jl
发帖数: 398 | 1 A random varible converge to -\infty in provability
是什么定义? 多谢!
A random varible converge to -\infty with probability 1 s
是什么定义? 多谢! |
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p***c
发帖数: 2403 | 2 只有一个r.v.，收敛个屁啊
如果是一列
1. for all K>0, \lim_n P(X_n<-K)=1
2. P({w|\lim_n X_n(w)=-\infty})=1 |
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b********e
发帖数: 58 | 3 \begin{equation}
\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty}
e^{-y^2}dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
e^{-(x^2+y^2)}dxdy =\int_{0}^{2\pi}\int_{0}^{\infty} e^{-\rho^2}
\rho d\rho d\theta = 2\pi /2 = \pi
\end{equation}
Notice:
\begin{equation}
\int_{-\infty}^{\infty} e^{-x^2}dx =\int_{-\infty}^{\infty}
e^{-y^2}dy
\end{equation}
we prove the conclusion that
\begin{equation}
\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}
\end{equation} |
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l******i
发帖数: 1404 | 4 直接说说那个扩展吧：
让S_k=能够扔出k个不同值的次数，k<=n。题目要求E[S_k]。
任意给定N>=k，现在先考虑P(S_k=N)。
P(S_k=N)
=1*P(任意给定第N次结果的值，前面N-1次结果构成与第N次不同的k-1个值)
=(C_{n-1}^{k-1})*((k-1)^(N-1))/(n^(N-1))
=(C_{n-1}^{k-1})*(((k-1)/n)^{N-1});
现在求期望：
E[S_k]
=sum_{N=k}^{\infty}[N*P(S_k=N)]
=sum_{N=k}^{\infty}[N*(C_{n-1}^{k-1})*(((k-1)/n)^{N-1})]
=(C_{n-1}^{k-1})*sum_{N=k}^{\infty}[N*(((k-1)/n)^{N-1})]
=(C_{n-1}^{k-1})*P
这里P=sum_{N=k}^{\infty}[N*(((k-1)/n)^{N-1})];
那么((k-1)/n)*P = sum_{N=k}^{\infty}[N*(((k-1)/n)^N)], denoted by (1)
另一方面： P = s... 阅读全帖 |
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x*****i
发帖数: 846 | 5 K(x)是个偶函数，
\int_{-\infty}^{\infty}K(x)dx=1
并且
\int_{-\infty}^{\infty}|x|K(x)dx<\infty
能不能推出
\int_{-\infty}^{\infty}x^2K(x)dx<\infty |
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x******a
发帖数: 6336 | 6 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 |
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M****i
发帖数: 58 | 7 My idea:
If the function c(t) verifies:
i) c(t)>=0;
ii) \int_0^s c(t) dt<\infty, for every s>=0;
iii) \int_0^\infty c(t) dt=\infty;
iv) \int_0^\infty c(t)^2 dt<\infty.
Then the integral
I(t)=\int_0^t c(s)*exp(-\int_s^t c(r)dr) dB(s)
converges to 0 in probability.
Example: c(t) = 1/(t+1).
Proof: The idea is to show that I(t) is L^2 convergent to 0, which implies
convergence in probability. For this purpose,
by Ito isometry (since iv) holds) we get
E(I(t)^2)
=\int_0^t c(s)^2*\exp(-2\int_s^t c(r) ... 阅读全帖 |
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k*****y
发帖数: 744 | 8 Thanks for noting that.
The integral on (-infty, infty) can be divided into two parts, (-infty, 0)
and (0, infty), which will be the same. That is where the 2 comes from.
