e****t
发帖数: 766 | 1 If X and Y are Bivariate normal distribution
P(X>a | Y>b) , does it have a closed form ?
I guess no. |
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a*****t
发帖数: 28 | 2 hello everyone
There is anyone can tell me how to generate bivariate random sample base on
the density function( given), by using SAS or other software.
Thank you |
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a*****t
发帖数: 28 | 3 bivariate distribution is not use the inverse method |
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n*****n
发帖数: 3123 | 4 不对。两个都是bivariate normal.
x,y 是独立normal, 那么(x,y)^T是bivariate normal. 所有线性变换都是normal, 维
数要看变换. 你的第一种情况（x,x+y)^T = A*(x,y)^T, A=(1,0;1,1). 所以(x,x+y)^T
是bivariate normal.
同样x,y,z是独立normal, 那个(x,y,z)^T是tri-variate normal. 你的第二种情况是(x
+y,y+z)^T=B*(x,y,z)^T, B=(1,1,0,0,1,1)。所以(x+y,y+z)是bivariate normal.
我之前看错了，以为你说A,B是normal, A,B的correlation是rho. 那么(A,B)不是
bivariate normal. 这里关键是
x,y是normal, (x,y)不一定是bivariate normal.
但x,y是独立normal, (x,y)是bivariate normal. 然后再从线性变换推。 |
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o******e
发帖数: 1001 | 5 第一种情况：假定x,y为独立正态分布，求概率P(x>a,x+y>b)时，我们可以转化为P(A>a
,B>b),A,B为bivariate normal distribution.
第二种情况：假定x,y，z为独立正态分布，求概率P(x+y>a,y+z>b)时，我们也可以转化
为P(A>a,B>b),A,B为bivariate normal distribution,这样看上去第二种情况并不必第
一种复杂。
但是如果我们不转化为bivariate normal distribution,而直接用积分计算，第一种情
况两重积分就可以解，而第二种情况需要三重积分，后者要比前者复杂很多，这是为什
么？ |
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o******e
发帖数: 1001 | 6 谢谢上面的各位大牛，尤其是hezhi和ningyan,我现在已经搞明白什么是bivariate
normal distribution.
假设x,y为正态分布，A=ax+by,B=cx+dy,则(A,B)为bivariate normal distribution.第
二种情况不符合上面的式子，所以不是bivariate normal distribution.
再次感谢！ |
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n*****n
发帖数: 3123 | 7 The statement is not correct. I didn't take a look at your proof, but it
should be wrong in somewhere.
Consider the case, X,Y,Z are all standard normal. Y is independent of X and
Z, but X and Z are not independent. So (X, Y) are bivariate normal, (Y,Z)
are bivariate normal, but (X,Z) may not be bivariate normal. Any examples
showing that marginal normal does not imply joint normal can be used here as
counterexample. |
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h***i
发帖数: 3844 | 8 What is true is that if the random variable pair (X,Y) follows the bivariate
normal distribution, and Cov(X,Y) = 0, then X and Y must be independent.
But what is not true is that if each of X and Y is normally distributed, and
Cov(X,Y) = 0, then X and Y must be independent.
two random variables are each normally distributed, that does not
necessarily mean that they jointly follow a bivariate normal distribution
bell |
|
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c******s
发帖数: 90 | 9 Please see this link below.
http://www.noise.cz/sbra/sibram02/2-Ses/Fegan.htm
In this paper, the author present a way to generated correlated uniform R.V.
Another result you may be able to use is:
If X and Y are bivariate-normal with correlation rho,
Let Ux=normcdf(X) and Uy=normcdf(Y), then
Ux and Uy are bivariate-uniform with correlation
(6/pi)*arcsin(rho/2). |
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A*******s
发帖数: 3942 | 10 对啊，搞bivariate analysis来screen掉一堆variable还是有风险的。有些
insignificant in bivariate analysis的，也有可能become significant after
adjusted by other covariates.
