Answer to Question #299693 in Statistics and Probability for bonke

Question

Y10with usual OLS assumptions.  Derive the expression for the error variance.

Answer 1

To derive variance for OLS estimator.

Solution:

$\begin{aligned} &\operatorname{Var}(\hat{\alpha})=E\left[(\hat{\beta}-\beta)^{2}(\bar{X})^{2}\right]+E\left(\bar{\varepsilon}^{2}\right)-E[2 \bar{\varepsilon}(\hat{\beta}-\beta) \bar{X}] \\ &\operatorname{Var}(\hat{\alpha})=(\bar{X})^{2} \operatorname{Var}(\hat{\beta})+\operatorname{Var}(\bar{\varepsilon})-2 \bar{X} \times E[\bar{\varepsilon}(\hat{\beta}-\beta)]...(i) \\ &E[\bar{\varepsilon}(\hat{\beta}-\beta)]=\frac{\sigma^{2}}{n^{2}}{\sigma_{r}^{2}} E[0]=0 \end{aligned}$

Substituting this in equation (i) ,

$\begin{gathered} \operatorname{Var}(\hat{\alpha})=(\bar{X})^{2} \operatorname{Var}(\hat{\beta})+\operatorname{Var}(\vec{\varepsilon})-2 \bar{X} \times 0 \\ \operatorname{Var}(\hat{\alpha})=(\bar{X})^{2} \operatorname{Var}(\hat{\beta})+\operatorname{Var}(\bar{\varepsilon}) \\ \operatorname{Var}(\hat{\alpha})=(\bar{X})^{2} \frac{\sigma^{2}}{n \widehat{\sigma_{x}^{2}}}+\operatorname{Var}\left(\frac{1}{n} \sum_{j=1}^{n} \varepsilon_{j}\right) \quad\left[\text { using } \operatorname{Var}(\hat{\beta})=\frac{\sigma^{2}}{\mathrm{n} \widehat{\sigma_{\mathrm{X}}^{2}}} \text { and } \bar{\varepsilon}=\frac{1}{n} \sum_{j=1}^{n} \varepsilon_{j}]\right. \end{gathered}$

$\operatorname{Var}(\hat{\alpha})=\frac{(\bar{X})^{2} \sigma^{2}}{n \widehat{\sigma_{x}^{2}}}+\left(\frac{1}{n}\right)^{2} \operatorname{Var}\left(\sum_{j=1}^{n} \varepsilon_{j}\right) \quad \text { [ Using } \operatorname{Var}(Y)=\operatorname{Var}(a X+b)=a^{2} \operatorname{Var}(X)]$

The error terms $\varepsilon 1, \varepsilon 2, \ldots, \varepsilon_{j}$  are independent. Thus, using property 2 B ,

$\operatorname{Var}(\hat{\alpha})=\frac{(\bar{X})^{2} \sigma^{2}}{n \widehat{\sigma_{x}^{2}}}+\frac{1}{n^{2}} \sum_{j=1}^{n} \operatorname{Var}\left(\varepsilon_{j}\right)$

Once again, by constant variance assumption, $\operatorname{Var}\left(\varepsilon_{i}\right)=\sigma^{2}$  (a constant),

$\begin{aligned} \therefore \operatorname{Var}(\hat{\alpha}) &=\frac{(\bar{X})^{2} \sigma^{2}}{n \widehat{\sigma_{x}^{2}}}+\frac{1}{n^{2}} \sum_{j=1}^{n} \sigma^{2} \\ \operatorname{Var}(\hat{\alpha}) &=\frac{(\bar{X})^{2} \sigma^{2}}{n \widehat{\sigma_{x}^{2}}}+\frac{n \sigma^{2}}{n^{2}} \\ \operatorname{Var}(\hat{\alpha}) &=\frac{(\bar{X})^{2} \sigma^{2}}{n \widehat{\sigma_{x}^{2}}}+\frac{\sigma^{2}}{n} \end{aligned}$

This is the required variance of OLS estimator.