B******t
发帖数: 40 | 1 Calculate the primary value of integral:
Integral{-Inf, +inf; F }
where F=1/[(1+x^2)*(x-x1)^2]
x1 is a real number.
我知道如果X1是实轴上的单极点，就是F=1/[(1+x^2)*(x-x1)]，结果是
i*2PI*Res[F,i]+i*PI*Res[x1]
但是现在x1 是实轴上二级奇点，这个积分存在吗？
请高手指点。 |
|
b********a
发帖数: 300 | 2 如果有一个GI/M/Inf queue, 然后我现在知道number of users in the queue is
Poisson distributed, 那么是否能说这个Queue一定是M/M/inf?
谢谢 |
|
n******t
发帖数: 4406 | 3 经典的Riemann广义积分一般定义成 lim_{x->inf}\int_0^x f(t)dt.
direct Riemann Integrable 是指直接在(0,\inf)上定义Riemann上和
和下和，然后要求上和和下和收敛且相等。
例子比如说sin(x)/x. |
|
a***n
发帖数: 3633 | 4 如果a(i)=lim_{n->inf}a_n(i)
那么是不是有
lim_{n->inf}{sum a_n(i)}<=sum a(i)
两个sum都是对i求和，从1到无穷。如果只有有限个a(i)那么很好办
<=肯定是等号成立。但是有无穷个a(i)时呢？ |
|
b***e
发帖数: 3337 | 5 Integral of exp{-(a*x^2+b*x+c)} from 0 to infinite
i know the answer of Integral of exp{-(a*x^2+b*x+c)} from -inf to inf
= sqrt(pi/a)*exp((b^2-4ac)/4a)
thanks!! |
|
b*********n
发帖数: 56 | 6 The formula could be true, for we know that int(exp(-x^2), x=-inf..inf) is
known to be Pi (or Pi/2).
This problem is just a calculus homework. Spend some time on transforming a*
x^2+b*x+c into a form like a*y^2+d, and then .... |
|
b****r
发帖数: 5889 | 7 int[0..inf,f(2x)] ?= int[0..inf, f(x)]
int is the Integrate Func. How to prove that? Why I thought the fommer is
half the latter? |
|
s*x
发帖数: 3328 | 8 1/x在0点不连续吧,左边趋于-inf,右边趋于+inf,应该是|1/x|才对 |
|
y**c
发帖数: 133 | 9 Is there anybody can help me to solve this question:
A sphere of radius=R0 has gas in it;
There has a potential V(R)=V0*exp[-(R-R0)/a].
The atoms are colliding in the sphere, and the potential acts on the particles.
The question is:
How to get the Fourier transform of V(t), i.e.
Integral(from -inf to +inf) [V(t)*V(t+tau)]*exp(-i omega t) d tau
you may consider potential is much smaller than the kinematic energy, if otherwise too difficult to solve.
Thank you! |
|
A****s
发帖数: 129 | 10 statement: in an open interval (a,b), f(x/2+y/2)<=1/2f(x)+1/2f(y)
f (a,b) to R continous --> f is convex.
sketch of proof: consider the function of t, t in (0,1), p,q constants in(a,
b):
g(t)=f(tp+(1-t)q)-tf(p)+(1-t)f(q). g is continuous.
g^-1((0,inf)) is open since (0,inf) is open, i.e., the points in (0,1)
which make g>0 are indeed a family of disjoint open intervals.
suppose for some p,q,and t' such that g(t')>0. then t' is in one of the open
intervals. then for those two endpoints of this op |
|
i********w
发帖数: 2223 | 11 【 以下文字转载自 Statistics 讨论区 】
发信人: icebergzjw (喝多感觉真好), 信区: Statistics
标 题: 请教:随机变量的分布函数问题
发信站: BBS 未名空间站 (Fri Feb 26 16:22:07 2010, 美东)
有一个随机变量 (例如 20xx年电汽车电池的大小).
在目前没有什么有用相关信息的情况下(只知道随机变量应属于(0,+inf)), 我需要为它
假设一个看似合理的分布函数.
请问有什么推荐么?
1)我想用正态分布,但是不满足随机变量范围.请问有没有什么modified normal
distribution 符合这个要求吗?
2)Log-normal倒是符合随机变量的要求,可是我没有办法解释为什么选log-normal,不选
其他的
分布函数.(我现在也不知道还有哪些常用的分布函数,它的随机变量是(0,+inf)的).
多谢! |
|
s*****e
发帖数: 115 | 12
F(x*,y*) >= min_y F(x*,y)
Don't see why this is true? Just think again!
