5
$\begingroup$

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

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

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,

\begin{equation} \text{var}(\tilde{u} - u) = \text{var}(u) + \text{var}(\tilde{u}) - 2\text{cov}(\tilde{u}, u), \end{equation}

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

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Dec 31, 2019 at 20:16
  • $\begingroup$ 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)). $\endgroup$
    – JLee
    Commented Dec 31, 2019 at 20:27
  • $\begingroup$ That's crucial contextual information, because it completely changes what you are asking! $\endgroup$
    – whuber
    Commented Dec 31, 2019 at 21:07
  • 1
    $\begingroup$ Okay yes, sorry. I have edited the question to add that piece of information. $\endgroup$
    – JLee
    Commented Dec 31, 2019 at 21:16

1 Answer 1

2
$\begingroup$

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, $$\mbox{cov}(u, \tilde u) = E \Bigl [ \bigl \{\tilde u - E(\tilde u) \bigr \} \, \bigl \{ \tilde u - E(\tilde u)\bigr \} \Bigr ] = \mbox{var}(\tilde u).$$

$\endgroup$
1
  • $\begingroup$ Using $E[\tilde{u}] = u$ and $E[u] = \tilde{u}$, should you not obtain $cov(u, \tilde{u}) = E\Big[\{ E(\tilde{u}) - \tilde{u}\}\{\tilde{u} - E(\tilde{u})\}\Big]$? $\endgroup$
    – JLee
    Commented Jan 1, 2020 at 21:40

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.