When x is in (0, infty), then y = (x-1/x) covers the range of (-infty, infty
). |
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d***a
发帖数: 13752 | 9 呵呵，能不能这样做？
0+0+0 = lim_{xtoinfty} 6/x
lim_{x to infty} (1+1+1)/x = lim_{x to infty} 6/x
lim_{x to infty} (2+2+2)/x = lim_{x to infty} 6/x
...
lim_{x to infty} (9+9+9)/x = lim_{x to infty} 6/x |
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M*****e
发帖数: 11621 | 10 我不知道你们用的什么书，有些词可能不一样。
你们应该在引进measure之后证明/quote了一些定理，其中一条是说
Measure has continuity from above.
Let \mu be the measure. If you have events A_i with A_i \downarrow A, then \
mu(A_i) \downarrow \mu(A).
In this case, let's consider A_i = (-\infty, x+1/i], then we have
A_i \downward (-\infty, x]. According to the theorem, we have
\lim_{i\to\infty}F(x+1/i)=\lim_{i\to \infty} P((-\infty,x+1/i]) = P((-\infty
, x]) = F(x)
Now we only have to argue it is true when we replace 1/i by positive h. |
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o**a
发帖数: 76 | 11 下面这个积分一开始是正的，但是分部积分后就变负的了: ( Re z > 0)
\int_0^{\infty} 2\eta/(\eta^2+z^2) * 1/(e^{2\pi\eta}-1) d\eta
(假如z是很小的正实数，上面每一项都是正的，所以积分是正的)
=1/\pi \int_0^{\infty} \eta/(\eta^2+z^2) * 2\pi/(e^{2\pi\eta}-1) d\eta
=1/\pi \int_0^{\infty} \eta/(\eta^2+z^2) * 2\pi e^{-2\pi\eta}/(1-e^{-2\pi\eta}
) d\eta
=1/\pi \int_0^{\infty} \eta/(\eta^2+z^2) d(log(1-e^{-2\pi\eta}))
=1/\pi \eta/(\eta^2+z^2) * log(1-e^{-2\pi\eta})|_{\eta from 0 to \infty}
-1/\pi \int_0^{\infty} log(1-e^{-2\pi\eta}) (\eta^2+z^2-2\eta^2)/(\eta^2 |
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b*k
发帖数: 27 | 12 证明级数 {Un/Sn} 发散 means
\sum_{n = 1}^{\infty} Un/Sn 发散
Assume U_n > 0 and
lim_{n->\infty} S_n = \infty
since Un = S_n-S_{n-1}
, intuitively, sum U_n/S_n = sum (S_n-S_{n-1})/S_n
-> \int d(S_n)/S_n = ln(S_n)-> \infty
Strict prove: Use Abel lemma:
Let b1 >= b2 >= ... >= bn >= 0
m <= a1 + a2 ... +an <= M
Then
m*b1 <= a1*b1 + a1*b2+ ...+an*bn <= b1*M
Let bk = 1/Sk, ak = Uk
then Sn = a1+...+an and m < Sn < M
the m*1/S1<= sum {Un/Sn}, when n -> \inft
m -> infty ==> sum(Un/Sn} ->\infty
, which diverges |
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o******e
发帖数: 1001 | 13 有三个变量x, y and z. both f(x,y) and f(y,z) are bivariate normal
distributions. x and z are independent. what is the probability P(x
z
Solution 1:
Given the conditions, f(x,y,z) is multivariate normal distribution with the
correlation of zero for x and z. Then P(x
_{\infty}^K2\int_{\infty}^K3 f(x,y,z)dxdydz