不过在业界，至少在银行，还是不少人直接这么干的。大家都不太在乎，呵呵。其实直
接搞个forward/backward也不费事。 |
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h***i
发帖数: 3844 | 11 第二种情况：假定x,y，z为独立正态分布，求概率P(x+y>a,y+z>b)时，我们也可以转化
为P(A>a,B>b),A,B为bivariate normal distribution,这样看上去第二种情况并不必第
一种复杂。
兄弟，谁告诉你
A,B为bivariate normal distribution
你自己先把这个证明了以后，再用在你的计算上，然后说计算简单了，你明白自己省略
了什么了吧。
_{ |
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o******e
发帖数: 1001 | 12 我有看了一些东西，觉得你说的有道理。
现在看来只要能做linear transformation的都可以转化成bivarate normal
distribution. 那bivariate normal distribution符合传递率吗？
比如，(A,B),(B,C)都是bivariate normal distribution, 那(A,C)也是吗？
^T
(x |
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t****r
发帖数: 702 | 13 Let Z=-X, then (X,Y) bivariate normal and (Y,Z) bivariate normal does not im
ply (X,-X) is normal. |
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d********m
发帖数: 3662 | 14 大哥，不要这么认真啊。
首先，椭圆就是bivariate normal，下将军又推出了mean response of y，这个不是回
归是什么？
另外，线性回归也算高大上？这种东西在多元分析里算是最底层的东西了好吧。现在鬼
才用这个东西呢。 |
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s****u
发帖数: 1433 | 15 问题在于这不是线性回归。所以你装过缩了！
开个玩笑，莫生气。
: 大哥，不要这么认真啊。
: 首先，椭圆就是bivariate normal，下将军又推出了mean response of y，这个
不是回
: 归是什么？
: 另外，线性回归也算高大上？这种东西在多元分析里算是最底层的东西了好吧。
现在鬼
: 才用这个东西呢。
|
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g****g
发帖数: 1828 | 16 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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l***n
发帖数: 4 | 17 Assume X is Uniformly distributed on [0,1]; Y is Uniformly
distributed on [-a,a] (0
then what is the p.d.f or distribution function of Z = X + Y?
Or what is Probability{X+Y
I tried to use convolution method, but in case of z>1-a & z>a,
it doesnt work.
Please help me.
Thanks |
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y*****w
发帖数: 1350 | 18 这篇文章在搞统计的看来就是一坨屎
娘希匹，Number of Nobel Laureates per 10 Million Population必然不是normally
distributed，而总国家数又不足30，不能用central limit theorem，所以那linear
correlation就是个意淫，只能是Spearman correlation，就甭谈用什么linear
regression做prediction了。而且NEJM这个档次的文章只做一个bivariate
correlation，不是侮辱读者的智商么。 |
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k**********g
发帖数: 989 | 19
f =
@(x, c, s) (0.5 * x - 0.5 * sqrt ((x - c) .^ 2 + s) + 0.5 * c)
Input: "x", scalar or vector. Operation will be applied element-wise.
Configuration parameters:
(Param #1)
Corner position - "c" - scalar, positive.
For inputs "x" smaller than "c", asymptotic behavior is y = x.
For inputs "x" larger than "c", asymptotic behavior is y = c.
(Param #2)
Corner smoothness - "s" - scalar, positive.
For smaller values of "s", the curve (x, y) will look like a sharp angle
near the point (c, c).
For lar... 阅读全帖 |
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q*******n
发帖数: 1334 | 20 Assume X = [X1, X2]^t is bivariate normal distributed with mean vector m = [
m1, m2]^t and covariance matrix C = [c1, r; r, c2].
1. Define s = [s1, s2]^t as the dumb variable vector for MGF, then the MGF
of the vector X is
M_X(s) = E[exp(s^t * X)] = exp(m^t s + 1/2*s^t*C*s). (eqn. 1)
2. Set s1 = s2 = u in (eqn. 2), then we can get the MGF of X1+X2
M_{X1+X2}(u) = E[exp( (X1+X2)u ) = exp((m1+m2)u + 1/2*(c1+c2+2r)*u^2) (eqn.
2)
3. Eqn. 2 is in the form of the MGF for Normal RV. Therefore, X1+X2 is |
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h***i
发帖数: 3844 | 21 第一步就错了，两normal， joint不一定是bivariate normal
[
. |
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h**********k
发帖数: 168 | 22 how to express the following integral in terms of bivariate normal CDF?
