(x)，tanh或者logit都行， [-infy,0] 用 - f(-x).
看你的目的。如果正巧跟我做的东西很像，那你的一阶导数很重要。 |
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C******a
发帖数: 115 | 24
我是这个意思。
想象一下，应该是不难的。
Let X＝[0,1]^\infty, f_n: X\to [0,1] maps x to the
n-th coordinate of x. Let B be the Borel algebra of [0,1].
Let F_n=f_n^{-1}(B). Then F_n is \sigma-field,
but \cup_n F_n is not field. |
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f*******d
发帖数: 339 | 25
it was a typo, should be
f(r)=1/r \int_0^\infty g(k)/k sin(kr) dk |
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B***y
发帖数: 83 | 26
I am not sure about what problem might occur in the numeric
integration. Just
some suggestions:
1) when r is quite large, sin(kr) is oscillating quite fast, thus
usually you
need to take smaller time-steps to take care of it.
2) since you choose g(k) =1, then \int_0^\infty 1/k sin(kr) dk is not
absolutely
integrable, thus there might be some problem when k is small, where
1/k is quite
large, causing some numeric problem.
good luck. |
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l****a
发帖数: 366 | 27 I take some time to type the following passage from a book
called "Concrete Mathematics: A foundation for Computer
Scienence"
written by three geniuses: Graham, Knuth, and Patashnik.
"A remakrable theorem was discovered independently by three
mathematicians - Bohl, Sierpinski, and Weyl - at about the
same time in 1909: if \alpha is irrational then the
fractional
parts {n\alpha} are very uniformly distributed between 0
and 1, as n \rightarrow \infty. One way to state this is
that:
\[ \lim_{n \ri |
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I***e
发帖数: 1136 | 28
I don't think the conclusion is right. Take a1=1 and ai=0 for i>1. Take bi=0
for all i.
THen An = 1 and Bn=0 for all n.
How can lim An = lim Bn ?
I guess you meant An= sum_{i=n}^infty ai. If so, you have to also have some
extra conditions like ai, bi are both positive sequences. Then you can use the
discrete version of BCT ( Bounded convergence theorem )
Hope this helps.
icare |
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f**n
发帖数: 401 | 29 Yes. Sorry about that.
Let me try to make it clearer:
Actually I am trying to emulate an integral by a summation:
You can imagine there is a interval(x axis) and I divide the interval into
n smaller intervals. After this decomposition, we index all intervals, let's
say, interval 1,2,...,n. For interval i, a_i and b_i are defined.
So if n->\infty, both series {a_i} and {b_i} have to evolve. |
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x******i
发帖数: 3022 | 30 define
f_j(t) := exp(-x_j*t)
define inner product
(a,b) = \int_0^{\infty} a(t)*b(t)*dt
then
the matrix is
C_ij = (f_i,f_j)
which is obviously positive definite, because
sum a_i*a_j*C_ij = (sum a_i*f_i, sum a_j*f_j) >=0 |
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d****a
发帖数: 193 | 31 usually n --> infty, se --> 0
sd is a constant |
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d******e
发帖数: 7844 | 32 这答得哪而跟哪儿啊？
SVM最大的优点是Minimize Maximum Margin的思想，让generalization能力极大的提高。
你说的很多predictor跟outcome不相关，准确的说应该是Margin最终只决定于其边缘和
内部的样本，这些样本被称为support vector，这种sample sparsity的结构让结果更
稳定。但这只是L2 SVM最大化geometric margin的结果。如果使用L1 SVM最大化L_{\
infty} margin,那么，得到的就是support feature了，也就是feature的sparsity.
我喜欢的另一个优点是：虽然使用surrogate loss，但是却是Fisher consistent的，
而且是convex的，有很多非常成熟的高速解法，比如cutting-plane，优化起来快速便
捷。
至于什么非线性，那是kernel的功劳，Logistic Regression一样也可以有非线性的版
本。 |
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y***s
发帖数: 23 | 33 1. I guess NO.
Suppose B_t is BM with 0<=t< infty
Let T=inf{t>0:B_t=1} and it is a stopping time.
Then B_T/T^0.5=1/T^0.5, which probably is not a normal r.v. |
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c*****l
发帖数: 135 | 34 Cumulative hazard can go to infty.
★ 发自iPhone App: ChineseWeb - 中文网站浏览器 |
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z****e
发帖数: 702 | 35 \sum_{k=0}^\infty beta(k+1,c+1), 其中c为正，
这个怎么证明等于1/c?我仿真是对的，但是不知道怎么证明。 |
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d******e
发帖数: 17 | 36 可以，直接regression, correlation就行了；
但结论仅对此人成立,且不一定和group-pooled correlation相近；
Denote Z= Individual No
因为你估计的是 corr(X1,X2 | Z=1),
Need corr(X1,X2 | Z=i)=corr(X1,X2 | Z=j), for any i, j
且 corr(X1,X2 | Z=i)！=corr(X1,X2）
所以结论是你的estimator coverge to corr(X1,X2 | Z=1) 且还需假设 n1 ->\infty.
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1、即使所有人都相同的 correlation within individual, group correlation is
not equal to correlation within individuals. 这和把一个longitudinal data进行
pool算一样；
2. 需要假设individual correlation equa... 阅读全帖 |
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s******h
发帖数: 539 | 37 The general result is, for any non-negative measurable function g,
E(g(X)) = \int_{[0, \infty)} Pr(g(X) >=t) dm(t), where m is the Lebesgue
measure. It can be quite useful sometimes. |
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l*******r
发帖数: 28 | 38 有限。等于 \int_{0}^{+\infty} t N(0,1) dt |
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d*****a
发帖数: 627 | 39 如果是标准正态分部，－infty到0的积分不是1／2？ |
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l******t
发帖数: 96 | 40 in (1) E log(X) could be - infty.
)<
/Y
LnY
by
=1 |
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