Covariance between OLS estimates in a non-standard linear model

Question

Consider a variant of the classical linear model:

$y_i=a + b\left(x_i-\bar{x} \right)+e_i $

Because $e_i \sim N \left (0, \sigma^2 \right)$, $y_i \sim N \left(a+ b \left(x_i -\bar{x} \right), \sigma^2 \right)$.

The OLS estimates of the slope and the intercept coefficient are:

$\hat{b}=\frac{\sum \left(x_i - \bar{x} \right) \left(y_i-\bar{y} \right)}{\sum \left(x_i - \bar{x} \right)^2} $ (as usual), but $\hat{a}=\bar{y}$

The two estimates, being a linear combination of normal variables are themselves normal with $\hat{b}\sim N \left( b ,\frac{ \sigma^2}{\sum \left(x_i - \bar{x} \right)^2}\right)$ and $\hat{a} \sim N \left( a, \frac{\sigma^2}{n}\right)$

I then need to show that the covariance between $\hat{a}$ and $\hat{b}$ is $0$

but I am stuck evaluating $$E \left[ \hat{a} \hat{b} \right]=\frac{1}{\sum \left( x_i-\bar{x} \right)^2} E \left[\bar{y} \sum \left(x_i -\bar{x} \right) y_i \right] $$

Could you please help me here?

Thanks

Answer 1

For compactness, denote $X^*\equiv X - \bar X$. Denote the $n\times 2$ regressor matrix by $\mathbf Z$ (that includes a series of ones and the series of $X^*$). Then in matrix notation, the variance-covariance matrix of the $2 \times 1$ OLS estimator $\hat \beta = (\hat a, \hat b)'$ is

$$\operatorname{Var}(\hat \beta) = \sigma^2_e\left(\mathbf Z'\mathbf Z\right)^{-1}$$

Write the matrix $\mathbf Z$ in block form and you get a $1 \times 2$ block matrix. This makes easy to write out $\mathbf Z'\mathbf Z$ which will give you your answer. Signal that you're done so that I can complete this answer.

ADDENDUM
The OP found his way so:
In block form, the regressors matrix is written as a $1\times 2$ matrix

$$\left[\begin{matrix} \mathbf 1 & \mathbf x^* \end{matrix}\right]$$

where $\mathbf 1$ is a $1\times n$ column vector of ones and $\mathbf x^*$ is the $1\times n$ column containing the data on $X^*$ the centered version of $X$. Then

$$\mathbf Z'\mathbf Z =\left[\begin{matrix} \mathbf 1' \\ \mathbf x^{*'}\\ \end{matrix}\right]\left[\begin{matrix} \mathbf 1 & \mathbf x^* \end{matrix}\right] = \left[\begin{matrix} \mathbf 1'\mathbf 1& \mathbf 1'\mathbf x^{*}\\ \mathbf x^{*'}\mathbf 1 & \mathbf x^{*'}\mathbf x^{*}\\ \end{matrix}\right]$$

But $$\mathbf 1'\mathbf x^{*} = \mathbf x^{*'}\mathbf 1 = \sum_{i=1}^nx_i^* =\sum_{i=1}^n(x_i-\bar x) = 0$$

So the off-diagonal elements of $\mathbf Z'\mathbf Z$ are zero and so will be the off-diagonal elements of $\left(\mathbf Z'\mathbf Z\right)^{-1}$, i.e. of the variance-covariance matrix of the OLS estimator. But these off-diagonal elements represent the covariance between the elements of the OLS estimator. QED.