The statement is TRUE even if "min" is replaced by "inf", and F(x,y) can be
any (even discontinuous) function, i.e.,
inf_{x,y} F(x,y) = inf_y { inf_x F(x,y) } = inf_x { inf_y F(x,y) }
The proof will be somewhat different from the "min" case since there may not
exist (x*,y*) such that F(x*,y*) = inf_{x,y} F(x,y). But still, this is just a very
simple exercise for "inf/sup" in intro to real analysis. |
|
g*****5
发帖数: 3285 | 13 【 以下文字转载自 Military 讨论区 】
发信人: applyfaculty (applyFacaoti), 信区: Military
标 题: 请问所男们，一维直线上的点和二维平面上的点一样多吗？
发信站: BBS 未名空间站 (Tue Aug 30 14:15:51 2011, 美东)
有几个可以回答
－－－－－
一样多。
64楼有正确答案
但是远非“都是无限所以一样多”这么简单，比如整数集和实数集都是无限，但可以用康托的对角线证明整数集的数量小于实数集：
如果你假设他们数量相同，则他们之间可建立一一对应的关系，所以任意一个实数都可以用整数来数，则，所有的（0，1）之间的实数都可以被编号写成q1,q2,...如下表：
q1: 0. a11 a12 a13 .............
q2: 0. a21 a22 a23 .............
q3: 0. a31 a32 a33 .............
....
其中a11,a12,等等都是0-9的自然数。所有的实数都应该能从此表找到了。但若取
x = 0.b1 b2 b3,...
且使b... 阅读全帖 |
|
z****e
发帖数: 702 | 14 thank you, 实际上这个问题还是要求实函数，因为它是基于以下case，
\int_0^\inf p^-1 e^-p 为无穷大，所以想把它分解成无穷个积分有限的项和：
如\int_0^\inf p^a e^-p (a>0)这样每一项积分就有限了。 |
|
f******h
发帖数: 104 | 15 这真的假的？哪个大牛给鉴定一下?
First letter:
-------------------------
成桐兄：
钟家庆自Austin回Harvard后与我谈及他Austin一行见闻。其中一事使我甚为困
扰。此事乃关乎你学生田刚在Kahler-Einstein Metric 问题上之结果。钟家庆对此
问题甚感兴趣。他来Harvard后曾阅读我在Columbia所讲之文章，即我托他带给你的
那篇。他在Austin时向田刚及曹怀东问及田刚Kahler-Einstein Metric的结果。他
主要是从曹怀东处大约获知田刚所寄Kahler-Einstein Metric文章之方法及结果。
他回来后将所知一一转告我。因他Austin之行短暂，所知自多有未尽之处。单从据
他所拿到的部分看，除田刚自信应该有办法将来可做出之结果不算外，田刚确实做
到的部分中他所用方法与我在Columbia讲的基本上一样，结果也未出我所能做出的
范围。该方法我四月在Columbia及Maryland并六月在巴黎都曾在演讲中讲述。我四
月在Columbia讲时你和田刚及其他数位你的学生均在座。若田刚有新方法，... 阅读全帖 |
|
f*****s
发帖数: 95 | 16 来自主题: Mathematics版 - 黎曼猜想 P(x) 和 f(s) 的关系是通过傅立叶变换，其实就是用到如下积分,
\int y^s/s ds = 0 if y < 1
= 2\pi i if y > 1
积分是在复平面上从 a-\inf i 到 a+\inf i for some a>1. 令 I(p
0 otherwise. 从以上公式我们有，
1/(2\pi i)\int (x/p)^s /s ds = I(p
所以
1/(2\pi i)\int f(s) x^s /s ds
= 1/(2\pi i)\int \sum_p p^{-s} x^s /s ds
= 1/(2\pi i)\sum_p \int (x/p)^s /s ds
= \sum_p I(p
= P(x)
这样，我们就大概有
P(x) = 1/(2\pi i) \int \log zeta(s) x^s/s ds
但如果 zeta(s) 算不出，上面也是空谈。黎曼的另一关键思想是把 zeta(s)从Re(s)>1
推... 阅读全帖 |
|
l*3
发帖数: 2279 | 17 (inf - 70,000,000)/(inf -2)
这是老张的工作所占的百分比. |
|
y**c
发帖数: 133 | 18 Is there anybody can help me to solve this question:
A sphere of radius=R0 has gas in it;
There has a potential V(R)=V0*exp[-(R-R0)/a].
The atoms are colliding in the sphere, and the potential acts on the particles.
The question is:
How to get the Fourier transform of V(t), i.e.
