LinearMixedModels.Rnw

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\title{Linear Mixed Models Summary}
\author{Shravan Vasishth (vasishth@uni-potsdam.de)}
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\begin{center}
     \normalsize{Linear Mixed Models Summary Sheet} \\
    \footnotesize{
    Compiled by: Shravan Vasishth (vasishth@uni-potsdam.de)\\
    Version dated: \today}
\end{center}

<<echo=F>>=
load("MAS473.RData")
library(lme4)
@

\section{Background and review of linear model theory}

In linear modelling we model the mean of a response $Y_1,\dots,Y_n$ as a function of a vector of predictors $x_1,\dots,x_n$. We assume that the $Y_i$ are conditionally independent given $\mathbf{x}, \mathbf{\beta}$. When $Y$'s are not marginally independent, we have $Cor(Y_1,Y_2)\neq 0$, or $P(Y_2\mid Y_1)\neq P(Y_2)$.

Linear mixed models are useful for correlated data where $\mathbf{Y}\mid \mathbf{X}, \mathbf{\beta}$ are not independently distributed. 

%\begin{equation}
%P(Y_1,\dots,Y_n\mid x_1,\dots,x_n,\beta) = 
%\prod P(Y_i,\dots,x_i,\beta)
%\end{equation}


\section{Basic specification of LMMs}

\begin{equation}
\explain{Y_i}{n\times 1} = \explain{X_i}{n\times p}~~\explain{\beta}{p\times 1} + \explain{Z_i}{n\times q}~~\explain{b_i}{q\times 1} + \explain{\epsilon_i}{n\times 1}
\end{equation}

where $i=1,\dots,m$, let $n$ be total number of data points.

Distributional assumptions:

$ b_i \sim N(0,D)$ and $\epsilon_i \sim N(0,\sigma^2 I)$. $D$ is a $q\times q$ matrix that does not depend on $i$, and $b_i$ and $\epsilon_i$ are assumed to be independent.

$Y_i$ has a multivariate normal distribution:

\begin{equation}
Y_i \sim N(X_i \beta, V(\alpha))
\end{equation}

where $V(\alpha)=Z_i D Z_i^T+\sigma^2 I$, and $\alpha$ is the variance component parameters.

Note:
\begin{enumerate}
\item
D has to be symmetric and positive definite. 
\item
The $Z_i$ matrix columns are a subset of $X_i$. In the random intercept model, $Z_i = 1_i$.
\item
In the varying intercepts and varying slopes model, $X_i = Z_i = (1_i, X_i)$. Then:

\begin{equation}
Y_i = X_i (\beta+b_i) +\epsilon_i
\end{equation}

or

\begin{equation}
Y_i = X_i \beta_i +\epsilon_i \quad \beta_i \sim N(\beta,D)
\end{equation}

\begin{equation}
\begin{split}
D=&
\begin{pmatrix}
d_{00} & d_{01}\\
d_{10} & d_{11}\\
\end{pmatrix}\\
=& 
\begin{pmatrix}
d_{00}=Var(\beta_{i0}) & d_{01}=Cov(\beta_{i0},\beta_{i1})\\
d_{10}=Cov(\beta_{i0},\beta_{i1}) & d_{11}=Var(\beta_{i1})\\
\end{pmatrix}
\end{split}
\end{equation}

\item
The conditional mean of $Y$ given the block effect $b_i$ is:

\begin{equation}
E(Y\mid b_i) = X_i \beta + Z_i b_i
\end{equation}

\item
The marginal mean for $Y$:

\begin{equation}
\begin{split}
E(Y_i) = & E(E(Y_i\mid b_i))\\
      = & E(X_i \beta + Z_i b_i) \\
      = & X_i \beta
\end{split}
\end{equation}

