由买买提看人间百态

boards

本页内容为未名空间相应帖子的节选和存档,一周内的贴子最多显示50字,超过一周显示500字 访问原贴
Mathematics版 - a question about random number generator
相关主题
计算器游戏关于二维布朗运动
Help on random integer generator关于pairwise correlation
请教:algebra VS. sigma algebra求一个联合分布
有用singular软件或者搞invariant polynomial theory的吗?求问一个markov chain的问题
question of orthotonal transformation and eigenvectors求助:矩阵微分
另一个统计方面的问题,有关假设检验的关于sencond fundamental form tensor的一个问题
Variance and covariance questioncovariant v.s. contravariant
A question about conditional pdf如何评价non-linear regression的好坏?
相关话题的讨论汇总
话题: random话题: covariance话题: cov话题: number话题: matrix
进入Mathematics版参与讨论
1 (共1页)
r****y
发帖数: 1437
1
e.g., given a covariance matrix
cov = [ 1 0.02
0.02 1]
how to generate two random number series that have such covariance?
let's make it more specific, two random numbers both follow normal
distribution.
I guess there must be an algorithm invented already by someone for
such thing.
H****h
发帖数: 1037
2
If you have iid rv X_1,...,X_n ~N(0,1), and a matrix A.
Let Y=(Y_1,...,Y_n)^t=AX, where X=(X_1,...,X_n)^t.
Then the covariance matrix for Y is A^tA.
g******n
发帖数: 339
3
use chlosky decomposition s.t. cov=d’d, then generate two independent x1~N(
0,1), x2~N(0,1), y=d'(x1,x1)~BVN(0,cov).

【在 r****y 的大作中提到】
: e.g., given a covariance matrix
: cov = [ 1 0.02
: 0.02 1]
: how to generate two random number series that have such covariance?
: let's make it more specific, two random numbers both follow normal
: distribution.
: I guess there must be an algorithm invented already by someone for
: such thing.
:

1 (共1页)
进入Mathematics版参与讨论
相关主题
如何评价non-linear regression的好坏?question of orthotonal transformation and eigenvectors
外行问个基本的统计问题 (转载)另一个统计方面的问题,有关假设检验的
a question on brownian motionVariance and covariance question
一句话,文兰当院士没有明显优势A question about conditional pdf
计算器游戏关于二维布朗运动
Help on random integer generator关于pairwise correlation
请教:algebra VS. sigma algebra求一个联合分布
有用singular软件或者搞invariant polynomial theory的吗?求问一个markov chain的问题
相关话题的讨论汇总
话题: random话题: covariance话题: cov话题: number话题: matrix