as the log odds ratio $\theta_{y}$ in the full data (i.e. before values in $x$ are set to missing). The log odds of $x=1$ for $y=0$ in the observed and missing data are given by $$\begin{aligned}
        \left[\theta_{0} \mid r=1\right] &= \theta_{0}^{\text{obs}} = \text{ln}\left(\frac{n_{10}p_{r_{1}}}{n_{00}p_{r_{0}}}\right); \\
        \left[\theta_{0} \mid r=0\right] &= \theta_{0}^{\text{mis}} = \text{ln}\left(\frac{n_{10}\left(1-p_{r_{1}}\right)}{n_{00}\left(1-p_{r_{0}}\right)}\right),
    \end{aligned}$$ respectively. This implies that the correct adjustment in the imputation model’s intercept should be $$\begin{aligned}
        \theta_{0}^{\text{mis}} - \theta_{0}^{\text{obs}} &= \text{ln}\left(\frac{\left(1-p_{r_{1}}\right)p_{r_{0}}}{\left(1-p_{r_{0}}\right)p_{r_{1}}}\right) \\
        &= \text{ln}\left(\frac{\text{exp}\left(\alpha_{0}\right)}{\text{exp}\left(\alpha_{0}+\alpha_{x}\right)}\right) \\
        &= -\alpha_{x},
    \end{aligned}$$ which is minus the log odds ratio of observing $x$ for $x = 1$ compared to $x = 0$ in .

$x$ is MNAR dependent on $x$ and $y$
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Under this missingness mechanism, the posited model for the response indicator $r$ of $x$ is given by $$\label{eq:selection_mnar_x_y}
        \text{logit}\left[p\left(r=1 \mid x,y\right)\right] = \alpha_{0} + \alpha_{x}x + \alpha_{y}y,$$ and the corresponding probabilities of observing $x$ are $$p\left(r=1 \mid x=j, y=k\right) = p_{r_{jk}} = \text{expit}\left(\alpha_{0} + \alpha_{x}x + \alpha_{y}y\right); \quad j,k = 0,1.$$ For imputation model , the log odds ratios of $x=1$ for $y=1$ compared to $y=0$ in the observed and missing data are $$\begin{aligned}
        \theta_{y}^{\text{obs}} &= \text{ln}\left(\frac{n_{00}p_{r_{00}}n_{11}p_{r_{11}}}{n_{01}p_{r_{01}}n_{10}p_{r_{10}}}\right); \label{eq:thetay_obs_mnar_x_y}\\
        \theta_{y}^{\text{mis}} &= \text{ln}\left(\frac{n_{00}\left(1-p_{r_{00}}\right)n_{11}\left(1-p_{r_{11}}\right)}{n_{01}\left(1-p_{r_{01}}\right)n_{10}\left(1-p_{r_{10}}\right)}\right). \label{eq:thetay_mis_mnar_x_y}       
    \end{aligned}$$ Again, it can be shown from and that 