\item
The conditional variance of $Y$ given the block effect $b_i$ is:

\begin{equation}
Cov(Y_i \mid b_i) = Cov(\epsilon_i) = \sigma^2 I
\end{equation}

\item The marginal variance of $Y$ averaged over the distributions of $b_i$ is

\begin{equation}
\begin{split}
Cov(Y_i) =& Cov(Z_i b_i) + Cov(\epsilon_i)\\
 =& Z_i Cov(b_i) Z_i^T + Cov(\epsilon_i)\\
 =& Z_i D Z_i^T + \sigma^2 I
\end{split}
\end{equation}

\end{enumerate}

\section{$\sigma_b^2$ describes both between-block variance, and within block covariance}

Consider the following model, a varying intercepts model:

\begin{equation}
Y_{ij} = \mu + b_i + e_{ij},
\end{equation}


with $b_i\sim N(0,\sigma^2_b)$, $e_{ij}~N(0,\sigma^2)$.

Note that variance is a covariance of a random variable with itself, and then consider the model formulation. If we have

\begin{equation}
Y_{ij} = \mu + b_i + \epsilon_{ij}
\end{equation}

where i is the group, j is the replication, if we \textit{define} $b_i\sim N(0, \sigma^2_b)$, and refer to $\sigma^2_b$ as the between group variance, then we must have


\begin{equation}
\begin{split}
Cov(Y_{i1}, Y_{i2}) =& Cov(\mu + b_i + \epsilon_{i1} , \mu + b_i + \epsilon_{i2})\\
=& \explain{Cov(\mu, \mu)}{=0} + 
 \explain{Cov(\mu, b_i)}{=0} +
 \explain{Cov(\mu, \epsilon_{i2})}{=0}+ \\
 ~~& \explain{Cov(b_i,\mu)}{=0} +
  \explain{Cov(b_i,b_i)}{+ve} \dots\\
  =&  Cov(b_i, b_i) = Var(b_i) = \sigma^2_b\\
\end{split}
\end{equation}

\section{Some basic types of linear mixed model and their variance components}

\subsection{Varying intercepts model}

\begin{fmpage}{\linewidth}
The model for a categorical predictor is:

\begin{equation}
Y_{ijk} = \beta_j + b_{i}+\epsilon_{ijk}
\end{equation}
\end{fmpage}

\noindent
$i=1,\dots,10$ is subject id, $j=1,2$ is the factor level, $k$ is the number of replicates (here 1).
$b_i \sim N(0,\sigma_b^2), \epsilon_{ijk}\sim N(0,\sigma^2)$.

\begin{fmpage}{\linewidth}
For a continuous predictor:

\begin{equation}
Y_{ijk} = \beta_0 + \beta_1 t_{ijk} + b_{ij} +\epsilon_{ijk}
\end{equation}
\end{fmpage}

The general form for any model in this case is:

\begin{equation}
\begin{pmatrix}
Y_{i1}\\
Y_{i2}
\end{pmatrix}
\sim
N\left(
\begin{pmatrix}
\beta_1\\
\beta_2\\
\end{pmatrix}
,
V
\right)
\end{equation}

where 

\begin{equation}
V =\begin{pmatrix}
\sigma_b^2 + \sigma^2 & \sigma_b^2\\
\sigma_b^2 & \sigma_b^2 + \sigma^2\\
\end{pmatrix}
=
\begin{pmatrix}
\sigma^2_{b} + \sigma^2  &  \rho\sigma_{b}\sigma_{b}\\
\rho\sigma_{b}\sigma_{b} & \sigma^2_{b}+\sigma^2  \\       
\end{pmatrix}
\end{equation}