Integral(from -inf to +inf) [V(t)*V(t+tau)]*exp(-i omega t) d tau
you may consider potential is much smaller than the kinematic energy, if
otherwise too difficult to solve.
Thank you! |
|
B*******t
发帖数: 135 | 19 Thank you for your reply!
integrability.
x)
It is very obvious that Jensen is the key to use since convex function is
the issue here.
However, what is tricky is how to find the localizing sequence. I considered
about the following
\tau_n = inf{t: sup|\phi(X_t)-\phi(X_0)| >= n, t >= n}
But I am stuck in proving
\tau_n -> inf almost surely.
I actually solved this problem by DCT instead of Fatou's lemma. Did you
actually mean DCT?
I have not seen how Fatou is involved here.
But still thank you for |
|
y**y
发帖数: 25 | 20 the problem is like
B(t) is a standard brownian motion,
what is the probability that B reaches b before hiting -a?
I know this kind of problem could be solved by martingale stopping time
thereom.
But I wonder if it still could be done by transition probability density(tpd
).
Since we know the tpd of B(t) is N(0,t),so we could calculate
P(-a
then I don't know how to continue to get the solution of my problem.
by limit(t=inf) or integrate t from 0 to inf?
or even the whole |
|
J*****n
发帖数: 4859 | 21
probabilit
that
Markov 链
解迭代方程：
a_n=0.45*a_{n-1}+0.55*a_{n+1} (1)
a_0=1
a_{inf}=0
目标求a_10
(1)可以化为：11(a_{n}-a_{n-1})=9(a_{n-1}-a{n-2})
设D_n=a_{n}-a{n-1},则D_n=9*D_{n-1}/11
a_{inf}=a_0+(a_1-a_0)+(a_2-a_1)+.....
i.e. 0=1+D_1+D_2+.....
则D_1=-2/11,a_10=D_1+D_2+....+D_10+a_0=(9/11)^10 |
|
j******n
发帖数: 271 | 22 I guess we need some continuity such as:
lim_{n->\inf} F(P_n) = F(Union_{n from 0 to \inf} P_n), for P_n subset of
P_{n+1}.
region
and P2
we |
|
n******r
发帖数: 1247 | 23 Is that P(max(X,Y)
Then use Y|X=x is N(/rho*x,1-/rho^2) and x
Finally integrate x from -inf to inf? |
|
t*******e
发帖数: 172 | 24 Is not this G can define a measure on the plane?
For any X we can define G(X)= inf \Sigma G(P_{i}), where inf with respcet to
all the coverings of X by polygons.
Maybe we need G is countable summable.
). |
|
f********y
发帖数: 278 | 25 这是Zhou Xinfeng书里的一段，书里令X-N(0,1)正态分布，
E[X|X>0]=\int_0^\inf xf(x)dx
f(x)正态分布的density function,
我觉得这似乎不对啊，E[X|X>0]是求在X>0条件下的期望值，\int_0^\inf xf(x)dx只是
计算X>0那一段的值。 |
|
c*****w
发帖数: 50 | 26 E[X|X>0]=\int_0^\inf xf(x)dx/(\int_0^\inf f(x)dx) |
|
x******a
发帖数: 6336 | 27 是无穷大
Let T_0=inf{n>0; S_n=0}, then
P(T_0<\infty) =2*min(p, q);
E(T_0|T_0<\infty)=1+ \frac{1}{|p-q|}.
main steps for the probability to go back to zero:
1. let T_1=inf{n>0; S_n=1}. verify P(T_1<\infty)= 1 if p>=q; p/q,
otherwise;
2. P(T_0<\infty)= P(S_1=1, T_0< \infty) + P(S_1=-1, T_0<\infty)
=pP(T_{-1}<\infty) +qP(T_1 |
|
a********e
发帖数: 508 | 28 actually the neccessary and sufficient condition is
c(inf)=0, \lim \int c(s)ds=inf
by applying the rule \lim f/g=\lim f'/g'on var(I) |
|
z*****n
发帖数: 165 | 29 \int_-\inf^\inf Sin(x)/x dx |
|
r*******y
发帖数: 1081 | 30 So what is the answer to
\int_{-inf}^{inf} e^{ix} / x dx ?
thanks.
Jordan' |
|
z*****n
发帖数: 165 | 31 Due to the ambiguity, the notation in the problem really means:
\int_\-inf^\-epsilon + \int_\epsilon^\inf
and then take limit of \epsilon -> 0+
Which is the "v.p." value of integration.
