f**********g
发帖数: 107 | 1 如果两个R.V.是independent，那么结果肯定是normal。如果dependent的话感觉不是，
但翻遍手头的所有书都没找到参考。麻烦知道的顺便告知参考文献。 |
f**********g
发帖数: 107 | 2 Good counter example, but can we also say 0 follows normal(0,0)? |
Q******g
发帖数: 607 | 3 following that example, you can make anything you like.
why has to be normal?
【在 f**********g 的大作中提到】
: Good counter example, but can we also say 0 follows normal(0,0)?
|
f**********g
发帖数: 107 | 4 I mean that if X & Y normal, and X = -Y, then Z=X+Y=0 is still a normal. |
d****r
发帖数: 135 | 5 independent 的话一定是normal。你可以随便设出两个normal distribution r.v.，然
后利用independence得到两个r.v.的乘积density function，进一步用积分计算他们的
和的density function，你会发现还是一个normal的density function。这个计算不难。
楼上给出的例子不是一个反例。因为nomarl distribution 有一个极限情况，可以被认
为是variance趋向无穷的情况。而那个反例恰好就是这样子的，它的density function
是在0点blow up，其它点值为0，这个在一些讲广义函数的书里被叫做delta 函数。因
为正规normal distribution实际上是热方程在一维情况下，初值为delta函数的基本解
的图像（似乎差个scalar），因此delta 函数恰好就是一个极限状态的normal
distribution。
【在 f**********g 的大作中提到】
: 如果两个R.V.是independent，那么结果肯定是normal。如果dependent的话感觉不是，
: 但翻遍手头的所有书都没找到参考。麻烦知道的顺便告知参考文献。
|
f******k
发帖数: 297 | 6 no.
X, Y normal, Y=-X if |X|>=1, Y=X if |X|<1.
【在 f**********g 的大作中提到】
: 如果两个R.V.是independent，那么结果肯定是normal。如果dependent的话感觉不是，
: 但翻遍手头的所有书都没找到参考。麻烦知道的顺便告知参考文献。
|
s*j
发帖数: 7 | 7 这个Y还是normal吗？
【在 f******k 的大作中提到】
: no.
: X, Y normal, Y=-X if |X|>=1, Y=X if |X|<1.
|
A*******r
发帖数: 768 | 8 http://www.springerlink.com/content/x453xu/?p=b2b422921b154f5c8a9974cb3c6b0e66&pi=1
Statistical Physics
Including Applications to Condensed Matter
Publisher Springer New York
DOI 10.1007/b106783
Copyright 2005
%%%%%%%%%%%%%%%%%%%%%%%%%%
Page 33
%%%%%%%%%%%%%%%%%%%%%%%%%%
Hence, the sum of two normal random variables is always (even if they are
mutually
dependent) another normal random variable. |
h***i
发帖数: 3844 | 9 强啊
【在 A*******r 的大作中提到】
: http://www.springerlink.com/content/x453xu/?p=b2b422921b154f5c8a9974cb3c6b0e66&pi=1
: Statistical Physics
: Including Applications to Condensed Matter
: Publisher Springer New York
: DOI 10.1007/b106783
: Copyright 2005
: %%%%%%%%%%%%%%%%%%%%%%%%%%
: Page 33
: %%%%%%%%%%%%%%%%%%%%%%%%%%
: Hence, the sum of two normal random variables is always (even if they are
|
b******k
发帖数: 58 | 10 The sum of two normally distributed R.V. is still normal, and the
correlation is integrated to the variance of the sum. HTH. |
|
|
A****e
发帖数: 44 | 11 I was wrong.thanks for pointing out.. :)
难。
function
【在 d****r 的大作中提到】
: independent 的话一定是normal。你可以随便设出两个normal distribution r.v.，然
: 后利用independence得到两个r.v.的乘积density function，进一步用积分计算他们的
: 和的density function，你会发现还是一个normal的density function。这个计算不难。
: 楼上给出的例子不是一个反例。因为nomarl distribution 有一个极限情况，可以被认
: 为是variance趋向无穷的情况。而那个反例恰好就是这样子的，它的density function
: 是在0点blow up，其它点值为0，这个在一些讲广义函数的书里被叫做delta 函数。因
: 为正规normal distribution实际上是热方程在一维情况下，初值为delta函数的基本解
: 的图像（似乎差个scalar），因此delta 函数恰好就是一个极限状态的normal
: distribution。
|
f******k
发帖数: 297 | 12 Yes, it is.