Note also that the mean response for the subject, i.e., \textit{conditional} mean of $Y_{ij}$ given the subject-specific effect $b_i$ is:

\begin{equation}
E(Y_{ij}\mid b_i)= X_{ij}^T \beta + b_i
\end{equation}

The mean response in the population, i.e., the marginal mean of $Y_{ij}$:

\begin{equation}
E(Y_{ij})= X_{ij}^T \beta
\end{equation}

The marginal variance of each response is:

\begin{equation}
\begin{split}
Var(Y_{ij})=& Var(X_{ij}^T \beta+b_i + \epsilon_{ij})\\
 =& Var(\beta+b_i + \epsilon_{ij})\\
 =& \sigma_b^2 + \sigma^2\\
\end{split}
\end{equation}

the covariance between any pair of responses $Y_{ij}$ and $Y_{ij'}$ is given by

\begin{equation}
\begin{split}
Cov(Y_{ij},Y_{ij'}) =& Cov(X_{ij}^T\beta + b_i + \epsilon_{ij},X_{ij'}^T\beta + b_i + \epsilon_{ij'})\\
=& Cov(b_i + e_{ij},b_i + e_{ij'})\\
=& Cov(b_i,b_i)=\sigma_b^2
\end{split}
\end{equation}

The correlation is

\begin{equation}
Corr(Y_{ij},Y_{ij'})= \frac{\sigma_b^2}{\sigma_b^2+\sigma^2}
\end{equation}

In other words, introducing a random intercept induces a correlation among repeated measurements.

$\hat{V}$ is therefore:

\begin{equation}
\begin{pmatrix}
\hat{\sigma}^2_{b} + \hat{\sigma}^2  &  \hat{\rho}\hat{\sigma}_{b}\hat{\sigma}_{b}\\
\hat{\rho}\hat{\sigma}_{b}\hat{\sigma}_{b} & \hat{\sigma}^2_{b}+\hat{\sigma}^2  \\       
\end{pmatrix}
\end{equation}

Note: $\hat{\rho}=1$. But this correlation is not \textit{estimated} in the varying intercepts model.

\subsection{Varying intercepts and slopes (with correlation)}

\begin{fmpage}{\linewidth}
The model for a categorical predictor is:

\begin{equation}
Y_{ij} = \beta_1+b_{1i}+(\beta_2+b_{2i})x_{ij}+\epsilon_{ij} \quad i=1,...,M, j=1,...,n_i
\end{equation}
\end{fmpage}

with $b_{1i}\sim N(0,\sigma_1^2), b_{2i}\sim N(0,\sigma_2^2)$, and $\epsilon_{ij}\sim N(0,\sigma^2)$.


\textit{Note: I have seen this presentation elsewhere}:

\begin{equation}
Y_{ijk} = \beta_j + b_{ij}+\epsilon_{ijk}
\end{equation}

\noindent
$b_{ij}\sim N(0,\sigma_b)$. The variance $\sigma_b$ must be a $2\times 2$ matrix:

\begin{equation}
\begin{pmatrix}
\sigma_1^2 & \rho \sigma_1 \sigma_2\\
\rho \sigma_1 \sigma_2 & \sigma_2^2\\
\end{pmatrix}
\end{equation}

The general form for the model is:

\begin{equation}
\begin{pmatrix}
Y_{i1}\\
Y_{i2}
\end{pmatrix}
\sim
N\left( 
\begin{pmatrix}
\beta_1\\
\beta_2\\
\end{pmatrix}
,
V
\right)
\end{equation}

where 

\begin{equation}
V =
%\begin{pmatrix}
%\sigma_b^2 + \sigma^2 & \sigma_b^2\\
%\sigma_b^2 & \sigma_b^2 + \sigma^2
%\end{pmatrix}=
\begin{pmatrix}
\sigma^2_{b,A} + \sigma^2  &  \rho\sigma_{b,A}\sigma_{b,B}\\
\rho\sigma_{b,A}\sigma_{b,B} & \sigma^2_{b,B}+\sigma^2  \\       
\end{pmatrix}
\end{equation}

\subsection{No varying intercepts, only slopes for each level}

\begin{fmpage}{\linewidth}
The model is

\begin{equation}
Y_{ijk} = \beta_j + b_{ij} + \epsilon_{ijk} 
\end{equation}
\end{fmpage}

The random effects are:

$b_{ij}=\begin{pmatrix}
b_{i1}\\
b_{i12}
\end{pmatrix}
\sim N(0,\sigma_b^2)$, where $\sigma_b^2=
\begin{pmatrix}
\sigma_1^2 & \rho\sigma_1 \sigma_2 \\
\rho\sigma_1 \sigma_2 & \sigma_2^2 \\
\end{pmatrix}$. 

