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Distribution of sample variances

If random samples \( X_i \), \( i = 1, \ldots, n \) are drawn from an \(
\mathrm{N}(\mu, \sigma^2) \) distribution, then \( X_i \sim
\mathrm{N}(\mu, \sigma^2) \) and, by the Central Limit Theorem, \(
\bar{X} \sim \mathrm{N}(\mu, \frac{\sigma^2}{n}) \). Standardising, we have \( \frac{X_i - \mu}{\sigma} \sim
\mathrm{N}(0, 1) \) and \( \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim
\mathrm{N}(0, 1) \).

Next, we notice that

\begin{displaymath}
x_i - \mu = x_i + \bar{x} - \bar{x} - \mu
\end{displaymath}

so that

\begin{displaymath}
(x_i - \mu)^2 = (x_i + \bar{x} - \bar{x} - \mu)^2 = (x_i - \bar{x})^2 +
(\bar{x} - \mu)^2 + 2\bar{x}(x_i - \bar{x})
\end{displaymath}

Hence, we have

\begin{displaymath}
\sum_i{(x_i - \mu)^2} = \sum_i{(x_i - \bar{x})^2} + \sum_i{(\bar{x} -
\mu)^2} + 2\bar{x}\sum_i{(x_i - \bar{x})}
\end{displaymath}

However,

\begin{displaymath}
\sum_i{(x_i - \bar{x})} = \sum_i{x_i} - \sum_i{\bar{x}} = \sum_i{x_i}
- n\bar{x} = 0
\end{displaymath}

since \( n\bar{x} = \sum_i{x_i} \). So

\begin{displaymath}
\sum_i{(x_i - \mu)^2} = \sum_i{(x_i - \bar{x})^2} + n (\bar{x} -
\mu)^2
\end{displaymath}

We now have an alternative expression for the sample variance,

\begin{displaymath}
S^2 = \frac{1}{n} \sum_i{(x_i - \bar{x})^2} = \frac{1}{n}
\sum_i{(x_i - \mu)^2} - (\bar{x} - \mu)^2
\end{displaymath}

and so

\begin{displaymath}
n \frac{S^2}{\sigma^2} = \sum_i\frac{(x_i - \bar{x})^2}{\sigma^2} - \frac{(\bar{x} - \mu)^2}{\sigma^2/n}
\end{displaymath}

Since \( \frac{X_i - \bar{X}}{\sigma} \sim \mathrm{N}(0, 1) \) and \( \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim
\mathrm{N}(0, 1) \), we have

\begin{displaymath}
\frac{(X_i - \mu)^2}{\sigma^2} \ensuremath{\sim \mathrm{\chi^2}(1)}\end{displaymath}

and

\begin{displaymath}
\frac{(\bar{X} - \mu)^2}{\sigma^2/n} \ensuremath{\sim \mathrm{\chi^2}(1)}\end{displaymath}

This means that

\begin{displaymath}
\sum_i\frac{(X_i - \mu)^2}{\sigma^2} \ensuremath{\sim \mathrm{\chi^2}(n)}\end{displaymath}

and so

\begin{displaymath}
\sum_i\frac{(X_i - \mu)^2}{\sigma^2} - \frac{(\bar{X} - \mu)^2}{\sigma^2/n}
\ensuremath{\sim \mathrm{\chi^2}(n-1)}\end{displaymath}

Therefore,

\begin{displaymath}
n\frac{S^2}{\sigma^2} \ensuremath{\sim \mathrm{\chi^2}(n-1)}\end{displaymath}

If, instead, we a dealing with the unbiased estimate of the population variance,

\begin{displaymath}
\hat{\sigma}^2 = \frac{1}{n-1} \sum_i (x_i - \bar{x})^2
\end{displaymath}

This is clearly equivalent to

\begin{displaymath}
\hat{\sigma}^2 = \frac{n}{n-1} S^2
\end{displaymath}

It now follows that

\begin{displaymath}
n S^2 = (n - 1) \hat{\sigma}^2
\end{displaymath}

Hence,

\begin{displaymath}
(n - 1) \frac{\hat{\sigma}^2}{\sigma^2} \ensuremath{\sim \mathrm{\chi^2}(n - 1)}\end{displaymath}


next up previous
Next: Copyright Up: Notes about the Chi-Squared Previous: Distribution of squares of
Alexander Frolkin 2001-02-01