# Parameter estimation in the linear mixed effects model

In Parameter estimation and inference in the linear mixed effects model, page 1923, the variance

\begin{aligned} \text{var}(\tilde{u} - u) & = \sigma^2G - \text{var}(\tilde{u}) \\ & = \text{var}(u) - \text{var}(\tilde{u}), \end{aligned}

where $$\tilde{u} = GZ^\top H^{-1}(y - X\hat{\beta})$$ is the best linear unbiased predictor (BLUP) for the random effects vector $$u$$, where $$G$$ and $$H$$ are covariance matrices, $$Z$$ is a design matrix, $$y$$ is a vector of observations, and $$\hat{\beta}$$ is the maximum likelihood (ML) estimate for $$\beta$$.

By definition,

$$$$\text{var}(\tilde{u} - u) = \text{var}(u) + \text{var}(\tilde{u}) - 2\text{cov}(\tilde{u}, u),$$$$

this must mean that $$\text{cov}(\tilde{u}, u) = \text{var}(\tilde{u})$$. How can one show that $$\text{cov}(\tilde{u}, u) = \text{var}(\tilde{u})$$?

• I strongly doubt your reference makes such an invalid general assertion about variances: are you sure you transcribed it correctly? Try as I might, I cannot find anything like it on p. 1923. – whuber Dec 31 '19 at 20:16
• On page 1923 (part of Lemma 1) it is stated that $\text{var}(\tilde{u} - u) = \sigma^2G - \text{var}(\tilde{u})$, and $\text{var}(u) = \sigma^2G$ (see page 1922, Equation (6)). – JLee Dec 31 '19 at 20:27
• That's crucial contextual information, because it completely changes what you are asking! – whuber Dec 31 '19 at 21:07
• Okay yes, sorry. I have edited the question to add that piece of information. – JLee Dec 31 '19 at 21:16

We have that $$\mbox{cov}(u, \tilde u) = E \Bigl [ \bigl \{u - E(u) \bigr \} \, \bigl \{ \tilde u - E(\tilde u)\bigr \} \Bigr ].$$
But $$E(\tilde u) = u$$ and $$E(u) = \tilde u$$. Note that expectations are here taken with respect to the posterior of the random effects, not the prior. Hence, $$\mbox{cov}(u, \tilde u) = E \Bigl [ \bigl \{\tilde u - E(\tilde u) \bigr \} \, \bigl \{ \tilde u - E(\tilde u)\bigr \} \Bigr ] = \mbox{var}(\tilde u).$$
• Using $E[\tilde{u}] = u$ and $E[u] = \tilde{u}$, should you not obtain $cov(u, \tilde{u}) = E\Big[\{ E(\tilde{u}) - \tilde{u}\}\{\tilde{u} - E(\tilde{u})\}\Big]$? – JLee Jan 1 '20 at 21:40