Variance of OLS First-Difference Estimator for Partial Linear Model I am reading Yatchew's (2003) Semiparametric Regression for the Applied Econometrician.
On page 2 and 3, he considers a partial linear model of the form:
$$y_{i} = z_{i} \beta + f(x_{i}) + \varepsilon_{i}$$
where he assumes that $x_{i}$ is uniformly distributed over the unit interval, and $\mathbb{E}[z_{i} | x_{i}] = g(x_{i})$. Yatchew shows that if the first derivatives of $f$ and $g$ are bounded then we can estimate $\beta$ using OLS on the first-difference. In particular he shows that, if $z_{i} = g(x_{i})+u_{i}$, then:
$$y_{i} - y_{i-1} \approx (u_{i} - u_{i-1})\beta + (\varepsilon_{i} - \varepsilon_{i-1})$$
So that we can estimate $\beta$ by
$$\hat{\beta}_{diff} = \frac{\frac{1}{n}\sum_{i=2}^{n} (y_{i}-y_{i-1})(z_{i}-z_{i-1})}{\frac{1}{n}\sum_{i=2}^{n} (z_{i} -z_{i-1})^{2}} \approx \beta + \frac{\frac{1}{n}\sum_{i=2}^{n} (\varepsilon_{i} - \varepsilon_{i-1})(u_{i} - u_{i-1})}{\frac{1}{n}\sum_{i=2}^{n} (u_{i} - u_{i-1})^{2}}\\$$
Now, Yatchew (2003) claims that the denominator in the fraction on the right converges to $2\sigma_{u}^{2}$ (and I agree), and that the numerator has mean $0$ (I agree) and variance $6 \sigma_{\varepsilon}^{2} \sigma_{u}^{2}$. 
How does how does he get that the variance is $6 \sigma_{\varepsilon}^{2} \sigma_{u}^{2}$? I have tried to derive this for a long time, but consistently get that the variance is $4 \sigma_{\varepsilon}^{2} \sigma_{u}^{2}$. Can anyone show it?
This result is needed to derive the asymptotic distribution of the estimator $\hat{\beta}_{diff}$.
Any help is much appreciated.
 A: \begin{align*}
&Var \left( \frac{1}{n}\sum_{i=2}^{n} (u_{i} - u_{i-1})(\varepsilon_{i} - \varepsilon_{i-1}) \right)\\
\\  &= Var \left( \frac{1}{n} \sum_{i=2}^{n} (u_{i}\varepsilon_{i} - u_{i}\varepsilon_{i-1}  - u_{i-1}\varepsilon_{i} + u_{i-1}\varepsilon_{i-1}) \right)\\
\\
& =\frac{1}{n^{2}} Var \left( u_{n}\varepsilon_{n}+2 \sum_{i=2}^{n-2} u_{i}\varepsilon_{i}  + u_{1}\varepsilon_{1} - \sum_{i=2}^{n}  u_{i}\varepsilon_{i-1}  - \sum_{i=2}^{n} u_{i-1}\varepsilon_{i} \right)\\
\\
& =\frac{1}{n^{2}} \left( \sigma_{u}^{2}\sigma_{\varepsilon}^{2} +4 \sum_{i=2}^{n-2} \sigma_{u}^{2}\sigma_{\varepsilon}^{2}  + \sigma_{u}^{2}\sigma_{\varepsilon}^{2} + \sum_{i=2}^{n}  \sigma_{u}^{2}\sigma_{\varepsilon}^{2}  + \sum_{i=2}^{n} \sigma_{u}^{2}\sigma_{\varepsilon}^{2} \right)\\
\\
& =\frac{1}{n^{2}} \left( \sigma_{u}^{2}\sigma_{\varepsilon}^{2} +4 (n-2) \sigma_{u}^{2}\sigma_{\varepsilon}^{2}  + \sigma_{u}^{2}\sigma_{\varepsilon}^{2} + (n-1) \sigma_{u}^{2}\sigma_{\varepsilon}^{2}  + (n-1)\sigma_{u}^{2}\sigma_{\varepsilon}^{2} \right)\\
\\
& =\frac{1}{n^{2}} (6n-8) \sigma_{u}^{2}\sigma_{\varepsilon}^{2} \\
\\
&\approx \frac{6 \sigma_{u}^{2}\sigma_{\varepsilon}^{2}}{n}  \\ 
\end{align*}
where the last line is for large $n$. 
