a default predictive random set ${\mathcal{S}}=\{v_1: |v_1| \leq V_1\}$, where $V_1 \sim {{\sf N}}(0,2)$. For the same singleton assertion, the C-step this time gives plausibility function $${\mathsf{pl}}_y(\theta) = 1 - |2\Phi\bigl(2^{-1/2}(y_1-2\theta)\bigr) - 1|.$$

The claim is that inference based on the latter IM formulation is more efficient than that based on the former. To check this, we consider the sampling distribution of ${\mathsf{pl}}_Y(0)$ in the case where $Y=(Y_1,Y_2)$ is an independent ${{\sf N}}(0,2)$ random vector. Figure \[fig:plaus.qq\] shows a quantile plot of the two simulated samples. By the validity theorem, the plausibilities are both stochastically no smaller than ${{\sf Unif}}(0,1)$. However, we see that plausibilities for the reduced, one-dimensional predictive random set are exactly ${{\sf Unif}}(0,1)$ distributed, while those based on the two-dimensional predictive random set tend to be considerably larger. The larger plausibility means less efficiency, e.g., wider plausibility intervals, so the IM based on the reduced one-dimensional predictive random set is preferred. This difference in efficiency is explained by the fact that the two-dimensional predictive random set for $(V_1,V_2)$ corresponds to a larger-than-necessary predictive random set for $V_1$; the conflict cases in the two-dimensional case have little to no effect on efficiency.

In the remainder of this section, 