feasible allocations ${\mathscr{F}}$ is SSF, any DSIC allocation rule $\allocs$ on a type space $\typespaces$, there is a DSIC extension $\extallocs$ of $\allocs$ on $\operatorname{Conv}(\typespaces)$.

The proofs for both and can be found in the supplementary materials. The main result of this paper is that without this condition a DSIC extension may not exist.

\[thm:rand\] There is a two agent type space $\randtypespaces$ with a DSIC allocation rule $\randallocs$, such that $\randallocs$ cannot be extended by a DSIC allocation rule to $\operatorname{Conv}(\randtypespaces)$.

We prove the theorem in two steps. We first present a setting with three-dimensional preferences for which we show the non-existence of *deterministic* extensions. We then build on the construction, lifting it to a higher dimension, where we strengthen the argument and show the non-existence of extensions that even allow randomization.

Non-existence of deterministic extensions
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We first present type space $\dettypespaces = \dettypespacei[1] \times \dettypespacei[2]$ and the allocation rule $\detallocs$, and then show that $\detallocs$ is DSIC and yet cannot be extended by any deterministic DSIC allocation rule on $\operatorname{Conv}(\dettypespaces)$.

The two agents have identical type spaces: for $i = 1, 2$, $\dettypespacei = \dettypespace \coloneqq \{A = [1, 0, 0], B = [0, 1, 0], C 