+ Reply to Thread
Results 1 to 2 of 2

Thread: Restricted Maximum Likelihood (REML) Estimate of Variance Component.

  1. #1
    Points: 4,908, Level: 44
    Level completed: 79%, Points required for next Level: 42

    Posts
    78
    Thanks
    30
    Thanked 2 Times in 2 Posts

    Restricted Maximum Likelihood (REML) Estimate of Variance Component.




    Let, \mathbf y_i = \mathbf X_i\mathbf\beta + \mathbf Z_i\mathbf b_i+ \mathbf\epsilon_i,

    where

    \mathbf y_i\sim N(\mathbf X_i\mathbf\beta, \Sigma_i=\sigma^2\mathbf I_{n_i}+\mathbf Z_i \mathbf G\mathbf Z_i'),

    \mathbf b_i\sim N(\mathbf 0, \mathbf G),

    \mathbf\epsilon_i\sim N(\mathbf 0, \sigma^2\mathbf I_{n_i})

    \mathbf y_i is a n_i\times 1 vector of response for i^{th} individual at 1,2,\ldots, n_i time points, \mathbf X_i is a n_i\times p matrix, \mathbf \beta is a p\times 1 vector of fixed effect parameters, \mathbf Z_i is a n_i\times q matrix, \mathbf b_i is a q\times 1 vector of random effects, \mathbf \epsilon_i is a n_i\times 1 vector of within errors, \mathbf G is a q\times q covariance matrix of between-subject, \sigma^2 is a scalar.

    Note that, \mathbf X_i, \mathbf Z_i, and \mathbf G do NOT involve \sigma^2.

    Now I have to find out the Restricted Maximum Likelihood (REML) Estimate of \sigma^2, that is,

    \hat\sigma^2_R = \frac{1}{N_0-p}\sum_{i=1}^{N}(\mathbf y_i-\mathbf X_i\mathbf\beta)'(\mathbf I_{n_i}+\mathbf Z_i \mathbf G\mathbf Z_i')^{-1}(\mathbf y_i-\mathbf X_i\mathbf\beta),\ldots (1)

    where N_0 = \sum_{i=1}^{N}n_i.

    So first I wrote the Restricted Maximum Log-Likelihood :

    l_R \propto -\frac{1}{2}\sum_{i=1}^{N}\log\det(\Sigma_i)-\frac{1}{2}\sum_{i=1}^{N}\log\det(\mathbf X_i'\Sigma_i^{-1}\mathbf X_i)-\frac{1}{2}\sum_{i=1}^{N}(\mathbf y_i-\mathbf X_i\mathbf\beta)'\Sigma_i^{-1}(\mathbf y_i-\mathbf X_i\mathbf\beta)

    Then I have to differentiate log-likelihood, l_R, with respect to \sigma^2 and equate it to zero, i.e.,

    -\frac{1}{2}\frac{\partial}{\partial\sigma^2}\{\sum_{i=1}^{N}\log\det(\Sigma_i)+\sum_{i=1}^{N}\log\det(\mathbf X_i'\Sigma_i^{-1}\mathbf X_i) +\sum_{i=1}^{N}(\mathbf y_i-\mathbf X_i\mathbf\beta)'\Sigma_i^{-1}(\mathbf y_i-\mathbf X_i\mathbf\beta)\}|_{\sigma^2=\hat\sigma^2_R}=0


    But basically I can't differentiate,

    \frac{\partial}{\partial\sigma^2}\log\det(\Sigma_i)=\frac{\partial}{\partial\sigma^2}\log\det(\sigma^2\mathbf I_{n_i}+\mathbf Z_i \mathbf G\mathbf Z_i')

    \frac{\partial}{\partial\sigma^2}\log\det(\mathbf X_i'\Sigma_i^{-1}\mathbf X_i)=\frac{\partial}{\partial\sigma^2}\log\det(\mathbf X_i'(\sigma^2\mathbf I_{n_i}+\mathbf Z_i \mathbf G\mathbf Z_i')^{-1}\mathbf X_i) and

    \frac{\partial}{\partial\sigma^2}\Sigma_i^{-1}= \frac{\partial}{\partial\sigma^2}(\sigma^2\mathbf I_{n_i}+\mathbf Z_i \mathbf G\mathbf Z_i')^{-1}.


    How can I differentiate the above derivatives and get the REML estimate \hat\sigma^2_R in equation (1) ?

  2. #2
    Points: 4,908, Level: 44
    Level completed: 79%, Points required for next Level: 42

    Posts
    78
    Thanks
    30
    Thanked 2 Times in 2 Posts

    Re: Restricted Maximum Likelihood (REML) Estimate of Variance Component.


    Quote Originally Posted by tnkvrbox View Post
    Thương hi‡u ZIDOO dần ‘ược khẳng ‘‹nh l* thương hi‡u Android TV Box .......bla,bla,bla

    .

    tnkvrbox, YOU ARE REALLY A STUPID GUY. ARE YOU ENEMY OF THE ZIDOO COMPANY, SO THAT YOU DID SUCH NONSENSE, IDIOTIC WORK TO MAKE PEOPLE ANGER OF THAT COMPANY?

+ Reply to Thread

           




Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats