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I'm considering an EM algorithm of correlated random effects model

$y_{it} = \beta x_{it} + \mu_i + \varepsilon_{it},(i=1,...,n;t=1,...,T)$

where $y_{it}$ and $x_{it}$ are observed, but $\mu_i$ and $\varepsilon_{it}$ are not observed. Assume that all the variables are random(not deterministic).

In a vector form,

$y_{i} = \beta x_{i} + \iota_T \mu_i + \varepsilon_{i},(i=1,...,n)$

where $y_i=(y_{i1},...,y_{iT})'$, $\iota_T=(1,...,1)'$ etc. Assume that $x_i$ is correlated with $\mu_i$ but uncorrelated with $\varepsilon_i$.

To derive the complete-data log likelihood function, assume that $\mu_i$ is observable. Then, the joint density can be written as

$f(y_{i},x_{i},\mu_{i} )=f(y_i,x_i|\mu_i) f(\mu_i)$

However, I'm wondering if this is further written as

$f(y_{i},x_{i},\mu_{i} )=f(y_i|x_i, \mu_i) f(x_i|\mu_i) f(\mu_i)=f(\varepsilon_i|x_i, \mu_i) f(x_i|\mu_i) f(\mu_i)$

and the comlete-data log likelihood for $i$th observation can be written as

$\log f(y_{i},x_{i},\mu_{i} ) = \log f(\varepsilon_i|x_i, \mu_i) + \log f(x_i|\mu_i) +\log f(\mu_i)$

Is this formula correct?

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  • $\begingroup$ It is incorrect to write $f(y_i|x_i, \mu_i)$ as $f(\varepsilon_i|x_i, \mu_i)$, it should be$$f_Y(y_i|x_i, \mu_i) =f_\varepsilon(y_i-\beta x_i-\iota_T \mu_i|x_i, \mu_i)$$with indices required to separate the different densities. $\endgroup$
    – Xi'an
    Commented Mar 8, 2021 at 9:35
  • $\begingroup$ Many thanks! Yes, different notation should be used for each density. $\endgroup$
    – user0131
    Commented Mar 9, 2021 at 0:16
  • $\begingroup$ I have another question. How can I derive $f(x_i|\mu_i)$? If both $(x_i, \mu_i)$ follows the multivariate normal distribution, can I simply apply the well known formula for $x_i|\mu_i$? Or are there any ways to derive the conditional distribution? $\endgroup$
    – user0131
    Commented Mar 11, 2021 at 1:06
  • $\begingroup$ If a vector as a whole is Gaussian, its marginal and conditional are Gaussian as well. $\endgroup$
    – Xi'an
    Commented Mar 11, 2021 at 7:36
  • $\begingroup$ Many thanks! I try to used the Gaussian results. $\endgroup$
    – user0131
    Commented Mar 11, 2021 at 22:25

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