Index: The Book of Statistical ProofsProbability DistributionsUnivariate continuous distributions ▷ Normal distribution ▷ Relationship to chi-squared distribution

Theorem: Let $X_1, \ldots, X_n$ be independent random variables where each of them is following a normal distribution with mean $\mu$ and variance $\sigma^2$:

$\label{eq:norm} X_i \sim \mathcal{N}(\mu, \sigma^2) \quad \text{for} \quad i = 1, \ldots, n \; .$

Define the sample mean

$\label{eq:mean-samp} \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$

and the unbiased sample variance

$\label{eq:var-samp} s^2 = \frac{1}{n-1} \sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2 \; .$

Then, the sampling distribution of the sample variance is given by a chi-squared distribution with $n-1$ degrees of freedom:

$\label{eq:norm-chi2} V = (n-1) \, \frac{s^2}{\sigma^2} \sim \chi^2(n-1) \; .$

Proof: Consider the random variable $U_i$ defined as

$\label{eq:Ui} U_i = \frac{X_i - \mu}{\sigma}$ $\label{eq:norm-snorm} U_i \sim \mathcal{N}(0,1) \; .$

Then, the sum of squared random variables $U_i$ can be rewritten as

$\label{eq:sum-Ui2-s1} \begin{split} \sum_{i=1}^{n} U_i^2 &= \sum_{i=1}^{n} \left( \frac{X_i - \mu}{\sigma} \right)^2 \\ &= \sum_{i=1}^{n} \left( \frac{(X_i - \bar{X}) + (\bar{X} - \mu)}{\sigma} \right)^2 \\ &= \sum_{i=1}^{n} \frac{(X_i - \bar{X})^2}{\sigma^2} + \sum_{i=1}^{n} \frac{(\bar{X} - \mu)^2}{\sigma^2} + 2 \sum_{i=1}^{n} \frac{(X_i - \bar{X})(\bar{X} - \mu)}{\sigma^2} \\ &= \sum_{i=1}^{n} \left( \frac{X_i - \bar{X}}{\sigma} \right)^2 + \sum_{i=1}^{n} \left( \frac{\bar{X} - \mu}{\sigma} \right)^2 + 2\frac{(\bar{X} - \mu)}{\sigma^2} \sum_{i=1}^{n} (X_i - \bar{X}) \; . \end{split}$

Because the following sum is zero

$\label{eq:Xi-Xb} \begin{split} \sum_{i=1}^{n} (X_i - \bar{X}) &= \sum_{i=1}^{n} X_i - n \bar{X} \\ &= \sum_{i=1}^{n} X_i - n \cdot \frac{1}{n} \sum_{i=1}^{n} X_i \\ &= \sum_{i=1}^{n} X_i - \sum_{i=1}^{n} X_i \\ &= 0 \; , \end{split}$

the third term disappears, i.e.

$\label{eq:sum-Ui2-s2} \sum_{i=1}^{n} U_i^2 = \sum_{i=1}^{n} \left( \frac{X_i - \bar{X}}{\sigma} \right)^2 + \sum_{i=1}^{n} \left( \frac{\bar{X} - \mu}{\sigma} \right)^2 \; .$

Cochran’s theorem states that, if a sum of squared standard normal random variables can be written as a sum of squared forms

$\label{eq:cochran-p1} \begin{split} \sum_{i=1}^{n} U_i^2 = \sum_{j=1}^{m} Q_j \quad &\text{where} \quad Q_j = \sum_{k=1}^{n} \sum_{l=1}^{n} U_k B^{(j)}_{kl} U_l \\ &\text{with} \quad \sum_{j=1}^{m} B^{(j)} = I_n \\ &\text{and} \quad r_j = \mathrm{rank}(B^{(j)}) \; , \end{split}$

then the terms $Q_j$ are independent and each term $Q_j$ follows a chi-squared distribution with $r_j$ degrees of freedom:

$\label{eq:cochran-p2} Q_j \sim \chi^2(r_j) \; .$

We observe that \eqref{eq:sum-Ui2-s2} can be represented as

$\label{eq:sum-Ui2-s3} \begin{split} \sum_{i=1}^{n} U_i^2 &= \sum_{i=1}^{n} \left( \frac{X_i - \bar{X}}{\sigma} \right)^2 + \sum_{i=1}^{n} \left( \frac{\bar{X} - \mu}{\sigma} \right)^2 \\ = Q_1 + Q_2 &= \sum_{i=1}^{n} \left( U_i - \frac{1}{n} \sum_{j=1}^n U_j \right)^2 + \frac{1}{n} \left( \sum_{i=1}^{n} U_i \right)^2 \end{split}$

where, with the $n \times n$ matrix of ones $J_n$, the matrices $B^{(j)}$ are

$\label{eq:sum-Ui2-s3-Bj} B^{(1)} = I_n - \frac{J_n}{n} \quad \text{and} \quad B^{(2)} = \frac{J_n}{n} \; .$

Because all columns of $B^{(2)}$ are identical, it has rank $r_2 = 1$. Because the $n$ columns of $B^{(1)}$ add up to zero, it has rank $r_1 = n-1$. Thus, the conditions of Cochran’s theorem are met and the squared form

$\label{eq:Q1} Q_1 = \sum_{i=1}^{n} \left( \frac{X_i - \bar{X}}{\sigma} \right)^2 = (n-1) \, \frac{1}{\sigma^2} \, \frac{1}{n-1} \sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2 = (n-1) \, \frac{s^2}{\sigma^2}$

follows a chi-squared distribution with $n-1$ degrees of freedom:

$\label{eq:norm-chi2-qed} (n-1) \, \frac{s^2}{\sigma^2} \sim \chi^2(n-1) \; .$
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Metadata: ID: P233 | shortcut: norm-chi2 | author: JoramSoch | date: 2021-05-20, 10:18.