Solution 2:
P(x
K3 f(y,z)dz
Which is right? But I guess both are wrong. A |
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o******e
发帖数: 1001 | 14 没有啊！请看，
P(x
P(x+y
-\infty}^{b-y}f(z)dz
这是不按照bivariate normal distribution,纯从独立变量积分的。
X+ |
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g****g
发帖数: 1828 | 15 In probability theory, the normal (or Gaussian) distribution, is a
continuous probability distribution that is often used as a first
approximation to describe real-valued random variables that tend to cluster
around a single mean value. The graph of the associated probability density
function is “bell”-shaped, and is known as the Gaussian function or bell
curve:[nb 1]
f(x) = \tfrac{1}{\sqrt{2\pi\sigma^2}}\; e^{ -\frac{(x-\mu)^2}{2\sigma^2}
},
where parameter μ is the mean (location of the pe... 阅读全帖 |
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w******d
发帖数: 1301 | 16 为了阐述基督信仰是“一种基于上帝是大能大爱全知的前提上的理性”，我只好再拿数
学引入无穷大概念作类比了。
我估计你应该用过latex, 所以我就用latex写相关的公式，这样交流比较方便。
（古狗大学的神童可以查http://omega.albany.edu:8008/Symbols.html）
By definition, the property of infinity is defined by:
\infty > x, \forall x \in R
现代数学的理性就是基于这个无穷大的定义上的理性。 根据这个定义，
数学家就可以接受
\infty +1 = \infty
这样的奇怪的等式。
基于这个无穷大的定义上的理性，数学家就不会强行将消元法应用到这个等式上来消元
无穷大。一切数学推导都很和谐，数学大厦不会崩溃。
如果某民科闭着眼睛就是不肯理解无穷大的定义，一看到数学家列出
\infty +1 = \infty
就强行要用消元法把无穷大消元，然后振振有词地说，你看，因为你们引进了无穷大这
样非理性的概念，数学推导就会导致1=0的非理性结果，这不是显然的么？ 你们为什
么睁眼说... 阅读全帖 |
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o**a
发帖数: 76 | 17 我知道了，我的无穷乘积是从-\infty到\infty
实际上
cos(\sqrt(z))=\prod_{1}^{\infty} (1-z/(n*pi-pi/2)^2)
所以我得到了含根号的式子：
cos(\sqrt(z))=\sqrt[\prod_{-\infty}^{\infty} (1-z/(n*pi-pi/2)^2)] |
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x******i
发帖数: 3022 | 18
this sum can probably be done analytically.
at least a simpler one can be (if a>0 & b>0)
\sum_{-\infty < n <+\infty} 1/((n-ia)^2-b^2)
=> \iint_{-\infty}^{+\infty} dn*dt exp(-2*pi*i*n*t) 1/((n-ia)^2-b^2)
=> \int_0^{+\infty} dt cos(2*pi*b*t)*exp(-2*pi*a*t)/b
=> something easy |
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c*****t
发帖数: 520 | 19 考虑f(x,t)在0 到\infty的积分I(t)。即：
I(t) = \int^{\infty}_{0} f(x,t)dx .
其中f(x,t)是x 和t 的连续函数。f(x,t)>0 for any x>0, t>0.
f(0,t)=0，for any t>0.
对于所有有限的t>0 ，积分I(t)收敛。特别地，I(0)也收敛。
目标是要求证I(t)有界。
已知f(x,t)可以对t 求偏导数。df/dt(x,t)=g(x,t)连续for any x>0, t>0.
f(x,t)对x 也连续可微。
而且在0 到\infty上对g(x,t)形式上做积分，如果积分收敛，那么利用f(0,t)=0做分部
积分的结果非正。即：
J(t)= \int^{\infty}_{0} g(x,t)dx \leq 0, for any t>0.