Here, V is

\begin{equation}
V =
%\begin{pmatrix}
%\sigma_b^2 + \sigma^2 & \sigma_b^2\\
%\sigma_b^2 & \sigma_b^2 + \sigma^2
%\end{pmatrix}=
\begin{pmatrix}
\sigma^2_{b,A} + \sigma^2  &  \rho\sigma_{b,A}\sigma_{b,B}\\
\rho\sigma_{b,A}\sigma_{b,B} & \sigma^2_{b,B}+\sigma^2  \\       
\end{pmatrix}
\end{equation}

\textbf{Note that here, a random effect is computed for each material separately.}


One insight is that V can be derived from the random effects variance components, and the error term's variance component:

\begin{equation}
V=
\begin{pmatrix}
\sigma^2_{b,A} &\rho\sigma_{b,A}\sigma_{b,B}\\
\rho\sigma_{b,A}\sigma_{b,B} & \sigma^2_{b,B}\\
\end{pmatrix}
+
\begin{pmatrix}
\sigma^2 & 0\\
0 & \sigma^2\\
\end{pmatrix}
\end{equation}

\subsection{Nested models (e.g., Worker/Machine)}

\begin{fmpage}{\linewidth}
The model is:

\begin{equation}
Y_{ijk} = \beta_j + b_i + b_{ij} + \epsilon_{ijk}
\end{equation}
\end{fmpage}

Here, we force all random effects to be independent.
Observations between workers are independent, but observations on the same worker are correlated.

$b_i \sim N(0,\sigma_1^2), b_{ij} \sim N(0,\sigma_2^2)$, and $\epsilon\sim N(0,\sigma^2)$. $i$ is Worker, $j$ is machine, and $k$ is replicate.  

<<>>=
fm1<-lmer(score~Machine-1+(1|Worker/Machine),
data=Machines)
@

The term Worker/Machine is estimating machine variance within worker:

\begin{verbatim}
xyplot(score~Machine|Worker,Machines)
\end{verbatim}

The variance components in fm1:

\small
\begin{tabular}{rlll}
  \hline
Comp.\ & Groups & Name & Var\\ 
  \hline
$\hat{\sigma}_2^2$ & Machine:Worker & (Int) & 13.909  \\ 
$\hat{\sigma}_1^2$  & Worker & (Int) & 22.858  \\ 
$\hat{\sigma}^2$& Res &  &  0.925 \\ 
   \hline
\end{tabular}

Number of obs: 54, groups: Machine:Worker, 18; Worker, 6.

\normalsize

For observations on Worker $i$, 

\begin{equation}
Var(Y_{ijk})= \sigma_1^2 + \sigma_2^2 + \sigma^2 
\end{equation}

Variance between machines within workers:
\begin{equation}
Covar(Y_{ijk},Y_{ijk'})= \sigma_1^2 + \sigma_2^2
\end{equation}

Variance between workers:
\begin{equation}
Covar(Y_{ijk},Y_{ij'k'})= \sigma_1^2
\end{equation}

Note:

1. $\hat{\sigma}_1^2$ all observations have the same variance;

2. $\hat{\sigma}_2^2$: the covariance between observations corresponding to the same worker using different machines is the same, for any pair of machines.


\begin{verbatim}
> ranef(fm1)
$`Machine:Worker`    $Worker
    (Intercept)
A:6     1.91609     6 -7.514666
A:2     1.55253     2 -1.375925
\end{verbatim}

In this model, the sum of the random effects for Worker 1 on Machine A is

$s_1 = b_1 + b_{11}$

\begin{verbatim}
> ranef(fm1)
...
$`Machine:Worker` $Worker
(Intercept)
s1 = A:1 -0.75012+1.044598=0.29448
\end{verbatim}

and for Worker 1 on machine B,

$s_2=b_1+b_{21}$.

\begin{verbatim}
> ranef(fm1)
...
$`Machine:Worker` $Worker
(Intercept)
s2 = B:1 1.50002 + 1.044598 = 2.5446
\end{verbatim}

For all Workers and machines, we can obtain these random effects $s$ from this matrix:

<<>>=
mat<-matrix(
unlist(ranef(fm1)$`Machine:Worker`),6,3) 
+ 
matrix(unlist(ranef(fm1)$Worker),6,3)
@
<<echo=F>>=
rownames(mat)<-c(6,2,4,1,3,5)
colnames(mat)<-LETTERS[1:3]
@

<<>>=
mat
@

Using lmer, we have $b_{i}$ and $b_{ij}$ independent, but $s_1$ and $s_2$ are
correlated via the common term $b_1$. We can recover the \textbf{correlations between machine} through the vcov matrix of the random effects (BLUPs) (\textbf{but note that we never see this in the lmer output---what's the significance of the fact that these are correlated?}):

<<>>=
var(mat)
@

%Correlations between subjects:
%var(t(mat))

\subsection{Varying slopes, no varying intercept}

\begin{fmpage}{\linewidth}
\begin{equation}
Y_{ijk} = \beta_j + b_{ij} + \epsilon_{ijk}
\end{equation}
\end{fmpage}

<<>>=
fm3<-lmer(score~Machine-1+
            (Machine-1|Worker),
          data=Machines)
@

<<>>=
head(ranef(fm3))
@

The random effects for Worker 1 on Machine A is

$s_1 = b_{11}=0.31199$

and for Worker 1 on Machine B,

$s_2 = b_{12}=2.55323$.

The `Machine independent' Worker random effect (varying intercept) $b_i$ has been dropped. 
We have $b_{11}$ correlated with $b_{12}$. We can see this when we recover the (co-)variances between machines from the random effects: 

\small
<<>>=
var(ranef(fm3)$Worker)
@
\normalsize

Also, the variances for each machine (16, 74, 18)
are also allowed to be different. Here are the variance components:

\tiny
\begin{tabular}{rlllll}
  \hline
Comp.\ & Groups & Name & Variance  & Corr$_{1,\cdot}$ &  Corr$_{2,\cdot}$  \\ 
  \hline
$\hat{\sigma}_{j=1}^2$ & Worker & A & 16.640   &    \\ 
$\hat{\sigma}_{j=2}^2$ &  & B & 74.395 & 
$\hat{\rho}_{1,2}$ 0.803 &    \\ 
$\hat{\sigma}_{j=3}^2$ &  & C & 19.268 &
$\hat{\rho}_{1,3}$ 0.623 & $\hat{\rho}_{2,3}$ 0.771   \\ 
$\hat{\sigma}^2$ & Res &  &  0.925 &  &    \\ 
   \hline
\end{tabular}
\normalsize


\begin{equation}
Var(Y_{ijk})= \sigma_j^2 + \sigma^2
\end{equation}

\begin{equation}
Covar(Y_{ijk},Y_{ijk'})= \sigma_j^2
\end{equation}

\begin{equation}
Covar(Y_{ijk},Y_{ij'k'})= \rho_{j,j'} \sigma_j\sigma_{j'}
\end{equation}



Note that the BLUPs' vcov matrix reflects the estimated values:

<<>>=
diag(var(ranef(fm3)$Worker))
cor(ranef(fm3)$Worker)
# look at the fm3 output 
## (the random effects table)
@