Thanks.
\int_-\inf_+\inf{N(b-ax) N(c-ax)n(x)dx} ?
where N(.) is CDF and n(.) is PDF of standard normal distribution. |
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n********r
发帖数: 84 | 23 Let $\rho(u,v)$ denote the correlation of $u$ and $v$, and $t_1, t_2$
increasing functions. Given a bivariate normal distribution (x,y) where $\
rho(x,y)>0$, how does one show
\[
\rho(t_1(x),t_2(y)) \leq \rho(x,y)?
\] |
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n*e
发帖数: 50 | 24 bivariate uniform distribution? |
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o******e
发帖数: 1001 | 25 【 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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d*********a
发帖数: 255 | 26 【 以下文字转载自 Statistics 讨论区 】
发信人: dragoninsea (X+Y=Z), 信区: Statistics
标 题: 帮查一篇文章
发信站: BBS 未名空间站 (Fri Apr 22 12:28:33 2011, 美东)
Some new constructions of bivariate Weibull models
Author Info
Jye Lu
Gouri Bhattacharyya
谢谢 |
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B****n
发帖数: 11290 | 27 怎麼確定一定存在 我直覺覺得很可能找不到 任意兩個bivariate random variables都
要滿足距離d是同一分布 不管其它的random variables的位置 這實在有點不太可能 |
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q***t
发帖数: 21 | 28 假设0-->1和1-->2年是两个bivariate gaussian
correlation只能从-1到1，所以可以求出forward start option
的vol的上下限。 |
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h**********k
发帖数: 168 | 29 How to express the following integral in bivariate normal CDF/PDF?
\int_-\inf_+\inf{N(b-ax)N(c-ax)n(x)dx}
where N(.) is CDF and n(.) is PDF of standard normal distribution. |
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n********r
发帖数: 84 | 30 Let $\rho(u,v)$ denote the correlation of $u$ and $v$, and $t_1, t_2$
increasing functions. Given a bivariate normal distribution of (x,y) where $
\rho(x,y)>0$. How does one show
\[
\rho(t_1(x),t_2(y)) \leq \rho(x,y)?
\] |
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s*********o
发帖数: 13 | 31 Assume you have collected ten or 15 variables to use to predict a single
dependent variable using Multiple Regression. Suppose the results show
that
one of the variables does not have a strong relationship with the dependent
variable in the regression equation, but you notice that the variable has a
high bivariate correlation with the dependent variable.
Why do you think this might happen? What is there about the relationships
among some of the independent variables that might affect their |
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s*********e
发帖数: 1051 | 32 the question is a typical situtation of multicolinearity. if a independent
variable has high bivariate relation with dependent variable but is not
shown significant given other independent variables in the model, it implies
that this independent variable might have a high correlation with one or
some other independent variables. In a regression setting, multicolinearity
needs to be fixed. Otherwise, it will bring up inconsistency issue in the
parameter estimates.
by the way, i will take this typ |
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b***k
发帖数: 2673 | 33 ☆─────────────────────────────────────☆
heisenlin (heisenlin) 于 (Sun Sep 7 13:52:46 2008) 提到:
我这个问题很白痴，希望大家见量~
因为在做BIVARIATE OPTION PRICING BY USING LEVY PROCESS，用的是TREE MODEL，所
以要用MATLAB画一个三棱锥，其实就是一个3-D的BINOMIAL TREE，GRAPH只要有4个
STEPS就可以了。每一个NODE，要标出我的INTRINSIC VALUE。
其实这个图就是一个interpretation的作用，老板非逼我用MATLAB做。。。。
不好意思啊，因为自己写程序不用MATLAB，不太懂，麻烦大家了。
☆─────────────────────────────────────☆
heisenlin (heisenlin) 于 (Sun Sep 7 16:56:17 2008) 提到:
貌似我的问题太白痴。。。。大家都不屑回答。。。。。SIGH
☆──────────────── |
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n******r
发帖数: 1247 | 34 for joint bivariate normal, correlation=0 => independent
This is true
and |
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t*******e
发帖数: 172 | 35 Totally do not know your point. What is your claim?
