Try the fastest way to create flashcards
Question

# 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 $\sigma^{2}$.

Solution

Verified
Step 1
1 of 2

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:

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

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 $\sigma^2$

## Recommended textbook solutions

#### An Introduction to Mathematical Statistics and Its Applications

6th EditionISBN: 9780134114217 (1 more)Morris L. Marx, Richard J. Larsen
1,098 solutions

#### Probability and Statistics for Engineers and Scientists

9th EditionISBN: 9780321629111 (4 more)Keying E. Ye, Raymond H. Myers, Ronald E. Walpole, Sharon L. Myers
1,204 solutions

#### The Practice of Statistics for the AP Exam

5th EditionISBN: 9781464108730 (1 more)Daniel S. Yates, Daren S. Starnes, David Moore, Josh Tabor
2,433 solutions

#### Statistics and Probability with Applications

3rd EditionISBN: 9781464122163Daren S. Starnes, Josh Tabor
2,555 solutions