What happens to kernel regression (Nadaraya–Watson estimator) at a kink point? Suppose $Y_i=g(X_i)+e_i$, where $g(\cdot)$ is a function unknown to the researcher, and $E(e_i|X_i)=0$. Suppose $X_i$ is a random variable in $[-1,1]$ with a density that is everywhere positive, and the true regression function satisfies $g(x)=-x$ if $x\leq 0$, $g(x)=x$ if $x>0$. Note that here 0 is a kink point at which $g(x)$ is not differentiable.
We have data $\{Y_i,X_i\}_{i=1}^{n}$ and estimate $g(x)$ using
$\widehat{g}(x)=\frac{\sum_{i=1}^{n} Y_i K_{h}(X_i-x)}{\sum_{i=1}^{n}  K_{h}(X_i-x)}$, where $x\in(-1,1)$ and $K_{h}(\cdot)=\frac{1}{h}K(\cdot)$ with $K(\cdot)$ being the kernel function.
Question:
Is $\widehat{g}(0)$ consistent for $g(0)$?
If it is consistent, how does the rate of convergence of $\widehat{g}(0)$ differ from that of $\widehat{g}(x)$ where $x\neq 0$ and is in the interior of $[-1,1]$?
Thanks!
 A: Of course this depends on the bandwidth and kernel you choose. To fix things, we take the rectangular Kernel (uniform on $[-1,1]$).
Despite the kink, your function is quite nice; it's just the absolute value function which is $1$-Lipschitz. The Nadaraya-Watson kernel nicely adapts to precisely this type of smoothness (while it does not do as well with higher order smoothness; in that case you should use higher order local linear regression).
Let us conduct a quick study of the uniform Kernel Nadaraya Watson estimator at $0$ with bandwidth $h$: $\hat{g}(0)$ is just the average of the $Y_i$'s corresponding to $X_i$'s that lie in the interval $[-h,h]$. Notice that for $X_i \in [-h,h]$ we have that: $\lvert\mathbb E[Y_i \mid X_i] - g(0)\rvert \leq h$ and so the bias of $\hat{g}(0)$ is of order $O(h)$. On the other hand, we will be averaging approximately $O(n \cdot h)$ points and so if $\text{Var}[Y_i \mid X_i]$ is bounded, we have that $\text{Var}(\hat{g}(0)) = O(1/(n \cdot h))$.
So the mean squared error (MSE) is $O(h^2) +  O(1/(n \cdot h))$. Thus as long as we choose $h$ such that $h \to 0$ and $n \cdot h \to \infty$, the estimator will be consistent. This answers your first question.
Optimizing over bandwidth will yield $h = n^{-1/3}$ and so the MSE will be of order $O(n^{-2/3})$. What happens away from $0$? If you use the same bandwidth as you did at $0$, then the rate will be the same. However you can repeat the above argument, with a more careful decomposition for the Bias, to get that the bias is of order $h^2$ (if you make some additional smoothness assumption on $p$). So away from $0$, you can pick a larger bandwidth, leading to a MSE decaying as $O(n^{-4/5})$.