1. $\hat{\sigma}_j^2$ the variance of an observation depends on the machine being used; 

2. $\rho_{j,j'} \sigma_j\sigma_{j'}$ the covariance between observations corresponding to the same worker using different
machines is different, for different pairs of machines.

\begin{verbatim}
> var(ranef(fm3)$Worker)
         MachineA MachineB MachineC
MachineA   16.347   28.239   11.146
MachineB   28.239   74.093   29.181
MachineC   11.146   29.181   18.972
\end{verbatim}

\begin{equation}
\begin{pmatrix}
\sigma_{A}^2 & Cov_{A,B}     & Cov_{A,C}\\  
               & \sigma_{B}^2 & Cov_{B,C} \\
              &               & \sigma_{C}^2\\
\end{pmatrix}
\end{equation}

Note that, for given machines $j$ and $j'$, say A, B: 

$Covar(Y_{ijk},Y_{ij'k'}) = Cov_{A,B}=28.239 \approx  \rho_{A,B} \sigma_{A} \sigma_{B}
= .803 \times \sqrt{16.347} \times \sqrt{74.093} = 27.946$.  

\subsection{Comparing fm1 and fm3}

The sum of fm1's (Worker/Machine) ranefs ($b_{ij}+b_i$) are roughly the same as fm3's (Machine-1$\mid$ Worker) random effects $b_{ij}$ for each machine. \textbf{In other words, the random effect $b_i$ is folded into $b_{ij}$ in fm3.}

<<>>=
fm1<-lmer(score~Machine-1+
            (1|Worker/Machine),
          data=Machines)
fm3<-lmer(score~Machine-1+
            (Machine-1|Worker),
          data=Machines)
@

fm1's ranefs summed up:

<<>>=
#fm1's ranefs summed up are 
## roughly the same as the fm3 ranefs:
matrix(unlist(ranef(fm1)$`Machine:Worker`),
       6,3) +
matrix(unlist(ranef(fm1)$Worker),6,3)
@

The fm3 ranefs:

<<>>=
ranef(fm3)
@



\section{Parameter estimation}

\subsection{Likelihood based model fitting procedure}

Recall:

\begin{enumerate}
\item If we have two continuous random variables $Y$ and $Z$, with density functions $f_Y(y)$ and $f_Z(z)$ and joint density $f_{Y,Z}(y,z)$, then

\begin{equation} \label{eq1}
f_Y(y) = \int f_{Y,Z}(y,z)dz.
\end{equation}

\item
The conditional density of $Y\mid Z$ is defined as 

\begin{equation} \label{eq2}
f_{Y\mid Z}(y\mid z) = \frac{f_{Y,Z}(y,z)}{f_Z(z)}
\end{equation}

\noindent
so we can write 

\begin{equation}
f_{Y,Z}(y,z) = f_{Y\mid Z}(y\mid z) \times f_Z(z).
\end{equation}

\item
Combining equations \ref{eq1} and \ref{eq2}, we have

\begin{equation} \label{eq3}
f_Y(y) = \int f_{Y\mid Z}(y\mid z) * f_Z(z)dz
\end{equation}
\end{enumerate}


Equation \ref{eq3}, where we condition on a second random variable Z (note that Z could be ``non-observable'') can be helpful in deriving $f_Y(y)$, if the two densities on the RHS are easy to write down, and the integral can be solved.