What I claimed: this problem not doable without joint normally assumption.
You try to convince me any two normal distribution are jointly normal(am I
right?), which I am quite doubt about. e.g. I give you two random variable
which have correlation 0, but not indepent. This example show that there are
two N(0,1) are not jointly normal.
On the page 8 of the following notes, it maybe give the exactly argument.
http://www.athenasc.com/Bivariate-Nor |
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j*****4
发帖数: 292 | 36 let X=Max(X1,X2) and Bivariate Normal CDF F(x1,x2)
then F_X (x)=F(x,x)
take the derivative to get PDF. |
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m******g
发帖数: 12 | 37 Let Z = (X+Y)/sqrt(2), then X, Z is bivariate standard normal with
correlation 1/sqrt(2). So conditioning on value of Z, X has variable 1-
rho^2
= 1/2.
Actually conditioning on any value of X+Y, the conditional variance is
1/2.
By the way, the conditional mean is given by rho Z. In this case,
conditional on Z=1/sqrt(2), the mean of X is 1/sqrt(2)*1/sqrt(2) = 1/2.
In general, if X and Z are marginally standard normal, with correlation
rho,
then conditional on Z=z, X is normal with mean rho*z, and |
|
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t*****e
发帖数: 38 | 39 约好2点， 1点55找了个没人的位置（很难找），等到2点15分，还没来电话。回到工作
岗位，突然打来了。没有道歉，直接开始。
1.what is your strongest area? math,statistics,然后从统计开始
2. give me 3 important result of prob theory. This was OK.
3. what is the prob that A
result after computation, any other way, I gave an intuitive way, any other
way, I said sorry, he said" I have a better way I thought you could have
used,I am a little bit disappointed"
4. How could you generate a normal from uniform? I said use sum... 阅读全帖 |
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j********t
发帖数: 97 | 41 Consider Z = exp(lnX + lnY). lnX, lnY are normal and correlated, I guess (
lnX, lnY) is a joint normal, then lnX+lnY is normal. Utilize moment
generating function of normal to calculate E(Z) and E(Z^2).
But I'm not sure how to get corr(lnX, lnY) from corr(X, Y)=\rho and prove (lnX, lnY) is a bivariate normal? Anyone can help? Thanks! |
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f*******g
发帖数: 377 | 42 尽管写的是corr(X, Y)=\rho
但这个rho应该指的就是corr(lnX, lnY)
要不然不太好办
(lnX, lnY) is a bivariate normal? Anyone can help? Thanks! |
|
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m******4
发帖数: 12 | 44 bivariate normal, correlation rho, mean 0 and s.d. 1.
what's the distribution of the max?
thanks |
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C******n
发帖数: 9204 | 45 本质上就是conditional normal dist。我面的时候也问过bivariate normal, given一
个值，另一个多少。 |
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f*********5
发帖数: 367 | 46 已知X和Y都是N(0,1)分布的随机变量，但不知道他们俩是否独立还是有怎么的相关性，
连是否bivariate normal distribution都不知道，问能否确定
E[X|X+Y=1]？
我自己简单验证当X=Y或者X和Y独立的时候答案是0.5，但是能否证明无论他俩满足什么
相关性答案都是0.5？或者有没有大牛给个答案不是0.5的反例？
百思不得其解，求解惑 |
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发帖数: 1 | 47 let X=w1, and Y=w2-w1
P(w2>0|w1>0)=P(Y>-X|X>0)
X and Y are bivariate normal with 0 correlation.
P(Y>-X|X>0)=P(Y>-X,X>0)/P(X>0)
P(X>0)=1/2
P(Y>-X|X>0)=3/8
P(w2>0|w1>0)=3/8/(1/2)=3/4 |
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s*********o
发帖数: 13 | 48 Assume you have collected ten or 15 variables to use to predict a single
dependent variable using Multiple Regression. Suppose the results show
that
one of the variables does not have a strong relationship with the dependent
variable in the regression equation, but you notice that the variable has a
high bivariate correlation with the dependent variable.
Why do you think this might happen? What is there about the relationships
among some of the independent variables that might affect their |
|
o******e
发帖数: 1001 | 49 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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