但是实际上我不知道J(t)是否收敛。只是形式上用分部积分，结果是非正函数的积分，
当然结果可能是-\infty 。
而且我不知道J(t)是否在某个区域[t1,t2]上一致收敛。因此无法保证对I(t)求偏导可... 阅读全帖 |
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d*e
发帖数: 843 | 20 给定任意的一个单调增函数满足
g(t)<=t/2 for all t
g'(t)<1 for all t
g(t)->\infty，as t->\infty
可以构造函数 b 满足以下两个条件：
(i) b(\infty)=\infty
(ii) b(t)-b(g(t))=0
C是常数
可假设g光滑性，g，b皆为正值函数。 |
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d*e
发帖数: 843 | 21 Consider a stochastic integral I(t) = int_0^t c(s)a(t,s)dW(s),
where
1) c(s) (>=0) is a function such that int_0^\infty c(s)ds = \infty and
int_0^\infty c^2(s)ds < \infty
2) a(t,s) is a deterministic function satisfying
|a(t,s)| <= e^{-\int_s^t c(r)dr}
Q: show that I(t)->0 almost surely as t -> \infty
Note: if a(t,s) = e^{-\int_s^t c(r)dr}, the problem can be solved by the
method given by "MathFi" in this post
http://www.mitbbs.com/article/Quant/31266465_0.html
But it seems that the same approac... 阅读全帖 |
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f**n
发帖数: 401 | 22 Define A_n=\sum_{i=1}^n a_i, B_n=\sum_{i=1}^n b_i in which a_i and b_i are
both 无穷小量 or equivalently:
\lim_{n->\infty}a_i=\lim_{n->\infty}b_i=0
Moreover for the same i, a_i and b_i are 等价无穷小 or equivalently:
\lim_{n->\infty}a_i/b_i=1, 1<=i<=n
Now we want to prove:
\lim_{n->\infty}A_n=\lim_{n->\infty}B_n if either of the limits exists.
The above equation seems intuitively true...to me. But
it seems that the proof requires the exchange of \lim and \sum.
I am not sure the requirements for this exch |
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g*********n
发帖数: 43 | 23 跟风， 做一个类C++ 的version.
typedef struct
{
bool can;
int mincount;
int endind;
} solelem;
if (s==null || s.len ==0 ) throw invalidinput;
soelem solutions[s.len] = {false, infty, -1};
for (startind=s.len-1; startind >=0; startind--)
{
if (dict.contains(s.substr(startind)))
{
sol[startind] = {true, 1, s.len-1};
}
else
{
mincount = infty;
for (i=startind; i<=s.len-2; i++)
{
if (dict.contains(s.substr(startind, i)))
... 阅读全帖 |
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a**e
发帖数: 5794 | 24 如果你相信耶稣的话，就会知道下面的说法至少是不精确的，
\infty + 1 = \infty
准确一点的是
o(\infty + 1) = o(\infty)
o(1) = o(2) 也没问题。 |
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M*****e
发帖数: 11621 | 25 你让我算的是 = infty
不是 < infty 阿
= infty 是 0 <=>
没错 |
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M*****e
发帖数: 11621 | 26 是啊，没错阿。
你看题目问的是 P(\tau = \infty), 我们算的是0
答案给的是P(\tau < \infty) = 1
P(\tau < \infty) = 1 - P(\tau = \infty)
对的 |
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M*****e
发帖数: 11621 | 27 跟他说
{\tau = \infty } = {\tau < \infty} 的补集
所以
P(\tau = \infty) = 1 - P(\tau < infty)
何况原题问得就是左边那个
如果他还说你错
你去找老师
找系主任
找院主任
找校长
找参议员
找first lady |
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T*******n
发帖数: 493 | 28 default in-line math $\sum_{i=0}^\infty x_i$
override in-line math $\displaystyle\sum_{i=0}^\infty x_i$
default display math \begin{equation} \sum_{i=0}^\infty x_i \end{equation}
override display math \begin{equation} \textstyle\sum_{i=0}^\infty x_i \end{
equation} |
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o**a
发帖数: 76 | 29 这里是tex文件，可以直接latex编译:
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\begin{align*}
&\int_0^{\infty} \frac{2\eta}{\eta^2+z^2} \cdot \frac{d\eta}{e^{2\pi\eta}-1} \
\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot \frac{2\pi}{e^{2
\pi\eta}-1} d\eta \\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot \frac{2\pi e^{-2
\pi\eta}}{1-e^{-2\pi\eta}} d\eta\\
&=\frac{1}{\pi} \int_0^{\infty} \frac{\eta}{\eta^2+z^2} \cdot d(\log(1-e^{-2\
pi\eta} |
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L********3
发帖数: 204 | 30 f(x):[0,\infty)->R^+, f is strictly increasing and strictly convex, i.e.
f>0, f'>0, f''>=0 on (0,\infty). Moreover f(0)=0, f'(0)=0.