【在 s*j 的大作中提到】
: 这个Y还是normal吗？
|
H****h
发帖数: 1037 | 13 书错了。
【在 A*******r 的大作中提到】
: http://www.springerlink.com/content/x453xu/?p=b2b422921b154f5c8a9974cb3c6b0e66&pi=1
: Statistical Physics
: Including Applications to Condensed Matter
: Publisher Springer New York
: DOI 10.1007/b106783
: Copyright 2005
: %%%%%%%%%%%%%%%%%%%%%%%%%%
: Page 33
: %%%%%%%%%%%%%%%%%%%%%%%%%%
: Hence, the sum of two normal random variables is always (even if they are
|
H****h
发帖数: 1037 | 14 对
【在 f******k 的大作中提到】
: no.
: X, Y normal, Y=-X if |X|>=1, Y=X if |X|<1.
|
p*****e
发帖数: 15 | 15 We treat the 0 variance/covariance case as normal.
X is a random vector with n components.
Then X is multivariate normal if and only if
LX is univariate normal, for every 1 x n vector L.
See any general multivariate statistics textbook.
Therefore, the sum of two normals (indep. or not) is still normal if
we choose L=(1, 1). |
q*******n
发帖数: 1334 | 16 The sum is still normal, regardless of the dependence of the two summands.
This can be easily proved by using moment generating function.
【在 f**********g 的大作中提到】
: 如果两个R.V.是independent，那么结果肯定是normal。如果dependent的话感觉不是，
: 但翻遍手头的所有书都没找到参考。麻烦知道的顺便告知参考文献。
|
h***i
发帖数: 3844 | 17 pls show here
【在 q*******n 的大作中提到】
: The sum is still normal, regardless of the dependence of the two summands.
: This can be easily proved by using moment generating function.
|
g****e
发帖数: 1829 | 18 A lot of people have posted their answers. I think there is a point that one
should always bear in mind, (my personal opinion):
If X & Y are not independent and both are Normal distribution, we can draw a
conclusion: it must be LINEAR dependence, which means you can represent Y
~~~~~~
per follow up posts, this could be wrong (and I didn't really prove it).
in a linear form of X. the question then reduces to:
If X & Y are linearly dependent, is their sum still a normal? An
【在 f**********g 的大作中提到】
: 如果两个R.V.是independent，那么结果肯定是normal。如果dependent的话感觉不是，
: 但翻遍手头的所有书都没找到参考。麻烦知道的顺便告知参考文献。
|
f**********g
发帖数: 107 | 19 not so easy. If they are dependent, you can not directly multiply their
moment generating functions to see what you get.
【在 q*******n 的大作中提到】
: The sum is still normal, regardless of the dependence of the two summands.
: This can be easily proved by using moment generating function.
|
f**********g
发帖数: 107 | 20 如果这两个normal随机变量X and Y 可以组成一个multivariate normal，当然choose
L=(1, 1)，the sum is a normal. 但问题是是不是随便两个normal随机变量的joint
ditribution都是multivariate normal。我记得LX is univariate normal好像是一个
必要条件。
【在 p*****e 的大作中提到】
: We treat the 0 variance/covariance case as normal.
: X is a random vector with n components.
: Then X is multivariate normal if and only if
: LX is univariate normal, for every 1 x n vector L.
: See any general multivariate statistics textbook.
: Therefore, the sum of two normals (indep. or not) is still normal if
: we choose L=(1, 1).
|
|
|
v***o
发帖数: 51 | 21 你的线性dependence的结论把pdf的定义搞错了吧。我想你是根据正态都有一致的bell
curve得出的结论。但对support做其他种transformation也可以生成正态。
可以看看这里：http://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent
那里的examples回答了包括楼主在内以上好多问题。
我也对这个问题蛮有兴趣的，不知哪位大牛给通过实分析讲讲里面的故事，最好通过几
何阐述一下。我直觉是和在哪个维度上积分有关，比如到底是概率空间还是support空
间上。一家之言还望高人指教。
one
a
yes.
X+
【在 g****e 的大作中提到】
: A lot of people have posted their answers. I think there is a point that one
: should always bear in mind, (my personal opinion):
: If X & Y are not independent and both are Normal distribution, we can draw a
: conclusion: it must be LINEAR dependence, which means you can represent Y
: ~~~~~~
: per follow up posts, this could be wrong (and I didn't really prove it).
: in a linear form of X. the question then reduces to:
: If X & Y are linearly dependent, is their sum still a normal? An
|
f**********g
发帖数: 107 | 22 I think this is a good counter example.
【在 f******k 的大作中提到】
: no.
: X, Y normal, Y=-X if |X|>=1, Y=X if |X|<1.
|
f**********g
发帖数: 107 | 23 应该不是必须linear dependent。比如这个例子X, Y normal, Y=-X if |X|>=1, Y=X
if |X|<1.
one
a
yes.
