# Strategy-proofness of social choice function in two dimensions

Suppose the allocation space is the unit square $A=[0,1] \times [0,1] \in \mathbb{R}^2$. The outcome is a single point $x\in A$. Assume all agents have single-peaked preferences. That is, each agent $i$ has a preference ordering $\succeq_i$ over the outcomes; there exists a point $p_i=(x_i, y_i) \in A$ such that for all $x\in A \setminus \{p_i\}$ and all $\lambda \in [0,1)$, we have $(\lambda x + (1 − \lambda)p_i) \succ_i x$.

The social choice $f$ takes in the agents' preferences $(\succeq_1,\dots,\succeq_n)$, and output $x\in A$.

Why is the function $f(\succeq_1,\dots,\succeq_n)=(x_i,y_i)$ where $x_i$ and $y_i$ are the medians of the $x$ and $y$ coordinates, respectively, not strategy-proof?

What I think: In one dimension, we know that the median scheme is strategy-proof. I don't see how taking the median over each dimension is not?

In more than one dimension, however, you might get a better result by shifting the result's $x$ coordinate further away from your peak's $x$ coordinate – while that causes the result to end up further away from your acutal peak, you're then on a different line $\lambda x + (1 − \lambda)p_i$, which might be more favourable to you, since nothing in the premise constrains the relative ordering of points along different lines through the peak, so a point on one line further from the peak might be preferable to a point on another line closer to the peak.