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I am trying to prove that the First Difference method is asymptotically more efficient than OLS when the error term follows a random walk. Assume the following model

$$\begin{align}y_i&= \mathbf{X}_i\beta+\varepsilon_i\end{align}$$ where $y_i,\varepsilon_i\in\mathbb{R}^T,\ \beta\in\mathbb{R}^K$ and $\mathbf{X}_i\in\mathbb{R}^{T\times k}$

The first difference estimator is obtained by pre-multiplying this system of equations with $$\underset{\mathbb{R}^{(T-1)\times T}}{\mathbf{Q}_T}=\begin{bmatrix} -1 & 1 & 0 & \cdots &\cdots & 0 \\ 0 & -1 & 1 & 0 & \cdots & 0 \\ \vdots & &\ddots & \ddots & & 0\\ 0 & \cdots & \cdots & \cdots & -1 & 1 \end{bmatrix} $$

Assume (i) $\mathbb{E}[\varepsilon_i|\mathbf{X}_i]=0$ (ii) $\mathbb{E}[\mathbf{Q}_T\varepsilon_i\varepsilon_i^T\mathbf{Q}_T^T|\mathbf{X}_i]=\sigma^2_\Delta\mathbf{I}_{T-1}$.

Then the variance of the original error term is not identified because $\mathbf{Q}_T$ is not invertible. My first way to conitnue was to use the generalised inverse, so that $\mathbb{E}[\varepsilon_i\varepsilon_i^T|\mathbf{X}_i]=(\mathbf{Q}_T^T\mathbf{Q}_T)^g$, then

\begin{align} &N\cdot Avar(\hat{\beta}_{OLS}|\mathbf{X}_i)=(\mathbf{X}_i^T\mathbf{X}_i)^{-1}(\mathbf{X}_i(\mathbf{Q}_T^T\mathbf{Q}_T)^g\mathbf{X}_i)(\mathbf{X}_i^T\mathbf{X}_i)^{-1}\\\ &N\cdot Avar(\beta_{FD}|\mathbf{X}_i)=(\mathbf{X}_i^T\mathbf{Q}_T^T\mathbf{Q}_T\mathbf{X}_i)^{-1} \end{align} For $$N\cdot Avar(\hat{\beta}_{OLS}|\mathbf{X}_i) - N\cdot Avar(\hat{\beta}_{FD}|\mathbf{X}_i)$$ to be positive semi-definite,

$$N\cdot Avar(\hat{\beta}_{FD}|\mathbf{X}_i)^{-1} - N\cdot Avar(\hat{\beta}_{OLS}|\mathbf{X}_i)^{-1}$$ must be too. Then, after some algebra \begin{equation} \mathbf{X}_i^T(\mathbf{Q}_T^T\mathbf{Q}_T-\mathbf{X}_i(\mathbf{X}_i^T(\mathbf{Q}_T^T\mathbf{Q}_T)^g\mathbf{X}_i)^{-1}\mathbf{X}_i^T)\mathbf{X}_i \end{equation} meaning that \begin{equation} \mathbf{Q}_T^T\mathbf{Q}_T-\mathbf{X}_i(\mathbf{X}_i^T(\mathbf{Q}_T^T\mathbf{Q}_T)^g\mathbf{X}_i)^{-1}\mathbf{X}_i^T\end{equation} must be PSD. But this is where I am stuck, since I cannot do anything with the generalised inverse (apart from decomposing it to, say, $\mathbf{C}^T\mathbf{C}$). Could anyone give me a hint or how to continue (or give a hint for an alternative proof?)

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