Show unbiased OLS estimator and expression for variance of OLS estimator 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?

$E(Y|X,Z) = X \beta$ 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 $\theta$.
