A mixed-model covariance does not imply MAR dropout

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.