STA258 Lecture 07

• Example:
  • We have $Y_1,\dots ,Y_5\stackrel{iid}{\sim }N(0,1)$
  • $Y_6\sim N(0,1)$ and $Y_6$ is independent from the others.
  • $W=\sum _{i=1}^5{Y}_i^2$
  • $U=\sum _{i=1}^5(Y_i-\overline{Y}{)}^2$
  • a) Find the distribution of $\frac{\sqrt{5}{Y}_6}{\sqrt{W}}$
    • Recall that if $Z\sim N(0,1)$ and $W\sim {\chi }_{(\nu )}^2$ then $\frac{Z}{\sqrt{\frac{W}{\nu }}}\sim t_{(\nu )}$
    • Recall that $W_1\sim {\chi }_{({\nu }_1)}^2$ and $W_2\sim {\chi }_{({\nu }_2)}^2$
      • This means that $\frac{\frac{W_1}{{\nu }_1}}{\frac{W_2}{{\nu }_2}}\sim F_{({\nu }_1,{\nu }_2)}$
    • $Y_i\sim N(0,1)⟹Y_i^2\sim {\chi }_{(1)}^2$
    • So $W=\sum _{i=1}^5{Y}_i^2\sim \sim {\chi }_{(5)}^2$
    • $\frac{\sqrt{5}{Y}_6}{\sqrt{W}}$
      • $Y_6$ is standard normal
      • Bottom is root chi squared.
      • How can we divide by degrees of freedom?
    • $\frac{Y_6}{\sqrt{\frac{W}{5}}}\equiv \frac{N(0,1)}{\sqrt{\frac{{\chi }_{(5)}^2}{5}}}\sim t_{(5)}$
  • Find the Dist of $\frac{2(5{\overline{Y}}^2+Y_6^2)}{U}$
    • Find the dist of top and bottom
    • Top
      • $Y_6^2\sim {\chi }_{(1)}^2$
      • $\overline{Y}\sim N(0,\frac{1}{5})$
        • $Y_1,\dots ,Y_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$
        • $\overline{Y}\sim N(\mu ,\frac{{\sigma }^2}{n})$
      • Scale
        • If $X\sim N(\mu ,{\sigma }^2)$
          • Scale by $k$
          • $kX\sim N(k\mu ,k^2{\sigma }^2)$
        • $\sqrt{5}\overline{Y}\sim N(0,\frac{5(1)}{5})\equiv N(0,1)$
      • Square it:
        • $(\sqrt{5}\overline{Y}{)}^2\sim {\chi }_{(1)}^2$
        • So $5\overline{Y}\sim {\chi }_{(1)}^2$
      • $(5{\overline{Y}}^2+Y_6^2)\equiv {\chi }_{(1)}^2+{\chi }_{(1)}^2\equiv {\chi }_{(2)}^2$
      • $2(5{\overline{Y}}^2+Y_6^2)\equiv 2{\chi }_{(2)}^2={\chi }_{(2)}^2+{\chi }_{(2)}^2={\chi }_{(4)}^2$
    • Bottom
      • $S^2=\frac{1}{n-1}\sum _{i=1}^n(Y_i-\overline{Y}{)}^2$
      • $U=\sum _{i=1}^5(Y_i-\overline{Y}{)}^2$
      • $U=(n-1)S^2$
      • $\frac{U}{{\sigma }^2}=\frac{(n-1)S^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2$
      • Since ${\sigma }^2=1$
      • $\frac{(n-1)S^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2\stackrel{\sigma =1}{⟹}U=\sum _{i=1}^5(Y_i-\overline{Y}{)}^2\sim {\chi }_{(4)}^2$
    • $\frac{2(5{\overline{Y}}^2+Y_6^2)}{U}\equiv \frac{2{\chi }_{(2)}^2}{{\chi }_{(4)}^2}\equiv \frac{\frac{{\chi }_{(2)}^2}{2}}{\frac{{\chi }_{(4)}^2}{4}}\sim F_{(2,4)}$
      • Divide the top and bottom by $4$
    • Similar to an F Distribution.
• Confidence Interval
• Standard Error
• Example:
  • The number of grain nucleation sites per unit volume is $\sim \mathrm{Poisson}(\lambda )$
  • $50$ samples are collected with the average of $20$ grains per unit volume
  • Estimate $\lambda$ and place a $2$ standard error bound on error of estimation.
  • Let $Y_i$ be a random variable denoting the number of grain nucleation sites.
  • Each $Y_i\sim \mathrm{Poisson}(\lambda )i=1,2,\dots ,50$
  • $\mathrm{E}[Y_i]=\lambda$
  • $\mathrm{Var}(Y_i)=\lambda$
  • $\mathrm{E}[\overline{Y}]=\mathrm{E}[Y_i]=\lambda$
  • $\mathrm{Var}(\overline{Y})=\frac{{\sigma }^2}{n}=\frac{\mathrm{Var}(Y_i)}{n}$
  • $\mathrm{Var}(\overline{Y})=\frac{\lambda }{n}=\frac{\lambda }{50}$
  • $\mathrm{SE}[\overline{Y}]=\sqrt{\frac{\lambda }{n}}=\sqrt{\frac{\lambda }{50}}$
    • Plug-in Principle
    • If we have an unknown parameter, replace the parameter with its estimate.
    • $\approx \sqrt{\frac{\hat{\lambda }}{50}}$
    • What is $\hat{\lambda }$'s estimate? Sample mean.
    • $\approx \sqrt{\frac{\overline{y}}{50}}$
    • We have $\overline{y}=20$ from the question.
    • $\approx \sqrt{\frac{20}{50}}=0.632455532033676$
  • Bound is $2$ Standard Error: $2\cdot 0.632455532033676=1.26491106406735$
    • $2$ corresponds to a $95$ percent confidence level.
  • We estimate the mean to be $20$ from the question.
  • We expect the error of estimation to be less than $1.26491106406735$ with probability $95\mathrm{\%}$
    • With probability $95\mathrm{\%}$, the unknown parameter is going to be $20\pm 1.26491106406735⟹(18.7350889359,21.2649110641)$
• Confidence Intervals is a Point Estimator $\pm$ our cutoff $\times$ $\mathrm{SE}$.
  • Our Point Estimator is $\overline{Y}$
  • Our cutoff is $2$
  • Our $\mathrm{SE}[\overline{Y}]$
  • You have a $100\mathrm{\%}$ of scoring between $0$ and $100\mathrm{\%}$ on the test.
  • You can have the upper and lower confidence limit:
    • ${\hat{\theta }}_L<\theta <{\hat{\theta }}_U$
  • $P({\hat{\theta }}_L<\theta <{\hat{\theta }}_U)=1-\alpha$
    • $1-\alpha$ is our confidence level.
    • $\alpha$ is our significance level.