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Suppose we use the Kernel Regression Estimator $$\hat{m}(c)=\frac{\sum_{i=1}^n K\left(\frac{x_i-c}{h}\right)y_i}{\sum_{i=1}^n K\left(\frac{x_i-c}{h}\right)}$$ where $h\to 0$ and $nh\to \infty$ as $n\to \infty$.

The true DGP has form $$y_i=\alpha +\beta x_i +\gamma z_i+\varepsilon_i$$

I assume $\{(y_i,x_i,z_i)\}$ is i.i.d. and all variables are absolutely continuous, have finite second moments, and have positive density over the whole real line. I assume strict exogeneity in both variables, i.e., $\mathbb{E}[\varepsilon_i|x_i;z_i]=0$, and that $x_i$ and $z_i$ are independent of each other.

I want to know to what object $\hat{m}(c)$ converges.

To this end, define $\hat{g}(c)=\frac{1}{nh}\sum_{i=1}^n K\left(\frac{x_i-c}{h}\right)y_i$ and $\hat{f}(c)=\frac{1}{nh}\sum_{i=1}^n K\left(\frac{x_i-c}{h}\right)$ such that $\hat{m}(c)=\hat{g}(c)/\hat{f}(c)$. It is easy enough to show $\hat{f}(c)\xrightarrow{p} f(c)$ where $f(c)$ is the pdf of $x$. Now, letting $p(z)$ be the pdf of $z$, \begin{align*} \mathbb{E}[\hat{g}(c)] &= \mathbb{E}\left[\frac{1}{nh}\sum_{i=1}^nK\left(\frac{x_i-c}{h}\right)y_i\right] \\ &= \frac{1}{h}\mathbb{E}\left[K\left(\frac{x_i-c}{h}\right)y_i\right] \\ &=\frac{1}{h}\mathbb{E}\left[K\left(\frac{x_i-c}{h}\right)(\alpha +\beta x_i +\gamma z_i+\varepsilon_i)\right] \\ &=\frac{1}{h}\mathbb{E}\left[K\left(\frac{x_i-c}{h}\right)(\alpha +\beta x_i +\gamma z_i)\right] \\ &= \frac{1}{h}\int_\Omega K\left(\frac{x-c}{h}\right)(\alpha +\beta x +\gamma z) \, d\mathbb{P} \\ &=\frac{1}{h}\iint K\left(\frac{x-c}{h}\right)(\alpha +\beta x +\gamma z)f(x)p(z) \, dx \, dz \\ &=\iint K(u)(\alpha +\beta (uh+c) +\gamma z) f(uh+c)p(z) \, du \, dz \\ &= (\alpha + \beta c+\gamma \mathbb{E}[z])f(c) +o(1) \\ &\xrightarrow{n\to \infty} (\alpha + \beta c+\gamma \operatorname{\mathbb{E}}[z])f(c) \end{align*}

I can also show that $\mathbb{V}[\hat{g}(c)]\to 0$ which shows $\hat{g}(c)\xrightarrow{p} (\alpha + \beta c+\gamma \operatorname{\mathbb{E}}[z])f(c)$. Therefore, I think $$\hat{m}(c)\xrightarrow{p}\alpha + \beta c+\gamma \operatorname{\mathbb{E}}[z]$$

However, the TA for the course I'm taking mentioned that $\hat{m}(c)$ should converge to something potentially nonlinear in $c$ which I clearly am not getting. I think the reason for this could be that I am assuming $x_i$ and $z_i$ are independent. However, without this assumption I'm not sure how to find the probability limit of $\hat{m}(c)$.

Could someone either point out where I have gone wrong in my working or explain how to find the limit without assuming independence?

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I spoke with my TA, and the nonlinearity comes from a $\mathbb{E}[z|x]$ term. A more general approach shows that $\hat{m}(c)\xrightarrow{p} \mathbb{E}[y|x]$.

Using the same notation as in the question and letting $r(\cdot,\cdot)$ be the joint pdf of $(x,y)$, \begin{align*} \mathbb{E}[\hat{g}(c)] &= \mathbb{E}\left[\frac{1}{nh}\sum_{i=1}^n K\left(\frac{x_i-c}{h}\right)y_i\right] \\ &= \frac{1}{h}\mathbb{E}\left[K\left(\frac{x_i-c}{h}\right)y_i\right] \\ &=\frac{1}{h}\int K\left(\frac{x-c}{h}\right)y ~\mathrm d\mathbb{P} \\ &=\frac{1}{h}\int K\left(\frac{x-c}{h}\right)y \ r( x,y)~\mathrm d(x,y) \\ &=\frac{1}{h}\iint K\left(\frac{x-c}{h}\right)y \ r( x,y)~\mathrm dx\mathrm dy \\ &=\iint K\left(u\right)y \ r( uh+c,y)~\mathrm du\mathrm dy \\ &=\iint K\left(u\right)y \ [r(c,y)+( r( uh+c,y)-r(c,y))]~\mathrm du\mathrm dy \\ &=\int y \ r(c,y)~\mathrm dy + \iint K(u)y[r( uh+c,y)-r(c,y)]~\mathrm du\mathrm dy \\ &=\int y \ r(c,y)~\mathrm dy + o(1) \\ &= f(c)\mathbb{E}[y|x=c] +o(1) \\ &\to f(c)\mathbb{E}[y|x=c], \end{align*}

where the little $o$ bound is shown by assuming the kernel density has bounded support, and that $r(x,y)$ is continuous.

Then, it is a standard (but longer) argument to show that $\mathbb{V}[\hat{g}(c)]\to 0$. Therefore, $\hat{g}(c)\xrightarrow{p} f(c)\mathbb{E}[y|x=c]$. In combination with $\hat{f}(c)\xrightarrow{p} f(c)$, this gives that $$\hat{m}(c)\xrightarrow{p} \mathbb{E}[y|x=c].$$

In the question, it is given that the DGP for $y_i$ is $$y_i=\alpha +\beta x_i +\gamma z_i +\varepsilon_i$$ and that $\mathbb{E}[\varepsilon_i|x_i;z_i]=0$. Therefore, by law of iterated expectations, $$\mathbb{E}[y|x=c]=\alpha +\beta c +\gamma \mathbb{E}[z|x=c]$$ and this may be nonlinear in $c$ if the mapping $c\mapsto \mathbb{E}[z|x=c]$ is nonlinear.

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    $\begingroup$ Appreciate you self-answered your query. $\endgroup$ Commented Apr 18 at 8:04

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