Returning to parameter estimation in LMMs, the model is:

\begin{equation}
Y_i = X_i \beta + Z_i \beta_i + \epsilon_i, \quad i=1,\dots,M
\end{equation}

where 
$b_i \sim N(0,\Psi), \epsilon_i \sim N(0,\sigma^2 I)$. Let $\theta$ be the parameters that determine $\Psi$.

\begin{equation}
\begin{split}
L(\beta,\theta,\sigma^2\mid y) =& p(y:\beta,\theta,\sigma^2)\\
=& \prod_i^M p(y_i:\beta,\theta,\sigma^2)\\
=& \prod_i^M \int p(y_i\mid b_i, \beta,\sigma^2)p(b_i: \theta,\sigma^2)\,db_i
\end{split}
\end{equation}


we want the density of the observations ($y_i$) given the parameters $\beta, \theta$ and $\sigma^2$ only. In this case, using equation \ref{eq3} above, with $Y=y_i$ and $Z=b_i$ is helpful for deriving the density for $y_i$, because $f(y_i\mid b_i)$ (or, in the notation of (4.9), $p(y_i\mid b_i,\beta,\sigma^2)$) has a simple form, and so we can get a closed form expression for the integral.

\subsection{REML estimation (REstricted/REsidual ML)}


To estimate variance parameters, \textbf{first fit fixed effects using least squares}, and then focus attention on  residuals. 

The \textbf{residuals}' distribution depends on $\sigma^2$ and variance parameters $\theta$ of random effects.  

A \textbf{likelihood} for these parameters is formed based on the residuals alone. Maximization of this \textbf{marginal likelihood} gives estimates of
$\sigma^2$ and the other variance-covariance parameters which are less biased than the full maximum likelihood estimates.

Once the REML variance-covariance estimates are obtained the \textbf{fixed effects are re-estimated by maximum likelihood assuming the random effects parameters are known}, a procedure equivalent to generalized least squares.

Alternatively, define a restricted likelihood:

\begin{equation}
L_R (\theta,\sigma^2\mid y) = \int L(\beta,\theta,\sigma^2\mid y)\,d\beta
\end{equation}

and maximize this to obtain estimates of these parameters.

Unlike full (max.) likelihood, restricted lik.\ is not invariant to parameterization, so we cannot compare models with different fixed effects. 

\subsection{How the random effects are 'predicted' when using the ranef() command}

In linear mixed models, we fit models like these (the Ware-Laird formulation--see Pinheiro and Bates 2000, for example):

\begin{equation} 
Y = X\beta + Zu + \epsilon
\end{equation}

Let $u\sim N(0,\sigma_u^2)$, and this is independent from $\epsilon\sim N(0,\sigma^2)$.  

Given $Y$, the ``minimum mean square error predictor'' of $u$ is the conditional expectation:

\begin{equation}
\hat{u} = E(u\mid Y)
\end{equation}

We can find $E(u\mid Y)$ as follows. We write the joint distribution of $Y$ and $u$ as:

\begin{equation}
\begin{pmatrix}
Y \\
u
\end{pmatrix}
= 
N\left(
\begin{pmatrix}
X\beta\\
0
\end{pmatrix},
\begin{pmatrix}
V_Y & C_{Y,u}\\
C_{u,Y} & V_u \\
\end{pmatrix}
\right)
\end{equation}

$V_Y, C_{Y,u}, C_{u,Y}, V_u$ are the various variance-covariance matrices. 
It is a fact that

\begin{equation}
u\mid Y ~N(C_{u,Y}V_Y^{-1}(Y-X\beta)), 
Y_u - C_{u,Y} V_Y^{-1} C_{Y,u})
\end{equation}

This allows you to derive the BLUPs:

\begin{equation}
\hat{u}= C_{u,Y}V_Y^{-1}(Y-X\beta))
\end{equation}

Substituting $\hat{\beta}$ for $\beta$, we get:

\begin{equation}
BLUP(u)= \hat{u}(\hat{\beta})C_{u,Y}V_Y^{-1}(Y-X\hat{\beta}))
\end{equation}

Here's an example with R:

<<>>=
## Calculate the predicted random 
## effects by hand for the 
## ergoStool data
fm1<-lmer(effort~Type-1 + 
            (1|Subject),ergoStool)

## Here are the BLUPs we will 
## estimate by hand:
head(ranef(fm1))

## this gives us all the 
## variance components:
## this could have been done 
## ``by hand'',
## or at least an approximation:
VarCorr(fm1)
@

