how to prove that $\hat \sigma^2$ is a consistent for $\sigma^2$?

Question

Consider a regression model $Y_n=X_n\beta +\varepsilon$, where $X_n$ is a $n \times p_n$ matrix, and $\varepsilon=(\varepsilon_1,...,\varepsilon_n)'$ consists of independent and identically distributed variables with $E(\varepsilon_1)=0$ and $Var(\varepsilon_1)=\sigma^2$. Suppose that $\varepsilon_1 \sim N(0, \sigma^2)$ and suppose that $\hat \sigma ^2$ is the estimator of $\sigma^2$. Let

$$\hat \sigma ^2=\frac{\left\|Y_n-X_n\hat\beta\right\|^2}{n-p_n}.$$

Then, $\hat \sigma^2$ consistent for $\sigma^2$.

I can prove that $\hat \sigma^2$ is unbiased estimator for $\sigma^2$.

How can I continue to prove that $\hat \sigma^2$ consistent for $\sigma^2$?

Answer 1

Hint: \begin{align*} P(|\hat{\sigma}^2 - \sigma^2| > \epsilon) \le \frac{E[(\hat{\sigma}^2 - \sigma^2)^2]}{\epsilon^2} \end{align*} by Markov's/Chebyshev's inequality. Then take the limit.

Answer 2

Alternative hint:

\begin{eqnarray} \hat{\sigma}^{2}=\frac{Y_n'(I-P_{X_n})Y_n}{n-p_n}=\frac{\varepsilon'(I-P_{X_n})\varepsilon}{n-p_n}. \end{eqnarray} So, \begin{align} \hat{\sigma}^{2}=\frac{\frac{\varepsilon'\varepsilon}{n}-\frac{\varepsilon'X_n}{n}\left(\frac{X_n'X_n}{n}\right)^{-1}\frac{X_n'\varepsilon}{n}}{\frac{n-p_n}{n}}. \end{align}

This is a continuous function of three arguments \begin{eqnarray*} \frac{X_n'X_n}{n}&\to_p&?\\ \frac{\varepsilon'\varepsilon}{n}&=&\frac{1}{n}\sum_{t=1}^{n}\varepsilon_{t}^{2}\to_p?\\ \frac{X_n'\varepsilon}{n}&\to_p&?. \end{eqnarray*}