Show that: 2x^2f''(x)-xf'(x)+f(x)>0 on (0,\infty).
Counterexample is wellcome as well.
You can also relax the condition on the domain
(instead of on (0,\infty), any interval (0, a), a>0 is ok).
You can also neglect the condition f'(0)=0.
Thanks. |
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q*d
发帖数: 22178 | 31 发信人: athos (citr), 信区: Physics
标 题: [转载] 请教一个数学问题，急
发信站: The unknown SPACE (Fri Jul 18 23:20:19 2003), 站内信件
有一个定理，据说来自
Bellman, R. and K. L. Cooke (1963). Differential-Difference Equations. Acade
mic Press, New York
说，对f(s) = 1 + r(s) exp(-L s)，f(s) = 0 的根渐进逼近 f0(s) = 0 的根，其中，
f0(s) = 1 + r(\infty) exp(-L s)。
请问，如果今有g(s) = 1 + r1(s) exp(-L1 s) + r2(s) exp(-L2 s)，是否还有类似的
结论，g(s) = 0 的解渐进逼近g0(s) = 0的解？
其中g0(s) = 1 + r1(\infty)exp(-L1 s) + r2(\infty)exp(-L2 s)
\infty指无穷大，r,r1,r2都是s的有理分式，L,L1,L2都是非 |
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l****o
发帖数: 2909 | 32 First of all, |W(t)| is F(t)-measurable, where F(t) is generated by W(t).
Secondly, E[\int_0^T|W_t|^2 dt] =\int_0^T E[|W_t|^2] dt=\int_0^T t dt=T^2/2
<\infty. This proves that it sufficiently satisfy ITO itegration
definition. Therefore it is a martingale. Here you can use standard machine and
L^P space definition for more rigorous proof on how to construct a general integrand from simple integrand.
Thirdly, E[[\int_0^T|W_t| dw(t)]^2]=E[\int_0^T|W_t|^2 dt] =\int_0^T E[|W_t|^
2] dt=\int_0^T t ... 阅读全帖 |
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l******i
发帖数: 1404 | 33 要用Monotone Convegence Theorem来证E[1_{\tau<\infty}]=1，
where \tau=the first passage time to hit 1 or -1.
1_{\tau<\infty}=1 if \tau<\infty; 1_{\tau<\infty}=0 otherwise.
具体过程：Shreve's Book, Chapter 3, Section 6有详细证明。
请问这是哪家的什么职位的面试题？谢谢楼主共享。 |
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s********r
发帖数: 529 | 34 我想到一个办法，但是不知道是不是最简单，可能也有纰漏：
这个积分是乘以\exp(-2\sqrt{ac})后，变成\exp{-(ax+\dfrac{c}{x})^2}dx
进行变量代换，令x=\dfrac{c}{au},u是新的积分变量，则原来的积分=\dfrac{-c}{au^
2}\exp{-(au+\dfrac{c}{u})^2},再把u换成x，这个积分和原积分相等
令原来的积分等于C，则2C=\int{-\infty}{+\infty}(1-\dfrac{c}{ax^2}\exp{-(ax+\d
frac{c}{x})^2})dx,可以将e指数前面的一项提出1/a后放入dx内，则积分问题退化为\i
nt_{-\infty}^{+\infty}exp{-x^2}dx
我这个方法可能存在积分上下限的问题，明天还要考试，就当抛砖引玉吧 |
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H*******s
发帖数: 106 | 35 tau be the stopping at -10 or +infty.
