We study piece-wise constant signals corrupted by additive Gaussian noise over a $d$-dimensional lattice. Data of this form naturally arise in a host of applications, and the tasks of signal detection or testing, de-noising and estimation have been studied extensively in the statistical and signal processing literature. In this paper we consider instead the problem of partition recovery, i.e.~of estimating the partition of the lattice induced by the constancy regions of the unknown signal, using the computationally-efficient dyadic classification and regression tree (DCART) methodology proposed by \citep{donoho1997cart}. We prove that, under appropriate regularity conditions on the shape of the partition elements, a DCART-based procedure consistently estimates the underlying partition at a rate of order $\sigma^2 k^* \log (N)/\kappa^2$, where $k^*$ is the minimal number of rectangular sub-graphs obtained using recursive dyadic partitions supporting the signal partition, $\sigma^2$ is the noise variance, $\kappa$ is the minimal magnitude of the signal difference among contiguous elements of the partition and $N$ is the size of the lattice. Furthermore, under stronger assumptions, our method attains a sharper estimation error of order $\sigma^2\log(N)/\kappa^2$, independent of $ k^*$, which we show to be minimax rate optimal. Our theoretical guarantees further extend to the partition estimator based on the optimal regression tree estimator (ORT) of \cite{chatterjee2019adaptive} and to the one obtained through an NP-hard exhaustive search method. We corroborate our theoretical findings and the effectiveness of DCART for partition recovery in simulations.
翻译:我们研究的是因添加高尔西安噪音而腐蚀成碎片的恒定信号。 这种形式的数据自然出现在一系列应用中, 并且在统计和信号处理文献中广泛研究了信号检测或测试、 去除和估算等任务。 在本文中, 我们考虑的是分区恢复的问题, 也就是说, 估计由未知信号的粘结区引发的拉蒂的隔断, 即: 使用计算高效的 dyadic 分类和回归树( DCART) 方法, 由\ ciep{ donoho1997cart 提议。 我们证明, 在分配元素形状上的适当常规条件下, 基于 DCART 的 程序持续地估计了根基部分的隔断, 以美元=2 k\ log (N) /\ kapppa2, 其中, 美元是使用支持信号隔断的累变( dyadic roupate), $\\\\\ ladealdeal deal disal discode, $\\ kmatial ladeal roal roal rodeal mess roup roup roup rode, 在我们的一个最起码的 方法, 。 。 美元=x 美元, 美元=xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx