Mixed effects and longitudinal models — Examples

For either visit, the random part is $b_j+{\epsilon }_{ji}$, so its variance is ${\tau }^2+{\sigma }^2$ because the two terms are independent and centered. Equivalently, in the marginal-covariance formula use random-intercept design row $1$, $G=[{\tau }^2]$, and residual diagonal entry ${\sigma }^2$.

For the two distinct visits, both observations contain the same $b_j$. The covariance formula gives $\mathrm{Cov}(Y_{j1},Y_{j2})={\tau }^2$; the residual cross-covariance is zero by the assumed independence.

By [A1], both marginal variances are positive. Dividing the covariance from step 1.2 by the square root of the product of the two equal variances from step 1.1 gives ${\tau }^2\mathrm{/}({\tau }^2+{\sigma }^2)$ by [A2]. If ${\tau }^2=0<{\sigma }^2$ the ICC is $0$; if ${\sigma }^2=0<{\tau }^2$ it is $1$; if both are zero the correlation is undefined.

The matrix $G_2$ is positive definite because for every $(x,y)\mathrm{\ne }(0,0)$, $\left(\begin{array}{cc}x& y\end{array}\right)G_2\left(\begin{array}{c}x\\ y\end{array}\right)=\frac{4}{5}{(x+\frac{1}{4}y)}^2+\frac{11}{20}{y}^2>0.$ Also $G_1=I_2$ and $R_1=I_2$ are positive definite, while $R_2=(6\mathrm{/}5)I_2$ is positive definite since $6\mathrm{/}5>0$. The design has determinant $1$, so [F4] makes it invertible and therefore full rank.

For the first tuple, $V_1=Z{G}_1{Z}^T+R_1=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)+I_2=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)\mathrm{.}$

Direct multiplication gives $Z{G}_2{Z}^T=\left(\begin{array}{cc}4\mathrm{/}5& 1\\ 1& 9\mathrm{/}5\end{array}\right),V_2=Z{G}_2{Z}^T+R_2=\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)=V_1\mathrm{.}$ The covariance parameter tuples are distinct, so the covariance map is not injective.

Under either Gaussian tuple, $Y-X\beta$ has every scalar projection normal: in direction $u$, the coefficient vector on $(b,\epsilon )$ is $(Z^{T}u,u)$ and its variance is $u^T{V}_{i}u$. Since $V_1=V_2$, these projections have identical characteristic functions, so [F3] gives equal response laws. The covariance-component tuples are distinct, so the parameter is not identifiable.

The signs are centered, square-integrable, and independent; $b$ is independent of $\epsilon$. Thus this is a moment-only random-intercept mixed model with $G=[1]$ and $R=I_2$. By [F2], its complete-response covariance is $\mathrm{Cov}\left(\begin{array}{c}{Y}_1\\ Y_2\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)+I_2=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)\mathrm{.}$

Let the pattern space be $\mathcal{R}=\{0,1{\}}^2$ with its power-set $\sigma$-algebra, and define the full observation-pattern map $O(y_1,y_2)=(1,\mathbf{1}\{y_2=2\})$. It is Borel measurable because its first coordinate is constant and its second is the indicator of a Borel set. The induced kernel $K(y,A)=\mathbf{1}\{O(y)\in A\}$ is the point mass at $O(y)$ for each $y$; for every $A\subseteq \mathcal{R}$, $y\mapsto K(y,A)$ is measurable as the indicator of the Borel set $O^{-1}(A)$. Thus [F4] applies. Each sign triple has probability $1\mathrm{/}8$. The two supported states $(1,-1,-1)$ and $(1,-1,1)$ both give observed first-visit value $Y_1=0$, but their second-visit values are $Y_2=0$ and $Y_2=2$, respectively; their patterns are $(1,0)$ and $(1,1)$.

For the missing-visit pattern $r=(1,0)$, the mechanism probability is $1$ at $(Y_1,Y_2)=(0,0)$ and $0$ at $(0,2)$. These supported full-data points have the same observed coordinates $Y_{\mathrm{o}\mathrm{b}\mathrm{s}}(r)=Y_1=0$, so this pattern probability cannot depend only on the observed data for $r$. By [F3], the mechanism is not MAR, despite the correct random-intercept covariance in step 1.1.