0
$\begingroup$

I'm trying to determine $Cov(\hat{\beta}_0,\hat{\beta}_1|W^*)$ for $$ Z=\beta_0+\beta_1W+\xi $$ where $\xi|W \sim N(0, \sigma^2)$ where $\hat{\beta}_0=\bar{W}-\hat{\beta}_1\bar{Z}$ and $\hat{\beta_1}=\frac{\sum_{i=1}^n(W_i-\bar{W})(Z_i-\bar{Z})}{\sum_{i=1}^n(W_i-\bar{W})^2}$. I know this can be done in matrix form using $Var(\hat{\beta})=\sigma^2(W^TW)^{-1}$; however, I was hoping to do it just for covariance, but I end up getting stuck very early on. $$ \begin{align*} Cov(\hat{\beta}_0,\hat{\beta}_1|W^*) &=\mathbb{E}\left[\hat{\beta}_0\hat{\beta}_1|W^*\right]-\beta_0\beta_1 \\ &=\mathbb{E}\left[(\bar{W}-\hat{\beta}_1\bar{Z}) \left(\frac{\sum_{i=1}^n(W_i-\bar{W})(Z_i-\bar{Z})}{\sum_{i=1}^n(W_i-\bar{W})^2}\right)\Bigg|W^*\right]-\beta_0\beta_1 \\ &=\mathbb{E}\left[\bar{W}\left(\frac{\sum_{i=1}^n(W_i-\bar{W})(Z_i-\bar{Z})}{\sum_{i=1}^n(W_i-\bar{W})^2}\right)-\hat{\beta}_1\bar{Z}\left(\frac{\sum_{i=1}^n(W_i-\bar{W})(Z_i-\bar{Z})}{\sum_{i=1}^n(W_i-\bar{W})^2}\right)\Bigg|W^*\right]-\beta_0\beta_1 \\ &=\left(\mathbb{E}\left[\bar{W}\left(\frac{\sum_{i=1}^n(W_i-\bar{W})(Z_i-\bar{Z})}{\sum_{i=1}^n(W_i-\bar{W})^2}\right)\Bigg|W^*\right]-\mathbb{E}\left[\hat{\beta}_1\bar{Z}\left(\frac{\sum_{i=1}^n(W_i-\bar{W})(Z_i-\bar{Z})}{\sum_{i=1}^n(W_i-\bar{W})^2}\right)\Bigg|W^*\right]\right)-\beta_0\beta_1 \end{align*} $$ Is there a trick I am missing early on here?

$\endgroup$
0
$\begingroup$

It's simpler to go from matrix notation. Let $X=[1\ \ W]_{n\times 2}$, we know that $V=\operatorname{var}(\hat{\beta})=\sigma^2(X^TX)^{-1}$. This is a 2x2 matrix and we just need the covariance entry of it, i.e. $V_{12}=V_{21}$. $$V=\sigma^2\left(\begin{bmatrix}1^T\\ W^T\end{bmatrix}\begin{bmatrix}1 & W\end{bmatrix}\right)^{-1}=\sigma^2\left(\begin{bmatrix}n&n\bar{W}\\n\bar{W}& W^TW\end{bmatrix}\right)^{-1}=\frac{\sigma^2}{nW^TW-n^2\bar{W}^2}\begin{bmatrix}W^TW&-n\bar{W}\\-n\bar{W}&n\end{bmatrix}$$ which means $$\operatorname{cov}(\hat \beta_0, \hat \beta_1)=\sigma^2\frac{-\bar{W}}{W^TW-n\bar{W}^2}=\frac{-\sigma^2\bar W}{\sum(W_i-\bar W)^2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.