$\begin{array}{ll}\hfill \sum {l}_{i}\left(x\right){\left({X}_{i}-x\right)}^{2}& =\sum \frac{{w}_{i}\left(x\right)\left[{c}_{2}\left(x\right)-{c}_{1}\left(x\right)\left({X}_{i}-x\right)\right]{\left({X}_{i}-x\right)}^{2}}{{c}_{2}\left(x\right)-{c}_{1}{\left(x\right)}^{2}}\hfill \\ =\frac{{c}_{2}{\left(x\right)}^{2}-{c}_{1}\left(x\right){c}_{3}\left(x\right)}{{c}_{2}\left(x\right)-{c}_{1}{\left(x\right)}^{2}}.\hfill \end{array}$

For any nonnegative integer r, using Chebychev’s inequality, it can be established that

$\begin{array}{l}{\left(n{h}_{n}^{s+1}\right)}^{-1}\sum K\left(\left(x-{X}_{i}\right)/{h}_{n}\right){\left({X}_{i}-x\right)}^{s}\hfill \\ ={\left(n{h}_{n}\right)}^{-1}\sum K\left(\left(x-{X}_{i}\right)/{h}_{n}\right){\left\{\left({X}_{i}-x\right)/{h}_{n}\right\}}^{s}\stackrel{P}{\to}\int {u}^{s}K\left(u\right)duf\left(x\right).\hfill \end{array}$

Noting that $\int {u}^{s}K\left(u\right)du=0$ for s = 1 and s = 3 (since K is symmetric about 0), we have

$\begin{array}{ll}\hfill {c}_{1}\left(x\right)& ={h}_{n}\left[\int uK\left(u\right)du+{o}_{P}\left(1\right)\right]={o}_{P}\left({h}_{n}\right),\hfill \\ \hfill {c}_{2}\left(x\right)& ={h}_{n}^{2}\left[\int {u}^{2}K\left(u\right)du+{o}_{P}\left(1\right)\right]={h}_{n}^{2}\left[{\sigma}_{K}^{2}+{o}_{P}\left(1\right)\right],\text{and}\hfill \\ \hfill {c}_{3}\left(x\right)& ={h}_{n}^{3}\left[\int {u}^{3}K\left(u\right)du+{o}_{P}\left(1\right)\right]={o}_{P}\left({h}_{n}^{3}\right).\hfill \end{array}$

Plugging in these approximations for ${c}_{1}\left(x\right)$, ${c}_{2}\left(x\right)$, and ${c}_{3}\left(x\right)$ in the expression ...

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