First, calculate the predicted random effect for subject 1:

<<>>=
## The variance for the random 
## effect subject is the term C_{u,Y}:
(covar.u.y<-VarCorr(fm1)$Subject[1])
@

Estimated covariance between $u_1$ and $Y_1$
make up a var-covar matrix from this:

<<>>=
(cov.u.Y<-matrix(covar.u.y,1,4))
@

Estimated variance matrix for $Y_1$:

<<>>=
(V.Y<-matrix(1.7755,4,4)+
   diag(1.2106,4,4))
@

Extract observations for subject 1:

<<>>=
(Y<-matrix(ergoStool$effort[1:4],4,1))
@

Estimated fixed effects:

<<>>=
(beta.hat<-matrix(fixef(fm1),4,1))
@

Predicted random effect:

<<>>=
cov.u.Y %*% solve(V.Y)%*%(Y-beta.hat)
@

Compare with ranef command:

<<>>=
ranef(fm1)$Subject[1,1]
@

Calculate predicted random effects for all subjects:
<<>>=
#t(cov.u.Y %*% solve(V.Y)%*%
#    (matrix(ergoStool$effort,4,9)-
#       matrix(fixef(fm1),4,9)))
#ranef(fm1)
@

\section*{Correlation of fixed effects}

For an ordinary linear model, the covariance matrix (from which we can get the correlation matrix) of $\hat{\beta}$ is:

\begin{equation}
\sigma^2 \times (X^T X)^{-1}.
\end{equation}

For a mixed effects model, the standard deviations (standard errors) and correlations for the fixed effects estimators are listed at the end of the lmer output. 

<<>>=
lm.full<-lmer(wear~material-1+
                (1|Subject), 
              data = BHHshoes)
@

\begin{verbatim}
Correlation of Fixed Effects:
          matrlA
materialB 0.988 
\end{verbatim}


%(Note that you need to include the -1 term in the lmer command to match the parameterisation used in the solutions. The estimated correlation between $\hat{beta}_1$ and $\hat{beta}_2$ is $0.988$.
%We're not going to cover how to derive the covariance matrix in general, as it's quite complicated, but as we have a balanced design, we have simple forms for the parameter estimators:

\begin{equation}
\hat{\beta}_1 = (Y_{1,1} + Y_{2,1} + \dots + Y_{10,1})/10
\end{equation}


\begin{equation}
\hat{\beta}_2 = (Y_{1,2} + Y_{2,2} + \dots + Y_{10,2})/10
\end{equation}

<<>>=
b1.vals<-subset(BHHshoes,
                material=="A")$wear
b2.vals<-subset(BHHshoes,
                material=="B")$wear

vcovmatrix<-var(cbind(b1.vals,b2.vals))

## get covariance from off-diagonal:
covar<-vcovmatrix[1,2]
sds<-sqrt(diag(vcovmatrix))
## correlation of fixed effects:
covar/(sds[1]*sds[2])

#cf:
covar/((0.786*sqrt(10))^2)  
@




\end{multicols}
\end{document}
How does this work for multiple factors?

<<>>=
m1<-lmer(effort~Type-1+(1|Subject),
         ergoStool)

T1.vals<-subset(ergoStool,
                Type=="T1")$effort
T2.vals<-subset(ergoStool,
                Type=="T2")$effort
T3.vals<-subset(ergoStool,
                Type=="T3")$effort
T4.vals<-subset(ergoStool,
                Type=="T4")$effort

vals<-cbind(T1.vals,T2.vals,
            T3.vals,T4.vals)

## compute varcov matrix:
vcovmat<-var(vals)
## get sd's of each level:
sds<-sqrt(diag(vcovmat))

## T1.T2 correlation, the sds come 
## from the model fit:
1.7222/(1.728*1.728)
@