What is the distribution of $\frac{1}{n}\sum_{n=1}^{m} (X_i - \mu)^2$

Question

$X_1,\cdots ,X_n $ random sample from $N(\mu,\sigma^2)$

I know, that distribution of the random variable $$\frac{(n-1)S^2}{\sigma^2}$$ has a chi-square distribution with $(n-1)$ degrees of freedom, where $$S^2=\frac{1}{n-1}\sum^{n}_{i=1}(X_i-\bar{X})^2.$$

Is it true, that $\frac{1}{n}\sum_{n=1}^{m} (X_i - \mu)^2$ has $\frac{\chi^2_n \sigma^2}{n}$ distribution?

Answer 1

Yes, it's true: The chi-squared distribution with $n$ degrees of freedom is defined as the distribution of a sum of squares of $n$ independent standard normal random variables $Z_i$. So the most basic defining result for chi-squared random variables is that for $X_1, ..., X_n \text{ ~ IID N}(\mu, \sigma^2)$ we have:

$$\sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2 = \sum_{i=1}^n Z_i^2 \text{ ~ } \chi_n^2.$$

Multiplying both sides by $\sigma^2/n$ gives the result:

$$\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \text{ ~ } \frac{\chi_n^2 \cdot \sigma^2}{n}.$$

Answer 2

See Cochran's Theorem...

Cochran's Theorem tells when a variable follows normal distribution then square of the same Normal distributed variable will follow Chi-Square distribution in some manner. Now here, the samples are taken from the normal distribution so squaring those samples and taking sum over those squared samples will follow Chi-Square distribution in some manner. Now, sum of squaring the mean deviated samples is nothing but the sample variance $(n-1)S^2$. Now, by Cochran's theorem, $y = \frac{\sum_{i=1}^n (X_i-\bar{X})^2}{\sigma^2} \sim \chi^2_{n-1}$. So $\sum_{i=1}^n (X_i-\bar{X})^2 \sim\chi^2_{n-1} \sigma^2$.

This is my explanation.