Suppose the joint distribution of \(\left(X_1,v\right)\) is bivariate normal, i.e., \[\left(\begin{array}{c} X_{1}\\ v \end{array}\right)\sim N\left(\left(\begin{array}{c} 1\\ 0 \end{array}\right),\left(\begin{array}{cc} 4 & 3\\ 3 & 9 \end{array}\right)\right). \] A random sample \(\{\left(X_{1t},v_t\right)\}\) was obtained from this joint distribution and \(Y_t\) was obtained by \(Y_t=2+3X_{1t}+v_t\).
No, \(Y_t=2+3 X_{1t}+v_t\) is not a linear regression model, because \(\langle v,X_1 \rangle =\mathsf{Cov}\left(v, X_{1}\right) \overset{\mathsf{IID}}{=} \mathsf{Cov}\left(v_t, X_{1t}\right)= 3\neq 0\).
Because we have a random sample \(\{\left(X_{1t},Y_t\right)\}\) from a bivariate normal distribution \[\left(\begin{array}{c} X_{1}\\ Y \end{array}\right)\sim N\left(\left(\begin{array}{c} 1\\ 5 \end{array}\right),\left(\begin{array}{cc} 4 & 15\\ 15 & 45 \end{array}\right)\right),\] the least squares fit of \(Y\) on \(X_1\) will recover the best linear predictor \(\beta^*_0+\beta_1^* X_{1t}= \dfrac{5}{4}+\dfrac{15}{4} X_{1t}\).
Because English is your second language, it requires extra time to understand or figure out what the question is asking.
Therefore, in an exam, you should give yourself an allowance for problems with English comprehension.
Because you have to “design” a solution, it also requires extra time to write down the components of a complete solution.
A detailed solution is preferable but you need to decide what details are most important for the presentation of the solution.
Do NOT waste your time writing extra detail that is irrelevant. You waste time for the problem you are solving and others you have NOT solved!
Therefore, in an exam, you should give yourself an allowance for designing your solution.
Conclusion: Time pressure is real.
This quiz is based on Lectures 1 to 5. Some ideas from Chapter 2 and Chapter 4 are needed to obtain the correct answer.
It is not necessary that you produce the joint bivariate normal distribution of \(X_1\) and \(Y\). What is important is to know show that you can find \(\beta_0^*\) and \(\beta_1^*\). In addition, you should know the asymptotic behavior of \(\widehat{\beta}_0\) and \(\widehat{\beta}_1\).
To calculate \(\beta_0^*\) and \(\beta_1^*\): \[ \begin{eqnarray} \beta_1^* &=&\dfrac{\mathsf{Cov}\left(X_1,Y\right)}{\mathsf{Var}\left(X_1\right)}=\dfrac{\mathsf{Cov}\left(X_1,2+3X_1+v\right)}{\mathsf{Var}\left(X_1\right)}\\ &=&3+\dfrac{\mathsf{Cov}\left(X_1,v\right)}{\mathsf{Var}\left(X_1\right)}=3+\dfrac{3}{4}=\dfrac{15}{4} \\ \beta_0^* &=& \mathbb{E}\left(Y\right)-\beta_1^* \mathbb{E}\left(X_1\right) \\ &=& 2+3\mathbb{E}\left(X_1\right)+\mathbb{E}\left(v\right)-\dfrac{15}{4}\mathbb{E}\left(X_1\right)\\ &=& 2+3*1+0-\dfrac{15}{4}*1=\dfrac{5}{4} \end{eqnarray} \]
Note that we can drop the subscript \(t\) because we are in IID world.
Is \(Y_t=\dfrac{5}{4}+\dfrac{15}{4} X_{1t}+u_t\) a linear regression model? Show why or why not.
Is \(v_t\) an error from the best linear prediction of \(Y\) using \(X_1\)? Show why or why not.
Is \(Y_t=\dfrac{5}{4}+\dfrac{15}{4} X_{1t}+u_t\) a correctly specified linear regression model? Show why or why not.
As practice and check of past AE1 knowledge, try to obtain the bivariate normal displayed in the previous slide. Why is the joint distribution of \(X_1\) and \(Y\) also bivariate normal? How do you get the entries of the covariance matrix?
In the next slide, you will see another version of the quiz. This version is one possible illustration of what some call omitted variable bias (pages 43, 104, Exercise 3.23) or \(X_1\) is endogenous (page 245 Example 7.5).
Suppose the joint distribution of \(\left(X_1,v\right)\) is bivariate normal, i.e., \[\left(\begin{array}{c} X_{1}\\ v \end{array}\right)\sim N\left(\left(\begin{array}{c} \mu_1\\ 0 \end{array}\right),\left(\begin{array}{cc} \sigma^2_1 & \sigma_{1v}\\ \sigma_{1v} & \sigma^2_v \end{array}\right)\right). \] A random sample \(\{\left(X_{1t},v_t\right)\}\) was obtained from this joint distribution and \(Y_t\) was obtained by \(Y_t=\beta_0+\beta_1X_{1t}+v_t\).