Question

Consider a simple linear regression model

• A. If the observed value of \((\hat{\beta}_0,\hat{\beta}_1)\) is \((0,-0.11)\), then \((a,b)=(-4.6,4.3)\)
• B. If \((a,b)=(0,1)\), then the observed value of \((\hat{\beta}_0,\hat{\beta}_1)\) is \((0.26,0.20)\)
• C. \(Cov(\overline{Y},\hat{\beta}_0)=7.2\), where \(\overline{Y}=\dfrac{1}{5}\sum_{i=1}^{5}Y_i\)
• D. \(Var(\hat{\beta}_0)=2Var(\hat{\beta}_1)\)

Hint:

When \(\bar{x}=0\) in simple linear regression, \(\hat{\beta}_0=\bar{Y}\) and the formula for \(\hat{\beta}_1\) simplifies to \(\hat{\beta}_1=\frac{\sum x_iY_i}{\sum x_i^2}\).

Answer 1

The Correct Option is A, C, D

Step 1: Compute useful quantities.
The given \(x\)-values are
\[ -2,-1,0,1,2 \] So,
\[ \bar{x}=\frac{-2-1+0+1+2}{5}=0 \] Also,
\[ S_{xx}=\sum_{i=1}^{5}(x_i-\bar{x})^2 \] \[ S_{xx}=(-2)^2+(-1)^2+0^2+1^2+2^2 \] \[ S_{xx}=10 \]

Step 2: Write \(\hat{\beta}_0\) and \(\hat{\beta}_1\).
Since \(\bar{x}=0\),
\[ \hat{\beta}_0=\bar{y} \] and
\[ \hat{\beta}_1=\frac{\sum x_iy_i}{\sum x_i^2} \] Now,
\[ \bar{y}=\frac{2.7+a-0.1+b-2.3}{5} \] \[ \bar{y}=\frac{a+b+0.3}{5} \] Also,
\[ \sum x_iy_i=(-2)(2.7)+(-1)a+0(-0.1)+1(b)+2(-2.3) \] \[ =-5.4-a+b-4.6 \] \[ =b-a-10 \] Therefore,
\[ \hat{\beta}_1=\frac{b-a-10}{10} \]

Step 3: Check option (A).
Given
\[ (\hat{\beta}_0,\hat{\beta}_1)=(0,-0.11) \] So,
\[ \frac{a+b+0.3}{5}=0 \] \[ a+b=-0.3 \] Also,
\[ \frac{b-a-10}{10}=-0.11 \] \[ b-a-10=-1.1 \] \[ b-a=8.9 \] Solving
\[ a+b=-0.3 \] and
\[ b-a=8.9 \] we get
\[ b=4.3,\qquad a=-4.6 \] Thus,
\[ (a,b)=(-4.6,4.3) \] Hence, option (A) is true.

Step 4: Check option (B).
If
\[ (a,b)=(0,1), \] then
\[ \hat{\beta}_0=\frac{0+1+0.3}{5} \] \[ \hat{\beta}_0=0.26 \] But
\[ \hat{\beta}_1=\frac{1-0-10}{10} \] \[ \hat{\beta}_1=-0.9 \] So, the observed value is
\[ (0.26,-0.9) \] not
\[ (0.26,0.20) \] Hence, option (B) is false.

Step 5: Check option (C).
Since
\[ \bar{x}=0, \] we have
\[ \hat{\beta}_0=\overline{Y} \] Therefore,
\[ Cov(\overline{Y},\hat{\beta}_0)=Cov(\overline{Y},\overline{Y}) \] \[ =Var(\overline{Y}) \] Since each error has variance \(36\) and the observations are uncorrelated,
\[ Var(\overline{Y})=\frac{36}{5} \] \[ Var(\overline{Y})=7.2 \] Hence, option (C) is true.

Step 6: Check option (D).
For simple linear regression,
\[ Var(\hat{\beta}_1)=\frac{\sigma^2}{S_{xx}} \] Here,
\[ \sigma^2=36,\qquad S_{xx}=10 \] So,
\[ Var(\hat{\beta}_1)=\frac{36}{10}=3.6 \] Also,
\[ Var(\hat{\beta}_0)=\sigma^2\left(\frac{1}{n}+\frac{\bar{x}^2}{S_{xx}}\right) \] Since \(\bar{x}=0\),
\[ Var(\hat{\beta}_0)=36\cdot \frac15 \] \[ Var(\hat{\beta}_0)=7.2 \] Now,
\[ 2Var(\hat{\beta}_1)=2(3.6)=7.2 \] Thus,
\[ Var(\hat{\beta}_0)=2Var(\hat{\beta}_1) \] Hence, option (D) is true.

Step 7: Final conclusion.
The true statements are
\[ \boxed{(A),\,(C)\text{ and }(D)} \]