Covariance between OLS estimates in a non-standard linear model 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
 A: 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.