Under this definition, there is no problem of cos(x)/x, because it's an
odd function, so both parts cancel out after integration. |
|
s****p
发帖数: 19 | 32 Laplace transformation.
Let F(y)=\int_0^\inf e^{-yx} sin(x)/x dx then
-F'(y)=\int_0^\inf e^{-yx}sin(x) dx, which has close form. |
|
z****g
发帖数: 1978 | 33 哦哦，公式有点错，是
lim(1+1/x)^x = e when x->inf. 数学分析里的内容，比较可以反应数轴性质的极限之
一。做过
吉米的人肯定都知道....
lim ( 1 + 1/x)^x = e when x -> inf =>
lim ( 1 - x)^(1/x) = 1/e when x->0
你再比较一下 (1-0.01)^100， 0.01和100不是随便给的，是给你凑好的
x = 0.01很接近0, 可以用这个极限近似。1/e = 0.3679 |
|
x********i
发帖数: 10 | 34 1. I guess the statement is mistaken. It should be "What is the expected
index T that makes ST <= 1?" Or else the answer is infinity. First we can
get P(S_i<=1)=1/i!. Basically it can be proved by inductive reasoning. The
derivation is a bit complicated (refer to the problem of 'sum of random
variables' in Chapter 4 of the book by Xinfeng Zhou). Then E(T)=\Sum_i=1_to
_\Inf(i*P(S_i<=1))=\Sum_i=1_to_\Inf((i-1)!)=e (to see this evaluate the
Taylor expansion of e^x at x=1)
2. I think the answe... 阅读全帖 |
|
s*****e
发帖数: 20 | 35 Let S_0=0 and for n \in N define
T_n = inf { t >= S_{n-1} | B_t >= 1} and S_n = inf{ t>= T_n | B_t <= -1}
Show that P(lim n -> infinity T_n = infinity) = 1 |
|
C***m
发帖数: 120 | 36 用LLN需要var有限，借这个帖子，再问一个，
定义 \tau = inf { t >= 0 | B_t >= 1}
求E(\tau)和var(\tau)，谢谢。
其实想算这个
\tau = inf { t >= 0 | B_t >= a or B_t <= -b}
的mean 和var. a,b>0 |
|
s*****e
发帖数: 20 | 37 Let S_0=0 and for n \in N define
T_n = inf { t >= S_{n-1} | B_t >= 1} and S_n = inf{ t>= T_n | B_t <= -1}
Show that P(lim n -> infinity T_n = infinity) = 1 |
|
C***m
发帖数: 120 | 38 用LLN需要var有限，借这个帖子，再问一个，
定义 \tau = inf { t >= 0 | B_t >= 1}
求E(\tau)和var(\tau)，谢谢。
其实想算这个
\tau = inf { t >= 0 | B_t >= a or B_t <= -b}
的mean 和var. a,b>0 |
|
C***m
发帖数: 120 | 39 不是大牛，谢谢。
假设从1出发，有两个boundary 0和 a>1，计算P_1,.因为(1/3)^(X_t)是个martingale,
所以 (1-P_1)(1/3)^a+P_1(1/3)^0= (1/3)^1，再让a->inf
P_1=1/3
假设从-1出发，有两个boundary 0和 a<-1，计算P_(-1).(1-P_(-1))(1/3)^a+P_1(1/3)
^0= (1/3)^(-1)，P_(-1)=(3^(-a)-3)/(3^(-a)-1)再让a->inf
P_(-1)=1
感觉还挺不严格的，求大牛证实。 |
|
e*******t
发帖数: 44 | 40 就是火车从左向右开， 很多个停车站，假设车站坐标是-inf ... -1 0 1 ... inf，火车每隔k站停一下
每次停的时候， 有个人去射击它 ，但是看不见火车，用什么方法去射到它
以前版上看到过讨论 但是忘了 大侠能不能指点一下 非常感谢 |
|
l**********e
发帖数: 336 | 41 一个高斯分布的随机数Z，a是一个给定实数且a>0，要求E(Z|Z>a)比较紧的上界和下届
。可以假定Z是0均值， 方差1。
-------
我的想法是E(Z|Z>a) = \int_{a}^{\inf} ( p(z|z>a) * z ) dz，其中p(z|z>a) = p(
z) / P(z>a)，
所以E(Z|Z>a) = \int_{a}^{\inf} ( p(z) * z ) dz / P(Z>a)。这样的话分子部分应
该有解析解直接算出来。分母部分 P(Z>a)是一个标准的高斯的概率积分，求这个部分
的上下界就好了。那这个有什么比较常用的紧的上下界呢？
--------
或者除了这个做法，有什么其他的方法吗？比如直接去逼近E(Z|Z>a)？
--------
谢谢！！！ |
|
n******m
发帖数: 169 | 42 integral is from 0 to F and F to inf, sum is from 0 to K, and K to inf,
so from K to F, you used put instead of call (or the opposite, don't
remember detail), the difference is int(K,F){call-put}, which is the the
extra term.
