STA260 Lecture 05

Pre-Lecture

Lecture

Main Results

1. Marginal Distribution of Sample Variance

   • Core Theorem: $\boxed{{\displaystyle \frac{(n-1)S^2}{{\sigma }^2}=\frac{\sum _{i=1}^n(X_i-\overline{X}{)}^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2}}$
   • Distribution of $\overline{X}$: $\boxed{{\displaystyle \overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})}}$

2. Independence of Sample Mean and Sample Variance

   • Under normal sampling ($X_i\sim N(\mu ,{\sigma }^2)$), we assume $\overline{X}\perp \perp S^2$.

Key Proof Components

• Sum of Squares Decomposition Identity

• We first establish the algebraic identity:

  $$\sum _{i=1}^n(X_i-\mu {)}^2=\sum _{i=1}^n(X_i-\overline{X}{)}^2+n(\overline{X}-\mu {)}^2$$

• Derivation:

  • $\sum (X_i-\mu {)}^2=\sum (X_i-\overline{X}+\overline{X}-\mu {)}^2$
  • Expand: $\sum [(X_i-\overline{X}{)}^2+2(X_i-\overline{X})(\overline{X}-\mu )+(\overline{X}-\mu {)}^2]$
  • The middle term vanishes because $\sum (X_i-\overline{X})=n\overline{X}-n\overline{X}=0$.
  • Result: $\sum (X_i-\overline{X}{)}^2+n(\overline{X}-\mu {)}^2$

• Chi-squared from Sum of Squared Standard Normals

  • Recall that $W=\sum _{i=1}^n{Z}_i^2=\sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2\sim {\chi }_{(n)}^2$.
  • The MGF of this ${\chi }_{(n)}^2$ variable is $M_W(t)=(1-2t{)}^{-\frac{n}{2}}$.

• Proof of Sample Variance Distribution

• Divide the decomposition identity by ${\sigma }^2$:

  • $$\underset{W\sim {\chi }_{(n)}^2}{\underbrace{\sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2}}=\underset{Q}{\underbrace{\frac{(n-1)S^2}{{\sigma }^2}}}+\underset{Z^2\sim {\chi }_{(1)}^2}{\underbrace{{\left(\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}\right)}^2}}$$

• By MGFs (exploiting independence of $\overline{X}$ and $S^2$):

  • $M_W(t)=\mathrm{E}[e^{t(Q+Z^2)}]=M_Q(t)\cdot M_{Z^2}(t)$
  • $(1-2t{)}^{-\frac{n}{2}}=M_Q(t)\cdot (1-2t{)}^{-\frac{1}{2}}$
  • $M_Q(t)=\frac{(1-2t{)}^{-n/2}}{(1-2t{)}^{-1/2}}=(1-2t{)}^{-\frac{n-1}{2}}$
  • This identifies $Q$ as following a Chi-squared distribution with $n-1$ degrees of freedom.

Worked Example

• Chi-squared Probability Calculations Example
• Problem: Find $b_1$ and $b_2$ such that $P(b_1\le S^2\le b_2)=0.9$ given $n=10$ and ${\sigma }^2=1$.
• Method:
  1. Standardize to Chi-square: $P(9b_1\le \frac{(n-1)S^2}{{\sigma }^2}\le 9b_2)=0.9$.
  2. Let $C\sim {\chi }_{(9)}^2$. We need $P(9b_1\le C\le 9b_2)=0.9$.
  3. Split $\alpha =0.10$ into equal tails ($0.05$ each) using STA260-Winter2026-Statistical_Tables.pdf.
• Solution:
  • Lower Bound ($a$): $P(C\le a)=0.05⟹a=3.32511$.
    • $9b_1=3.32511⟹b_1\approx 0.369$.
  • Upper Bound ($b$): $P(C\le b)=0.95⟹b=16.919$.
    • $9b_2=16.919⟹b_2\approx 1.880$.

Statistical Tables

• STA260-Winter2026-Statistical_Tables.pdf
  • Contains the Chi-squared quantiles used in the example above.
  • Required for Confidence Intervals and Hypothesis Tests.

Post-Lecture

• STA260 Post-Lecture 05
• STA260 Lecture 05 Flashcards