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  • STA258 Lecture 06

STA258 Lecture 06 Raw

• Random samples like: $Y_1,Y_2,\dots ,Y_n\stackrel{iid}{\sim }N(\mu ,{\sigma }^2)$.
  • You can standardize to get a normal: $\frac{\overline{Y}-\mu }{\frac{\sigma }{\sqrt{n}}}\sim N(0,1)$
  • You can use the Distribution of S^2 $\frac{(n-1)S^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2$
    • $\frac{S^2}{{\sigma }^2}=\frac{W}{n-1}$ where $W\sim {\chi }_{(n-1)}^2$
• What if ${\sigma }^2$ is unknown?
  • Then we can consider $\frac{\overline{Y}-\mu }{\frac{?}{\sqrt{n}}}$
  • Since $S^2\to {\sigma }^2$, then $S\to \sigma$
  • $T=\frac{\overline{Y}-\mu }{\frac{S}{\sqrt{n}}}=\frac{\sqrt{n}(\overline{Y}-\mu )}{S}$
  • $T\sim t_{(\nu )}=f_T(t)=\frac{\mathrm{\Gamma }\left(\frac{\nu +1}{2}\right)}{\mathrm{\Gamma }\left(\frac{\nu }{2}\right)\sqrt{\pi \nu }}{(1+\frac{t^2}{\nu })}^{-\frac{\nu +1}{2}}$
    • T Distribution
  • This is defined for $-\mathrm{\infty }<t<\mathrm{\infty }$
  • Looks similar to a Normal Distribution.
  • As $\nu \to \mathrm{\infty }$ then $t_{(\nu )}\stackrel{D}{\to }N(0,1)$
  • Theorem:
    • If we have $Z\sim N(0,1)$ and $W\sim {\chi }_{(\nu )}^2$
      • We need $Z,W$ to be independent.
    • Then $\frac{Z}{\sqrt{\frac{W}{\nu }}}\sim t_{(\nu )}$
    • T Distribution
    • $T=\frac{\sqrt{n}(\overline{Y}-\mu )}{S}\cdot \frac{\frac{1}{\sigma }}{\frac{1}{\sigma }}=\frac{\frac{\sqrt{n}(\overline{Y}-\mu )}{\sigma }}{\frac{S}{\sigma }}$
    • $T=\frac{Z}{\frac{S}{\sigma }}$
    • $T=\frac{Z}{\sqrt{\frac{S^2}{{\sigma }^2}}}$
    • $T=\frac{Z}{\sqrt{\frac{W}{n-1}}}\sim t_{(n-1)}$
    • Why do we lose a degree of freedom?
      • Reality is the unknown $\sigma$.
      • We proxy it with $S$, so we have a model, not reality.
      • Using a model loses us a degree of freedom.
  • Theorem:
    • $Z\sim N(0,1)$
    • $W\sim {\chi }_{(n-1)}^2$
    • Then $\frac{Z}{\sqrt{\frac{W}{\nu }}}\sim t_{(\nu )}$
• Question
  • Assume $T\sim t_{(\nu )}$ and $W\sim {\chi }_{(\nu )}^2$
  • Find the $\mathrm{E}[T]$ and $\mathrm{Var}(T)$
  • $\mathrm{E}[T]$
    • If $Z\sim N(0,1)$ then $\mathrm{E}[Z]=0$
      • $\mathrm{Var}(Z)=\mathrm{E}[Z^2]-\mathrm{E}[Z{]}^2$
      • $\mathrm{Var}(Z)=\mathrm{E}[Z^2]$
      • Since $Z\sim N(0,1)$
      • Then
      • $1=\mathrm{E}[Z^2]$
    • $\mathrm{E}[T]=\mathrm{E}\left[{\displaystyle \frac{Z}{\sqrt{\frac{W}{\nu }}}}\right]$
    • We can do $\mathrm{E}[XY]=\mathrm{E}[X]\mathrm{E}[Y]$ if and only if $X,Y$ are independent.
    • So we have independency
    • $\mathrm{E}[T]=\mathrm{E}[Z]\mathrm{E}\left[\frac{1}{\sqrt{\frac{W}{\nu }}}\right]$
    • $\mathrm{E}[T]=0\mathrm{E}\left[\frac{1}{\sqrt{\frac{W}{\nu }}}\right]=0$
    • So the T Distribution has a mean of $0$
  • $\mathrm{Var}(T)$
    • $\mathrm{Var}(T)=\mathrm{E}[T^2]-\mathrm{E}[T{]}^2$
    • Since we know $\mathrm{E}[T]=0$
    • $\mathrm{Var}(T)=\mathrm{E}[T^2]-0$
    • $\mathrm{Var}(T)=\mathrm{E}[T^2]$
    • $\mathrm{Var}(T)=\mathrm{E}\left[{\frac{Z}{\sqrt{\frac{W}{\nu }}}}^2\right]$
    • $\mathrm{Var}(T)=\mathrm{E}\left[\frac{Z^2}{\frac{W}{\nu }}\right]$
    • $\mathrm{Var}(T)=\mathrm{E}[Z^2]\mathrm{E}\left[\frac{\nu }{W}\right]$
    • Sub $\mathrm{E}[Z^2]$
    • $\mathrm{Var}(T)=1\mathrm{E}\left[\frac{\nu }{W}\right]$
    • $\mathrm{Var}(T)=\mathrm{E}\left[\frac{\nu }{W}\right]$
    • $\mathrm{Var}(T)=\nu \mathrm{E}\left[\frac{1}{W}\right]$
      • If we have $W\sim {\chi }_{(\nu )}^2⟹\mathrm{E}[W]=\nu$
      • $\mathrm{E}\left[\frac{1}{X}\right]=\frac{1}{\mathrm{E}[X]}$?
