Let $Y_{1}, Y_{2}, \ldots, Y_{n}$ be a random sample from a normal distribution. Use the statement to prove that $S^{2}$ is consistent for $\sigma^{2}$ .

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Step 1

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Let $S_n^2$ be a sample variance from a sample of size $n$ from $N(\mu,\sigma^2)$ distribution.

Previous exercise gives:

\begin{equation*} \mathrm{Var}\left ( S_n^2\right )=\dfrac{2\sigma^4}{n-1} \end{equation*}

Also $S_n^2$ is unbiased estimator for $\sigma^2$ , by definition this means:

\begin{equation*} \mathrm{E}S_n^2 =\sigma ^2 \end{equation*}

We need to prove consistency of $\left ( S_n^2 \right )_{n\in \Bbb{N}}$ for $\sigma^2$ , that is, for all $\epsilon >0$ :

\begin{equation*} \lim \limits _{n\to \infty} \mathrm{P} \left ( |S_n^2 - \sigma^2 | <\epsilon \right ) = 1 \end{equation*}

As written above for all $n\in \Bbb{N}$ , $S_n$ is a variable with finite variance and mean so we can apply $\textbf{Chebyshev's inequality}$ (Theorem 5.7.1.) for all $S_n,\ n\in \Bbb{N}$ :

\begin{equation*} \forall \epsilon >0 \qquad \mathrm{P} \left ( |S_n^2 - \mathrm{E}S_n^2 | <\epsilon \right ) \geq 1 - \dfrac{\mathrm{Var}S_n^2}{\epsilon^2} \end{equation*}

Now substitute known mean and variance:

\begin{equation} \forall \epsilon >0 \qquad \mathrm{P} \left ( |S_n^2 - \sigma ^2 | <\epsilon \right ) \geq 1 - \dfrac{2\sigma^4}{(n-1)\epsilon^2} \end{equation}

Since this inequality holds for all $n$ and limes is monotonous, from (1) follows:

\begin{align*} \forall \epsilon >0 \qquad \lim \limits _{n\to \infty} \mathrm{P} \left ( |S_n^2 - \sigma ^2 | <\epsilon \right ) &\geq \lim \limits _{n\to \infty} \left (1 - \dfrac{2\sigma^4}{(n-1)\epsilon^2}\right ) \\ &= 1-\dfrac{2\sigma^4}{\epsilon^2}\lim \limits _{n\to \infty} \dfrac{1}{n-1}\\ &=1 \end{align*}

And since limes on the left is a limes of probabilities, it is always smaller than 1.

Thus we have:

\begin{equation*} \forall \epsilon >0 \qquad \lim \limits _{n\to \infty} \mathrm{P} \left ( |S_n^2 - \sigma^2 | <\epsilon \right ) = 1 \end{equation*}

So by definition $\left ( S_n^2 \right )_{n\in \Bbb{N}}$ is consistent for $\sigma^2$

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