1
$\begingroup$

Consider the usual linear mixed model: $$Y=X \beta+ZB+\epsilon $$ where Y and $\epsilon$ are $n$-dimensional random variables and $B$ is a $q$-dimensional random variable independent of $\epsilon$ so we have: $B \sim N_q(0,\Sigma)$ and $\epsilon \sim N_n(0, \sigma^2 I_n)$. The matrices $X$ and $Z$ are model matrices of dimensions $n \times p $ and $n \times q$ and $\beta \in R^p$.

Now we consider the ordinary least squares (OLS) estimator for: $$\tilde{\beta}=(X^TX)^{-1}X^TY.$$ But I think this is not the ML estimator in the linear mixed model. Now I have to show that $\tilde{\beta}$ is an unbiased estimator of $\beta$ and that $$\text{Var}(\tilde{\beta})=(X^T X)^{-1} X^T(Z \Sigma_{\theta}Z^T)X(X^TX)^{-1}+\sigma^2(X^TX)^{-1}.$$ I'm not sure how to show it but I think that $\tilde{\beta}$ is an unbiased estimator of $\beta$ if its expected value is equal to the true value of the parameter. But how can I show that? And what can I do to show the expression for $\text{Var}(\tilde{\beta})$? What do they mean with the $\theta$ in $\Sigma_{\theta}$? I hope anyone can help me?

$\endgroup$
4
  • $\begingroup$ Did you delete an identical post and posted it anew? Or is this question different from the one posted earlier today? $\endgroup$ Dec 12, 2021 at 19:24
  • $\begingroup$ It's just the same, I just hope more helpers is online now or that more people will see the question beacuse I use this profile there is more active than the other profile $\endgroup$
    – Lifeni
    Dec 12, 2021 at 19:28
  • 1
    $\begingroup$ You have found a way to game the system, but it is not really ethical... $\endgroup$ Dec 12, 2021 at 20:33
  • $\begingroup$ Is this a homework problem? $\endgroup$
    – user277126
    Dec 13, 2021 at 2:26

1 Answer 1

1
$\begingroup$

$E(Y|X,Z) = X \beta$ because $B$ and $\varepsilon$ have mean $0$, so $$E(\tilde{\beta}) = E((X^T X)^{-1} X^T Y) = (X^T X)^{-1} X^T X \beta = \beta$$ For the variance we have $$Var(Y | X,Z) = Z \Sigma_{\theta} Z^T + \sigma I_n$$ because $B$ and $\varepsilon$ are independent. Then use a similar idea to show the desired expression.

The $\theta$ in $\Sigma_{\theta}$ reflects the fact that the covariance of the random effects is unknown and is parameterized by $\theta$. I.e., we are trying to estimate $\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.