Displaying 3 results from an estimated 3 matches for "sigma_k".
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2004 Jun 07
2
MCLUST Covariance Parameterization.
...ley and A. Raftery. My problem is trying to figure out how the (model) identifier (e.g, EII, VII, VVI, etc.) relates to the covariance matrix. The parameterization of the covariance matrix makes use of the method of decomposition in Banfield and Rraftery (1993) and Fraley and Raftery (2002) where
Sigma_k = lambda_k*D_k*A_k*D_k^'
where Sigma_k is the covariance matrix for the kth (k=1,...,G), lambda_k is the kth groups constant of proportionality, D_k is the orthogonal matrix of eigenvectors for the kth group, and A_k is a diagonal matrix whose elements are proportional to the eigenvalues. The...
2011 Aug 01
3
formula used by R to compute the t-values in a linear regression
Hello,
I was wondering if someone knows the formula used by the function lm to compute the t-values.
I am trying to implement a linear regression myself. Assuming that I have K variables, and N observations, the formula I am using is:
For the k-th variable, t-value= b_k/sigma_k
With b_k is the coefficient for the k-th variable, and sigma_k =(t(x) x )^(-1) _kk is its standard deviation.
I find sigma_k = sigma * n/(n*Sum x_{k,i}^2 -(sum x_{k,i}^2))
With sigma: the estimated standard deviation of the residuals,
Sigma = sqrt(1/(N-K-1)*Sum epsilon_i^2)
With:
N: n...
2006 May 20
1
(PR#8877) predict.lm does not have a weights argument for newdata
Dear R developers,
I am a little disappointed that my bug report only made it to the
wishlist, with the argument:
Well, it does not say it has.
Only relevant to prediction intervals.
predict.lm does calculate prediction intervals for linear models from
weighted regression, so they should be correct, right?
As far as I can see they are bound to be wrong in almost all cases, if
no weights