        • Doesn't hold.
      • This is Jensen's Inequality
      • $W\sim {\chi }_{(\nu )}^2⟹\mathrm{E}[W^{\ast }]=\frac{2^k\mathrm{\Gamma }(k+\frac{\nu }{2})}{\mathrm{\Gamma }\left(\frac{\nu }{2}\right)}$
      • So $\mathrm{E}\left[\frac{1}{W}\right]=\mathrm{E}[W^{-1}]$
      • Meaning $k=-1$
      • $\mathrm{E}\left[\frac{1}{W}\right]$
      • $={\displaystyle \frac{2^{-1}\mathrm{\Gamma }(\frac{\nu }{2}-1)}{\mathrm{\Gamma }\left(\frac{\nu }{2}\right)}}$
        • Gamma Function
          • $\mathrm{\Gamma }(\alpha )=(\alpha -1)\mathrm{\Gamma }(\alpha -1)$
        • Gamma Distribution
      • $=\frac{1}{2}\frac{\mathrm{\Gamma }(\frac{\nu }{2}-1)}{(\frac{\nu }{2}-1)\mathrm{\Gamma }(\frac{\nu }{2}-1)}$
      • $=\frac{1}{\nu -2}$
    • So $\mathrm{Var}(T)=\frac{\nu }{\nu -2}$
• Example:
  • Compare two classes
  • $S_1^2$ and $S_2^2$
  • You can say the two classes have performed the same if $S_1^2=S_2^2$
  • To compare these two quantities, you can create a ratio: $\frac{S_1^2}{S_2^2}=1$
  • Use an F Distribution.
  • If $X\sim F_{{\nu }_1,{\nu }_2}$
  • $f_X(x)=\frac{\mathrm{\Gamma }\left(\frac{{\nu }_1+{\nu }_2}{2}\right){\left(\frac{{\nu }_1}{{\nu }_2}\right)}^{\frac{{\nu }_1}{2}}}{\mathrm{\Gamma }\left(\frac{{\nu }_1}{2}\right)\mathrm{\Gamma }\left(\frac{{\nu }_2}{2}\right)}\frac{x^{\frac{{\nu }_1}{2}-1}}{{(1+\frac{{\nu }_{1}x}{{\nu }_2})}^{\frac{{\nu }_1+{\nu }_2}{2}}}$
    • What the fuck #tk
  • Takes two degrees of freedom.
  • $S_1^2,S_2^2$ are the sample variances of independent samples of size $n_1,n_2$ respectively. They come from a Normal Distribution. Where variances are ${\sigma }_1^2,{\sigma }_2^2$ respectively.
  • Then $\frac{\frac{S_1^2}{{\sigma }_1^2}}{\frac{S_2^2}{{\sigma }_2^2}}\sim F_{n_1-1,n_2-1}$
• If we have two independent chi-squared random variables.
  • The ratio of $F=\frac{\frac{W_1}{{\nu }_1}}{\frac{W_2}{{\nu }_2}}\sim F_{{\nu }_1,{\nu }_2}$
  • $W_1\sim {\chi }_{({\nu }_1)}^2$
  • $W_2\sim {\chi }_{({\nu }_2)}^2$
  • Proof:
    • $\frac{(n_1-1)S_1^2}{{\sigma }_1^2}\sim {\chi }_{(n_1-1)}^2$
    • $\frac{(n_2-1)S_2^2}{{\sigma }_2^2}\sim {\chi }_{(n_2-1)}^2$
    • Create a ratio
    • $\frac{\frac{\frac{(n_1-1)S_1^2}{{\sigma }_1^2}}{(n_1-1)}}{\frac{\frac{(n_2-1)S_2^2}{{\sigma }_2^2}}{(n_2-1)}}$
    • $=\frac{\frac{S_1^2}{{\sigma }_1^2}}{\frac{S_2^2}{{\sigma }_2^2}}\sim F_{(n_1-1,n_2-1)}$
  • Example:
    • $n_1=6$
    • $n_2=10$
    • Both samples from normal.
    • With ${\sigma }_1^2={\sigma }_2^2$
    • Find $b$ where $P(\frac{S_1^2}{S_2^2}\le b)=0.95$
    • $F=\frac{\frac{S_1^2}{{\sigma }_1^2}}{\frac{S_2^2}{{\sigma }_2^2}}=\frac{S_1^2}{S_2^2}\sim F_{(n_1-1,n_2-1)}$
    • Sub in $n_1,n_2$
    • $F\sim F_{(5,9)}$
    • $P(F\le b)=0.95$
    • $P(F>b)=0.05$
    • Use the table STA260-Winter2026-Statistical_Tables.pdf
      • $\alpha =0.05$ right tail
      • $3.48$ is our $b$
    • So $b=F_{(0.05;5,9)}=3.48$