a****m
发帖数: 693 | 1 （Y_mean-theta)^T *Sigma*(Y-Y_mean)是不是等于0；
Y_mean: sample mean of Y(y1, y2, y3,....)
theta: true mean of Y,
Y: multiple dimension, y1,...yn
Sigma: variance^-1.
thanks | A*******s
发帖数: 3942 | 2 need 1 (1 is a n*1 vector with all one's) to be an
eigenvector of sigma
then （Y_mean-theta)^T *Sigma*(Y-Y_mean)
=（Y_mean-theta)*lambda*1'*(Y-Y_mean)=0
【在 a****m 的大作中提到】
: （Y_mean-theta)^T *Sigma*(Y-Y_mean)是不是等于0；
: Y_mean: sample mean of Y(y1, y2, y3,....)
: theta: true mean of Y,
: Y: multiple dimension, y1,...yn
: Sigma: variance^-1.
: thanks
| a****m
发帖数: 693 | 3
any reference?
【在 A*******s 的大作中提到】
: need 1 (1 is a n*1 vector with all one's) to be an
: eigenvector of sigma
: then （Y_mean-theta)^T *Sigma*(Y-Y_mean)
: =（Y_mean-theta)*lambda*1'*(Y-Y_mean)=0
| A*******s
发帖数: 3942 | 4 first i have to be sure i read your notation right.
Ymean and theta should be a scaler but in your post they seem to be n by 1
vectors (Ymean, Ymean, ...) and (theta, theta, ...). right?
So (Ymean-theta) should be a vector with all cells equal.
(Ymean-theta)'*(Y-Ymean)=0 since sum(Yi)=N*Ymean
so need （Y_mean-theta)' *Sigma at the same direction with （Y_mean-theta)',
that's the definition of eigenvectors.
【在 a****m 的大作中提到】
:
: any reference?
| a****m
发帖数: 693 | 5 they are multivariate parameter:
theta: true mean
Y_mean: sample mean,
Y: y1,y2...yn, n dimensional.
thanks
1
theta)',
【在 A*******s 的大作中提到】
: first i have to be sure i read your notation right.
: Ymean and theta should be a scaler but in your post they seem to be n by 1
: vectors (Ymean, Ymean, ...) and (theta, theta, ...). right?
: So (Ymean-theta) should be a vector with all cells equal.
: (Ymean-theta)'*(Y-Ymean)=0 since sum(Yi)=N*Ymean
: so need （Y_mean-theta)' *Sigma at the same direction with （Y_mean-theta)',
: that's the definition of eigenvectors.
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