on
replicated
maturity
formular
But
2/ |
|
m******4
发帖数: 15 | 43 Let k = Tn, then I think P(k) = n!/(n-k)!/n^k * k/n
To study the behavior of P(k) in the limit n --> inf,
the ratio k/n should be finite, otherwise, given any fixed k,
P(k) --> 0 when n --> inf.
Now, n and n-k are very large and we can use Stirling formula.
Here, we study the ratio r=k/sqrt(n), I use Mathematica and find
Limit[Sqrt[n]*P( k->r*sqrt(n) ), n->Infinity] --> exp(-r^2/2)*r
Very interesting, the most probable k is sqrt(n), perphas this
is the reason we choose r=k/sqrt(n). |
|
l***e
发帖数: 33 | 44 这篇文章很长(很大部分是引文啦), 没兴趣的话请跳过...
首先, 感谢unix的原文, 很清楚.
其次, 我想在他的基础上多说两句...
我们用I代表单位方阵, P~代表P的共轭转置. P^T代表P的转置.
O代表实正交方阵: O*O^T = O^T*O = I,
U代表酉方阵: U*U~ = I.
||A||代表普遍意义下的范数. ||A||^代表2范数(cf. unix文).
Q1) ||I|| = 1?
Q2) ||U|| = 1?
Q3) ||A|| = ||UAU~||?
Q4) ||A||^ = ||UAU~||^?
A1) a) 如果只把n*n矩阵看成一个n*n维的线性空间的话, ||I||可以是任何非零的数.
b) 如果||A||是consistend, 很容易证明 ||I|| = 1.
注: 以下只考虑consistend范数.
A2) 有反例: 让 O = [ [0.6, 0.8], [-0.8, 0.6] ]. 考虑无穷范数: ||A||inf.
(cf. unix文) 则: O*O~ = O~*O = I, ||O||inf = 1.4 > 1 |
|
a***a
发帖数: 434 | 45 比如默认是Inf 1.0, 0.9, 0.8, ...
怎么改为Inf, 1.0, 0.99, 0.98, 0.97, ...
求教~~thx！！ |
|
y*****t
发帖数: 1367 | 46 E(ln(X))=-inf because X=0 (i.e., ln(X)=-inf) with non-zero probability mass
of exp(-lambda) |
|
i********w
发帖数: 2223 | 47 有一个随机变量 (例如 20xx年电汽车电池的大小).
在目前没有什么有用相关信息的情况下(只知道随机变量应属于(0,+inf)), 我需要为它
假设一个看似合理的分布函数.
请问有什么推荐么?
1)我想用正态分布,但是不满足随机变量范围.请问有没有什么modified normal
distribution 符合这个要求吗?
2)Log-normal倒是符合随机变量的要求,可是我没有办法解释为什么选log-normal,不选
其他的
分布函数.(我现在也不知道还有哪些常用的分布函数,它的随机变量是(0,+inf)的).
多谢! |
|
P****D
发帖数: 11146 | 48 随机变量分布函数也好，世界上任何函数也好，在现实应用中不可能的那段你根本不用
管。比如你工作的公司宣布，凡是给海地捐款的，公司1:1 match。那这就是一个函数
，y（公司的捐款钱数）=x（你自己的捐款钱数）。y=x这个函数本身的定义域是(-inf,
+inf)没错。你觉得公司会操心如果你的捐款额是负数（在现实应用中不可能的情况）
怎么办？ |
|
|
a***r
发帖数: 420 | 50 在写一个小script（in R），有一段是：
sum=0
for (i in 1:n) {
temp=function(i)
sum=sum+temp
}
即是把n个function运算得到的值加起来，本来是很常规的步骤而已
问题是这个function（）得到的值是很小的，<10e-400，小于我OS能handle的了，值会
被记为O
于是我考虑log transform，问题是，transform之后，怎样求这些值的和呢？？
我找到一个解决类似问题的source code，它是这样写的：
logsum=-Inf
for (i in 1:n) {
temp=log(function(i))
if (temp>-Inf) {
logsum=temp+log(1+exp(logsum-temp))
}
}
我愚钝，没太整明白为啥可以这样近似。。。恳求达人讲解，或者其他可行的变换，非
常感谢~~！ |
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