So shouldn't it be M_0=E[M_{tau}]=p*a^{-10}+(1-p)*a^{+infty} ? (The last one
is +infty, not -infty)
S |
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s***e
发帖数: 911 | 36
以电动力学为例最好讲. 从Maxwell方程组可以导出真空的电磁波方程(也就是通常
的波动方程比如弦振动方程). 这方程的普遍解可以被证明具有如下形式:
f(x-c*t), 或者f(x+c*t). 正负号表示波的传播方向不同. 以下以右传波f(x-c*t)
为例. p(t)=x-c*t 就是这个波的"相". 相的传播条件是p(t)=constant, 于是就得到
相的传播速度: (d/dt)p(t)=0=>(d/dt)x=c. 因此c就是相速度. 对真空中的波动方程来
说, 速度是被良好定义了的, 无所谓什么群速度相速度的差别, 可以统统称为波速.
下面对f(x-v*t)作傅立叶变化:
f(x-v*t)=(1/Sqrt[2 Pi])*Integrate[g(k)*Exp[I*k*(x-v*t)],{k,-Infty,Infty}]
=(1/Sqrt[2 Pi])*Integrate[g(k)*Exp[I*(k*x-w*t)],{k,-Infty,Infty}]
其中w=k*c, 而c是常数. 频率w和波矢k的函数关系称为色散关系. 这里是线性关系,
意味着各个不同频率 |
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f**********d
发帖数: 4960 | 37 今天上午发个构造序列的帖子，为逗逗闷子。
因为在外面用手机发帖，书写上不太严谨，就此给参与讨论的各位道歉。
现重写如下。
这个例子对分析一些随机过程在其测度空间上的采样有关，所以有重要的应用意义。
想构造有限长序列 a = [a_{1}, a_{2}, ..., a_{k}, ..., a_{K}], 共有K个元素, K<
+\infty，满足以下要求：
1. sum_{k=1}^{K} a_{k} = 1, 且在K->+\infty时也成立。
2. lim_{K->+\infty} max_{1 \le k \le K} a_{k} = 0
一个简单的这样的例子是[1/K, ..., 1/K] |
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M*****e
发帖数: 11621 | 38 你可以这样理解，但不能这么写。
对于任意y_1 > y_2..... >0, with lim_{i\to \infty} y_i = 0
(-\infty, x]是 所有(-\infty, x+y_i]的交集
但是你不能把 y_i直接换成h, 因为在mearuable space的定义里面
the intersection of countably many events is an event
but the intersection of arbitrarily many events is not necessarily so
所以你的证明里引用的那个continuity from above property也是只适用于countably
many events的。
靠，老子中英文夹杂真恶心。 |
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T*******x
发帖数: 8565 | 39
Suppose r,s are in E, and 0
For any t \in (r,s), we want to show f \in L^t.
Let F0={x: |f(x)|<1}, F1={x: |f(x)|>=1},
then \int_X |f|^t = (\int_{F0} + \int_{F1}) |f|^t
\int_{F0} |f|^t <= \int_{F0} |f|^r dx < \infty,
\int_{F1} |f|^t <= \int_{F1} |f|^s dx < \infty,
so \int_X |f|^t dx < \infty, so f \in L^t.
这个连通性是Rudin习题上要证明的，不然也不容易往这方面想。 |
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d****n
发帖数: 12461 | 40 来自主题: Mathematics版 - 赌徒问题续 大家还是局限在数学范围内讨论吧，数学版么。
开始我认为这个和零常返类似，即P(赢)=1，E(赌博次数)=\infty
现在发现不仅P(赢)=1，而且只要每次赌徒有大于正常数概率赢，即P(赢一次)>c>0，则E(
赌博次数)<\infty，所以P(赌无穷次)=0，“不可能”出现无穷赌资的情况。但E(赌资)=\
infty
大家在下结论之前最好事先自己算算。 |
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d********f
发帖数: 8289 | 41 F(p) = int_0^\infty e^{pt} f(t) dt
the initial value and final value theorem say
pF(p) -> f(0) as p -> \infty
pF(p) -> f(\infty) as p ->0
but in the proof, f is assume to be differentiable,
I was wondering if we only assume f is continuous,
do the two theorems still hold?
how to prove them? |
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b****d
发帖数: 1311 | 42
Suppose x_n --> x. For each n there is y_{n,m}--> x_n and f_m(y_{n,m})-->f(x
_n) as m-->infty. Choose k_n>k_{n-1} (inductively) such that
|y_{n,k_n}-x_n|< 2^{-n} and |f_{k_n}(y_{n,k_n})-f(x_n)| < 2^{-n}.
