# Estimation via projecting onto a convex body

Assume that $$\theta$$ is in a convex body $$K \in \mathbb{R}^n$$ and we observe $$y = \theta + z$$, where $$z$$ is a noise term (following, say, the normal distribution). Consider an estimator of $$\theta$$ by projecting $$y$$ onto $$K$$ and denote it by $$\hat\theta_K$$. How does the risk $$\mathbb{E} \|\hat\theta_K - \theta\|^2_2$$ depend on $$K$$? Is it true that $$\mathbb{E} \|\hat\theta_K - \theta\|^2_2 \le \mathbb{E} \|\hat\theta_{K'} - \theta\|^2_2$$ if $$K'$$ is a superset of $$K$$?

Thanks!

• Should this inequality is reversed? If $K’$ includes $K$, I would expect the errors of projection to be larger with $K$.
– user44143
Commented Mar 7, 2022 at 3:48
• @MattF. : This inequality will hold if e.g. $K$ and $K'$ are concentric balls centered at $\theta$. The distances here are, not from $y$, but from $\theta$. Commented Mar 7, 2022 at 4:15
• Do you have a response to the answer below? Commented Mar 9, 2022 at 21:56

$$\newcommand\th\theta\newcommand\R{\mathbb R}$$The answer is no in general if the noise term is allowed to have, say, any distribution symmetric about the origin with finite second moments.
Indeed, let e.g. $$n=2$$, $$K':=[-40,10]\times[-18,3],$$ $$K:=\{(x_1,x_2)\in K'\colon\, 2x_1+x_2-2\le0\},$$ and $$\th:=(0,0)$$, so that $$\th\in K\subset K'$$. Let $$Y$$ be a random point in $$\R^2$$ such that $$P(Y=y)=1/2=P(Y=-y)$$, where $$y:=(40,-3)$$, so that $$-y\in K\subset K'$$. Then for the respective projections $$\hat\th_K$$ and $$\hat\th_{K'}$$ of $$Y$$ onto $$K$$ and $$K'$$ we have $$P(\hat\th_K=(10, -18))=1/2=P(\hat\th_K=-y),$$ $$P(\hat\th_{K'}=(10, -3))=1/2=P(\hat\th_{K'}=-y),$$ which implies $$E\|\hat\th_K-\th\|^2_2=\frac{2033}{2}>859=E\|\hat\th_{K'}-\th\|^2_2,$$ so that the inequality in question fails to hold.
This counterexample is illustrated by the following picture, showing the rectangle $$K'$$, the darker region $$K$$, $$\th=(0,0)$$ (red), points $$\pm y$$ (black), and the projections $$(10,-18)$$ and $$(10,-3)$$ of the point $$y$$ onto $$K$$ (green) and onto $$K'$$ (blue).