Beta hat conditional variance - Hansen Econometrics

Question

I'm working through Econometrics by Bruce Hansen, and I'm not sure how to get to his conditional variance proof on page 90.

Hansen says: For any $n \times r$ matrix $\mathbf{A} = \mathbf{A}(\mathbf{X})$: $\text{var}(\mathbf{A}^T\mathbf{y}|\mathbf{X}) = \text{var}(\mathbf{A}^T\mathbf{e}|\mathbf{X}) = \mathbf{A}^T\mathbf{D}\mathbf{A}$, where $D=\text{diag}(\sigma_1^2, ..., \sigma^2_n)$.

Why is this the case? I can see this step, but I'm not sure how to justify it from the full definition: $\text{var}(\mathbf{A}^T\mathbf{e}|\mathbf{X}) = \mathbf{A}^T\text{var}(\mathbf{e}|\mathbf{X})\mathbf{A} = \mathbf{A}^T\mathbf{D}\mathbf{A}$.

I.e. how do I get there with this: $\text{var}(\mathbf{Z}|\mathbf{X}) = \mathbb{E}[(\mathbf{Z} - \mathbb{E}[\mathbf{Z}|\mathbf{X}])(\mathbf{Z} - \mathbb{E}[\mathbf{Z}|\mathbf{X}])^T|\mathbf{X}], \mathbf{\hat\beta} = \mathbf{A}^T\mathbf{y}, \mathbf{A} = \mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}$

I tried but I kept on getting matrix dimensions that don't match up.

Answer 1

This step uses the bilinearity property of the variance operator. For any random vector $\mathbf{e}$ and any conformable constant matrix $\mathbf{A}$ you can derive the general bilinearity rule:

$$\begin{align} \mathbb{V}(\mathbf{A}^\text{T} \mathbf{e}) &= \mathbb{E}[(\mathbf{A}^\text{T} \mathbf{e}-\mathbb{E}(\mathbf{A}^\text{T} \mathbf{e})) (\mathbf{A}^\text{T} \mathbf{e}-\mathbb{E}(\mathbf{A}^\text{T} \mathbf{e}))^\text{T}] \\[6pt] &= \mathbb{E}[(\mathbf{A}^\text{T} \mathbf{e}-\mathbb{E}(\mathbf{A}^\text{T} \mathbf{e})) ((\mathbf{A}^\text{T} \mathbf{e})^\text{T}-\mathbb{E}(\mathbf{A}^\text{T} \mathbf{e})^\text{T})] \\[6pt] &= \mathbb{E}[(\mathbf{A}^\text{T} \mathbf{e}-\mathbb{E}(\mathbf{A}^\text{T} \mathbf{e})) (\mathbf{e}^\text{T} \mathbf{A}-\mathbb{E}(\mathbf{e}^\text{T} \mathbf{A}))] \\[6pt] &= \mathbb{E}[(\mathbf{A}^\text{T} \mathbf{e}-\mathbf{A}^\text{T} \mathbb{E}(\mathbf{e})) (\mathbf{e}^\text{T} \mathbf{A}-\mathbb{E}(\mathbf{e})^\text{T} \mathbf{A})] \\[6pt] &= \mathbb{E}[\mathbf{A}^\text{T} (\mathbf{e}-\mathbb{E}(\mathbf{e})) (\mathbf{e}^\text{T}-\mathbb{E}(\mathbf{e})^\text{T}) \mathbf{A}] \\[6pt] &= \mathbf{A}^\text{T} \mathbb{E}[(\mathbf{e}-\mathbb{E}(\mathbf{e})) (\mathbf{e}-\mathbb{E}(\mathbf{e}))^\text{T}] \mathbf{A} \\[6pt] &= \mathbf{A}^\text{T} \mathbb{V}(\mathbf{e}) \mathbf{A}. \\[6pt] \end{align}$$

(Note that in this working I have removed the conditioning element and just assumed that $\mathbf{A}$ is a constant. This simplifies the presentation and is analogous to your situation once you bring in the conditioning.) In the working in Hansen (2014) (p. 90) the matrix $\mathbf{A}$ is constant under the conditioning statement and the variance of the random vector is $\mathbb{V}(\mathbf{e}) = \mathbf{D} = \text{diag}(\sigma_1^2,...,\sigma_n^2)$ which gives the result of interest to you.