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  • STA260 Lecture 04

STA260 Lecture 04 Raw

STA260H5S_Winter2026_Chapter6_Distributions_Derived_Normal-Annotated.pdf

• Standard Normal Samples are Similar to Chi-squared Distribution
  • Let $X_1,\dots ,X_n$ be a Random Sample from $N(\mu ,{\sigma }^2)$. Then $Z_i=\frac{X_i-\mu }{\sigma }$ are iid random variables from the standard normal distribution.
    • Define samples from a normal, and then standardize.
  • $Z_i^2\sim {\chi }_{(1)}^2$ for $i=1,\dots ,n$
    • $Z_i^2\sim \mathrm{\Gamma }(\alpha =\frac{1}{n},\beta =2)$
    • By definition
    • So the standardized variables squared are similar to a Chi-squared Distribution with $1$ degree of freedom.
  • By convention we call each $Z_i^2$ as $W_i$
  • So $W=\sum _{i=1}^n{W}_i=\sum _{i=1}^n{Z}_i^2=\sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2\sim {\chi }_{(n)}^2$
  • Our $n$ samples determine how much degrees of freedom we have.
• Example:
  • $Z_1,\dots ,Z_6$ are Random Samples from the Standard Normal Distribution.
  • Find $b$ such that $P(\sum _{i=1}^6{Z}_i^2\le b)=0.95$
  • $P(W\le b)=0.95$
  • $P(\sum _{i=1}^n{\left(\frac{X_i-\mu }{\sigma }\right)}^2\le b)=0.95$
  • $P({\chi }_{(6)}^2\le b)=0.95$
  • $P({\chi }_6^2>b)=0.05$
  • On the table we get $12.5916$
• Marginal Distribution of Sample Variance
  • Let $X_1,\dots ,X_n$ be a Random Sample from a Normal Distribution with $N(\mu ,{\sigma }^2)$
  • $\frac{(n-1)S^2}{{\sigma }^2}=\frac{\sum _{i=1}^n(X_i-\overline{X}{)}^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2$
  • Since $S^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\overline{X}{)}^2$
    • $\frac{S^2}{{\sigma }^2}=\frac{1}{{\sigma }^2(n-1)}\sum _{i=1}^n(X_i-\overline{X}{)}^2$
    • $\frac{(n-1)S^2}{{\sigma }^2}=\frac{\sum _{i=1}^n(X_i-\overline{X}{)}^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2$
  • Example:
    • Suppose we have $X\sim N(?,{\sigma }^2=1)$
    • $n=10$
    • Find $b_1$ and $b_2$ where $P(b_1\le S^2\le b_2)=0.9$
    • $P(b_1\le S^2\le b_2)$
    • $P(\frac{(n-1)b_1}{{\sigma }^2}\le \frac{(n-1)S^2}{{\sigma }^2}\le \frac{(n-1)b_2}{{\sigma }^2})$
    • $P(\frac{(n-1)b_1}{{\sigma }^2}\le {\chi }_{(n-1)}^2\le \frac{(n-1)b_2}{{\sigma }^2})=0.9$
    • $P(9b_1\le {\chi }_{(n-1)}^2\le 9b_2)=0.9$
    • Let $a_1=9b_1$ and $a_2=9b_2$
    • $P(a_1\le {\chi }_{(n-1)}^2\le a_2)=0.9$
    • We want $90\mathrm{\%}$ to be shaded.
    • Point 1 as $0.95$
      • $a_2=3.32511$
    • Point 2 as $0.05$
      • $a_1=16.9190$
    • $16.919=9b_1$
    • $b_1=1.87988888888889$
    • $3.32511=9b_2$
    • $b_2=0.369456666666667$
• T Distribution
• F Distribution