X+
【在 g****e 的大作中提到】
: A lot of people have posted their answers. I think there is a point that one
: should always bear in mind, (my personal opinion):
: If X & Y are not independent and both are Normal distribution, we can draw a
: conclusion: it must be LINEAR dependence, which means you can represent Y
: ~~~~~~
: per follow up posts, this could be wrong (and I didn't really prove it).
: in a linear form of X. the question then reduces to:
: If X & Y are linearly dependent, is their sum still a normal? An
|
f**********g
发帖数: 107 | 24 这本书里的结论好像错了。
【在 A*******r 的大作中提到】
: http://www.springerlink.com/content/x453xu/?p=b2b422921b154f5c8a9974cb3c6b0e66&pi=1
: Statistical Physics
: Including Applications to Condensed Matter
: Publisher Springer New York
: DOI 10.1007/b106783
: Copyright 2005
: %%%%%%%%%%%%%%%%%%%%%%%%%%
: Page 33
: %%%%%%%%%%%%%%%%%%%%%%%%%%
: Hence, the sum of two normal random variables is always (even if they are
|
A*******r
发帖数: 768 | 25 懒得看
X+Y is normal if the disjoint distribution of X and Y is normal
【在 f**********g 的大作中提到】
: 这本书里的结论好像错了。
|
A*******r
发帖数: 768 | 26 我记得有个苏联人写的
Counterexamples in probability theory ?
里面有具体的 X, Y 是正态的 joint不是正态的例子
这本书有online version
【在 f**********g 的大作中提到】
: 这本书里的结论好像错了。
|
g****e
发帖数: 1829 | 27 Make sense.
【在 f**********g 的大作中提到】
: 应该不是必须linear dependent。比如这个例子X, Y normal, Y=-X if |X|>=1, Y=X
: if |X|<1.
:
: one
: a
: yes.
: X+
|
h***i
发帖数: 3844 | 28 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
【在 v***o 的大作中提到】
: 你的线性dependence的结论把pdf的定义搞错了吧。我想你是根据正态都有一致的bell
: curve得出的结论。但对support做其他种transformation也可以生成正态。
: 可以看看这里：http://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent
: 那里的examples回答了包括楼主在内以上好多问题。
: 我也对这个问题蛮有兴趣的，不知哪位大牛给通过实分析讲讲里面的故事，最好通过几
: 何阐述一下。我直觉是和在哪个维度上积分有关，比如到底是概率空间还是support空
: 间上。一家之言还望高人指教。
:
: one
: a
|
q*******n
发帖数: 1334 | 29 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
【在 h***i 的大作中提到】
: pls show here
|
h***i
发帖数: 3844 | 30 第一步就错了，两normal， joint不一定是bivariate normal
[
.
【在 q*******n 的大作中提到】
: 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
|
|
|
p*****e
发帖数: 15 | 31 It is true that normal marginals do not imply the joint normal or
dependence structure. Feedback has given the counter example.
The condition that every linear combination is univariate normal
is sufficient and necessary.
See http://en.wikipedia.org/wiki/Multivariate_normal_distribution
choose
【在 f**********g 的大作中提到】
: 如果这两个normal随机变量X and Y 可以组成一个multivariate normal，当然choose
: L=(1, 1)，the sum is a normal. 但问题是是不是随便两个normal随机变量的joint
: ditribution都是multivariate normal。我记得LX is univariate normal好像是一个
: 必要条件。
|
s******h
发帖数: 539 | 32 Give any two INDEPENDENT r.v's X, Z such that
X~N(0,1), P(Z=1)=P(Z=-1)=1/2
Let Y=Z*X
Then
X,Y~N(0,1)
Cov(X,Y)=0
Because
a) P(Y<=y)=P(-X<=y|Z=-1)*P(Z=-1)+P(X<=y|Z=1)*P(Z=1)
=1/2*(P(X>=-y)+P(X<=y))
=P(X<=y)
b)
E(XY)=E[E(XY|Z)]=E[ZE(X^2|Z)]=E[ZE(X^2)]=EZ*E(X^2)=0
EX=0=EY=0
So Cov(X,Y)=0.
But
1) |X|=|Y|=> X and Y are not independent.
2) Let U=X+Y, then U is not Normally distributed.
P(U<=u)=P(U<=u|Z=-1)*P(Z=-1)+P(U<=u|Z=1)P(Z=1)
=1/2*I{u>=0}+1/2*P(X<=u/2) |
j********3
发帖数: 9 | 33 Here's another counter example from
http://risktheory.narod.ru/papers/sumOfDep.pdf
Take X and Z to be independent N(0,1) random variables. Define Y as follows:
Y = |Z| if X >= 0
Y = -|Z| if X < 0
Then Y is a N(0,1) random variable but X + Y is not normal.
(Edit: Oops, sorry for the repost, just realized that this is essentially the same example as above) |
