In case of Heteroscedasticity, proof of $E(\hat{\sigma}^2)\ne \sigma^2$
In this post, we will prove mathematically that $E(\hat{\sigma}^2)\ne \sigma^2$ when there is some presence of hetero in the data.
For the proof of $E(\hat{\sigma}^2)\ne \sigma^2$, consider the two-variable linear regression model in the presence of heteroscedasticity,
\begin{align}
Y_i=\beta_1 + \beta_2 X+ u_i, \quad\quad (eq1)
\end{align}
where $Var(u_i)=\sigma_i^2$ (Case of heteroscedasticity)
as
\begin{align}
\hat{\sigma^2} &= \frac{\sum \hat{u}_i^2 }{n-2}\\
&= \frac{\sum (Y_i – \hat{Y}_i)^2 }{n-2}\\
&=\frac{(\beta_1 + \beta_2 X_i + u_i – \hat{\beta}_1 -\hat{\beta}_2 X_i )^2}{n-2}\\
&=\frac{\sum \left( -(\hat{\beta}_1-\beta_1) – (\hat{\beta}_2 – \beta_2)X_i + u_i \right)^2 }{n-2}\quad\quad (eq2)
\end{align}
Noting that
\begin{align*}
(Y_i-\hat{Y}_i)&=0\\
\beta_1 + \beta_2 X + u_i\, – \,\hat{\beta}_1 – \hat{\beta}_2X &=0\\
-(\hat{\beta}_1 -\beta_1) – X(\hat{\beta}_2-\beta_2) – u_i & =0\\
(\hat{\beta}_1 -\beta_1) &= – X (\hat{\beta}_2-\beta_2) + u_i\\
\text{Applying summation on both side}&\\
\sum (\hat{\beta}_1-\beta_1) &= -(\hat{\beta}_2-\beta_2)\sum X + \sum u_i\\
(\hat{\beta}_1 – \beta_1) &= -(\hat{\beta}_2-\beta_2)\overline{X}+\overline{u}
\end{align*}
Substituting it in (eq2) and taking expectation on both sides:
\begin{align}
\hat{\sigma}^2 &= \frac{1}{n-2} \left[ -(-(\hat{\beta}_2 – \beta_2) \overline{X} + \overline{u} ) – (\hat{\beta}_2-\beta_2)X_i + u_i \right]^2\\
&=\frac{1}{n-2}E\left[(\hat{\beta}_2-\beta_2)\overline{X} -\overline{u} – (\hat{\beta}_2-\beta_2)X_i-u_i \right]^2\\
&=\frac{1}{n-2} E\left[ -(\hat{\beta}_2 – \beta_2)(X_i-\overline{X}) + (u_i-\overline{u})\right]^2\\
&= \frac{1}{n-2}\left[-\sum x_i^2 Var(\hat{\beta}_2) + E[\sum(u_i-\overline{u}]^2 \right]\\
&=\frac{1}{n-2} \left[ -\frac{\sum x_i^2 \sigma_i^2}{(\sum x_i^2)} + \frac{(n-1)\sum \sigma_i^2}{n} \right]
\end{align}
If there is homoscedasticity, then $\sigma_i^2=\sigma^2$ for each $i$, $E(\hat{\sigma}_i^2)=\sigma^2$.
The expected value of the $\hat{\sigma}^2=\frac{\hat{u}_i^2}{n-2}$ will not be equal to the true $\sigma^2$ in the presence of heteroscedasticity.
Read about more on Remedy of Heteroscedasticity
More on heteroscedasticity on Wikipedia
