# Show that $\hat{\delta}_1=\hat{\beta}_1+(X_1^T X_1)^{-1} X_1^TX_2\hat{\beta}_2$

Let $\hat{\beta}=(\hat{\beta}_1,\hat{\beta}_2)^T$ be the least squares estimator in the regression model $Y=X_1\beta_1+X_2\beta_2+u$. Let $\hat{\delta}_1$ be the least squares estimator of the regression of $Y$ on $X_1$. Show that $$\hat{\delta}_1=\hat{\beta}_1+(X_1^T X_1)^{-1} X_1^TX_2\hat{\beta}_2.$$

Hello, to be honest I do not understand what do to here resp. what is meant with $\hat{\delta}_1$.

To my knowledge it is $$\hat{\beta}=(X^TX)^{-1}X^TY.$$

Is here meant that $X=(X_1,X_2)$ with $X_1=(1,...,1)^T$?

And what is meant with $\hat{\delta}_1$? Maybe $Y=\delta_1 X_1+u$ and $$\hat{\delta}_1=(X_1^TX_1)^{-1}X_1^TY?$$

## 1 Answer

Yes you are right in understanding what is $\hat \delta_1$. So now it is rather simple. Just replace in the last formula $Y=X_1\hat \beta_1 +X_2\hat \beta_2 + u$ and take in account that $X_1^T u=0$,

• But it is $Y=X_1\beta_1+X_2\beta_2+u$ and not $Y=X_1\hat{\beta}_1+X_2\hat{\beta}_2+u$ - or what do you mean? – math12 May 28 '14 at 20:27
• After estimation $\beta_{1,2}$ you get $u$ which is unobservable but derived from least squares such that $u$ is orthogonal to $X_{1,2}$ – Alexander Vigodner May 28 '14 at 21:25