# Partitioned regression model: estimator of beta 1

below is an exercise that is really giving me a hard time, I believe that there is a simple way around it but I can not find it:

Assume the correct regression model is Y = X$$\beta$$ + $$\epsilon$$ for E($$\epsilon$$) = 0 and var($$\epsilon$$) = $$\sigma^2$$I.

Assume the matrix X of dimensions n × m with m < n has full rank. Denote by $$\hat\beta$$ the ordinary least squares estimator of $$\beta$$. Assume as known that the upper left corner of the inverse of $$\begin{bmatrix}\Sigma_{11}&\Sigma_{12}\\\Sigma_{21}&\Sigma_{22} \end{bmatrix}$$

$$\Sigma_{11} = \Sigma_{11}^{-1}+\Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}\Sigma_{21}\Sigma_{11}^{-1}$$

and the lower right corner is $$\Sigma_{22} = (\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}$$

a. Assume that we forget some independent variables and fit the regression model $$Y = X_1\beta_{1\ast}+\epsilon_{\ast}$$

where X = [X1; X2] and E($$\epsilon_{\ast}$$) = 0 and var($$\epsilon_{\ast}$$) = $$\sigma^2$$I.

Write $$\beta$$=$$\begin{bmatrix}\beta_{1}\\\beta_{2}\end{bmatrix}$$

Assuming the “wrong” model we estimate $$\beta_{1}$$ by $$\hat\beta_{1\ast}$$=$$(X_1^TX_1)^{-1}X_1^TY$$

Let $$\hat\beta_1$$ be the best unbiased linear estimator of $$\beta_1$$ in the correct model. Show that var($$\hat\beta_1$$) − var($$\hat\beta_{1\ast}$$) = $$AB^{-1}A^{T}$$

where A = $$(X_1^TX_1)^{-1}X_1^TX_2$$

B=$$X_2^TX_2-X_2^TX_1A$$

Now, my first idea was to express the $$\hat\beta_1$$ from the partitioned model, and using Takeshi Amemiya textbook,

$$\hat\beta_1$$=$$(X_1^TM_2X_1)^{-1}X_1^TM_2Y$$

Where $$M_2=I-X_2(X_2^TX_2)^{-1}X_2^T$$

Where $$M_2$$ is idempotent, and $$M_2X_2=0$$

Therefore I finally get $$\hat\beta_1$$=$$\beta_1$$+$$(X_1^TM_2X_1)^{-1}X_1^TM_2\epsilon$$. The expected value of $$\hat\beta_1$$ is $$\beta_1$$ because the expected value of $$\epsilon$$ is 0, and the term that multiplies the $$\epsilon$$ is nonstochastic.

To calculate the variance of $$\hat\beta_1$$ I take the formula E(($$\hat\beta_1$$-E($$\hat\beta_1$$)($$\hat\beta_1$$-E($$\hat\beta_1)^T$$) which gets me to $$\sigma^2(X_1^TM_2X_1)^{-1}$$.

For $$\hat\beta_{1\ast}$$ I plug in the 'wrong' model and get $$\beta_{1\ast}$$+$$(X_1^TX_1)^{-1}X_1^T\epsilon_{\ast}$$. I get the expected value of $$\hat\beta_{1\ast}$$ to be $$\beta_{1\ast}$$ because the expected value of $$epsilon_{\ast}$$ is zero, so the variance I get using the same approach as above is $$\sigma^2(X_1^TX_1)^{-1}$$.

Now, when I subtract the variances as requested in the exercise, I don't get nearly the same thing they got in the text, I still have $$\sigma^2$$, and I can't get rid of the bracket because the inverse of the entire product is defined but the inverse of individual matrices might not be. Additionally, I have no idea what to do with the clue I was given about the upper left and the lower right corner of the inverse. Would there be a soul kind enough to give me a hint if I am even going in the proper direction? Thank you very very much

• Please add the "self-study" tag to your question. – Peter Flom Feb 22 at 12:34
• @PeterFlom Thank you for your comment, I just edited the post. – Ivana Feb 22 at 12:49