Then by assumption, y_{n,k_n}-->x and f_{k_n}(y_{n,k_n})-->f(x) as n-->infty
.
This implies that f(x_n)-->f(x) as n-->infty. |
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x******n
发帖数: 24 | 43 $\infty$ is a removable singularity of z*f(z) and you can give value 0 i
n Riemaanian sphere C\cup {\infty}, which give you $\infty$ is a removab
le singulary of f(z), then your result comes from Cauchy's integral theo
rem.
z) |
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h***l
发帖数: 3048 | 44 现在有这样一个函数f(x(t)),t 在[0,\infty)上定义。
要求这样的性质：
如果x(t)恒为0，则f(x(t))恒为0。
如果x(0)=0,但是中间x(t)变化，最后x又回到0,则f(x(t))不为0。
这个似乎要求f()具有记忆功能。像积分：\int_0^\infty x(t)dt
和 \max_0^\infty x(t) 具有这样的功能，大家能不能说说还有哪些
函数具有这样的功能？ |
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w**a
发帖数: 1024 | 45 in physics, we know:
the integration of \exp{i*k*x}dx, i=\sqrt{-1}, on the real line from -\infty
to +\infty
is a delta function I=2*\pi*\delta(k).
The usual way to show this identity is to use
Fourier transform of a delta function then study its inverse transform.
Here is my question: integrate the same integrand, but from x=0 to +\infty.
Shall we get one-half of I (above value)? How to show this? I really want to
use distribution technique to show this. Thank you. |
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o******e
发帖数: 1001 | 46 【 以下文字转载自 Statistics 讨论区 】
发信人: onehouse (whitehouse), 信区: Statistics
标 题: 问一个概率问题
发信站: BBS 未名空间站 (Tue Jan 6 21:43:42 2009)
有三个变量x, y and z. both f(x,y) and f(y,z) are bivariate normal
distributions. x and z are independent. what is the probability P(x
z
Solution 1:
Given the conditions, f(x,y,z) is multivariate normal distribution with the
correlation of zero for x and z. Then P(x
_{\infty}^K2\int_{\infty}^K3 f(x,y,z)dxdydz
Solution 2:
P(x |
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m*******s
发帖数: 3142 | 47 能否详细说一下什么地方会遇到 \infty - \infty？
我只记得有些时候会限制集合的测度有限，有时候又说 \infty等式也成立 |
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t******g
发帖数: 1136 | 48 Can anyone give me any hint to following problem?
Thanks.
Suppose that $f(x)$ is continuous at $x_0$. And suppose $\{a_n\}$
and $\{b_n\}$ are any two sequences such that $a_n
$\lim_{n\rightarrow \infty}a_n=\lim_{n\rightarrow \infty}b_n=x_0$.
And
\[
\lim_{n\rightarrow \infty} \frac{f(b_n)-f(a_n)}{b_n-a_n}=L.
\]
Can we approve $f'(x_0)$ exists, and equal to $L$? Or find out a
counter example to show $f'(x_0)$ does not exist. |
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jl
发帖数: 398 | 49 \sum_{i=2 to \infty} 1/i 不收敛.
\sum_{i=2}^{\infty} 1/ [i(log i)^1.0001] < \infty 对吗?
1/[i(log i)^1.0001] 比 1/i 收敛的快一点点, 这个级数和收敛吗?
多谢! |
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