Theory and Methods of Statistics

$\begin{array}{ll}\hfill \sum l_i\left(x\right){\left(X_i-x\right)}^2& =\sum \frac{w_i\left(x\right)\left[c_2\left(x\right)-c_1\left(x\right)\left(X_i-x\right)\right]{\left(X_i-x\right)}^2}{c_2\left(x\right)-c_1{\left(x\right)}^2}\hfill \\ =\frac{c_2{\left(x\right)}^2-c_1\left(x\right)c_3\left(x\right)}{c_2\left(x\right)-c_1{\left(x\right)}^2}.\hfill \end{array}$

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 ...