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Consider a simple linear mixed effect model (LMM) $$Y_{ij}=b_{i}+X_{ij}\beta+\epsilon_{ij},$$ where $b_i\sim N(0,\sigma_b^2)$, $\epsilon_{ij}\sim N(0,\sigma_e^2)$. Typically, one can estimate $\sigma_b^2,\sigma_e^2$, then use the Hessian matrix to calculate the confidence interval of $\beta$.

I am thinking another way, can we first calculate $\hat{b}_i=E(b_i|\{Y_{ij},X_{ij}\})$ and $Y^{*}_{ij}=Y_{ij}-\hat{b}_i$. And then, we infer $\beta$ based on simple linear regression between $\{X_{ij}\}$ and $\{Y^{*}_{ij}\}$?

I think this estimate should be consistent since it is a natural estimation based on EM algorithm. But I am not sure about the validity of the confidence interval.

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  • $\begingroup$ From your notation it seems like $\beta$ is a scalar. In this case $\beta=\left(Y_{ij}-\epsilon_{ij}-b_i\right)/X_{ij}$, which would make it normally distributed, since it is a linear combination of normally distributed variables. Why not estimate sample mean and sample variance? The underlying mean and variance will follow t- and $\chi^2$ distirbutions. If $\beta$ is a vector in your case, same logic will apply, but one would have to use appropriate vector generalizations $\endgroup$
    – Cryo
    Commented Apr 1 at 1:57
  • $\begingroup$ Yes, that's my understanding. But typical approach estimating variance of $beta$ is to first calculate the variance of the random effect and then use the property of MLE to estimate the variance of $\beta$. $\endgroup$
    – Fangzhi Luo
    Commented Apr 1 at 13:21
  • $\begingroup$ if by variance of random effect you mean the variance of $\epsilon$ then it makes sense. Not sure you need much else after that since corresponding properties of $\beta$ will follow. Nothing wrong however with skipping this and going straight for $\beta$. As long as underlying $\epsilon$ distribution is assumed to be normal it should not matter $\endgroup$
    – Cryo
    Commented Apr 2 at 19:56
  • $\begingroup$ Sorry for the confusion. By variance of random effect I mean the variance of $b$. $\endgroup$
    – Fangzhi Luo
    Commented Apr 3 at 0:04
  • $\begingroup$ my bad - got it mixed up $\endgroup$
    – Cryo
    Commented Apr 3 at 5:24

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You can recast your system of linear equations into form:

$$ \left(\begin{array}\\ 1 & 0 & 0 & 0 & \dots & 0 & 0 & X_{11} \\ 1 & 0 & 0 & 0 & \dots & 0 & 0 & X_{12} \\ \dots \\ 0 & 1 & 0 & 0 & \dots & 0 & 0 & X_{21}\\ \dots 0 & 0 & 0 & 0 & \dots & 0 & 1 & X_{nm} \\ \end{array}\right)\left(\begin{array}\\ b_1 \\ b_2 \\ \dots \\ b_n \\ \beta \end{array} \right)= \left(\begin{array}\\ Y_{11}-\epsilon_{11} \\ \dots\end{array} \right) $$

Use the following notation to denote corresponding terms: $$ \mathbf{Z}.\mathbf{c}=\mathbf{d} $$

Where $\mathbf{d}$ is a vector of independent normally distributed variables with means given by $Y_{ij}$, and with the same variance. You can then write

$$ \mathbf{c}=\left(\mathbf{Z}^T\mathbf{Z}\right)^{-1}\mathbf{Z}^T.\mathbf{d} $$

So the distribution of any coefficient on the left will be normal, with known mean and known variance, if you chose to marginalize other degrees out. in general, it will be multilinear normal

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