STA260 Practice 02

• 6.3.9

  • Find mean and variance of $S^2$
  • Mean
    • $S^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\overline{X}{)}^2$
    • $S^2=\frac{1}{n-1}\sum _{i=1}^n[X_i^2-2X_i\overline{X}+{\overline{X}}^2]$
    • $S^2=\frac{1}{n-1}[\sum _{i=1}^n{X}_i^2-2\sum _{i=1}^n{X}_i\overline{X}+n{\overline{X}}^2]$
    • $S^2=\frac{1}{n-1}[\sum _{i=1}^n{X}_i^2-2n\overline{X}\sum _{i=1}^n{X}_i+n{\overline{X}}^2]$
    • $\overline{X}=\frac{1}{n}\sum _{i=1}^n{X}_i$
    • $n\overline{X}=\sum _{i=1}^n{X}_i$
    • $S^2=\frac{1}{n-1}[\sum _{i=1}^n{X}_i^2-2n{\overline{X}}^2+n{\overline{X}}^2]$
    • $S^2=\frac{1}{n-1}[\sum _{i=1}^n{X}_i^2-n{\overline{X}}^2]$
    • $S^2=\frac{1}{n-1}[\sum _{i=1}^n{X}_i^2-n{\overline{X}}^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}\mathrm{E}[\sum _{i=1}^n{X}_i^2-n{\overline{X}}^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[\mathrm{E}[X_1^2+X_2^2+\cdots +X_n^2]-n\mathrm{E}[{\overline{X}}^2]]$
    • $\mathrm{Var}(\overline{X})=\mathrm{E}[{\overline{X}}^2]-\mathrm{E}[\overline{X}{]}^2$
    • $\mathrm{E}[{\overline{X}}^2]=\mathrm{Var}(\overline{X})+\mathrm{E}[\overline{X}{]}^2$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[\mathrm{E}[X_1^2+X_2^2+\cdots +X_n^2]-n\mathrm{E}[{\overline{X}}^2]]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[\mathrm{E}[X_1^2+X_2^2+\cdots +X_n^2]-n\mathrm{Var}(\overline{X})-\mathrm{E}[\overline{X}{]}^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[\mathrm{E}[X_1^2+X_2^2+\cdots +X_n^2]-\frac{n{\sigma }^2}{n}-n{\mu }^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[\mathrm{E}[X_1^2+X_2^2+\cdots +X_n^2]-{\sigma }^2-n{\mu }^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[\sum _{i=1}^n\mathrm{E}[X_i^2]-{\sigma }^2-n{\mu }^2]$
    • $\mathrm{Var}(X_i)=\mathrm{E}[X_i^2]-\mathrm{E}[X_i{]}^2$
    • $\mathrm{E}[X_i^2]=\mathrm{Var}(X_i)+\mathrm{E}[X_i{]}^2$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[n[\mathrm{Var}(X_i)+\mathrm{E}[X_i{]}^2]-{\sigma }^2-n{\mu }^2]$
    • From problem $\mathrm{E}[X_i]=\mu$ and $\mathrm{Var}(X_i)={\sigma }^2$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[n[\mathrm{Var}(X_i)+\mathrm{E}[X_i{]}^2]-{\sigma }^2-n{\mu }^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[n{\sigma }^2+n{\mu }^2-{\sigma }^2-n{\mu }^2]$
    • $\mathrm{E}[S^2]=\frac{1}{n-1}[n{\sigma }^2-{\sigma }^2]$
    • $\mathrm{E}[S^2]=\frac{{\sigma }^2}{n-1}(n-1)={\sigma }^2$
  • Variance
    • $\mathrm{Var}(S^2)=\mathrm{E}[S^4]-\mathrm{E}[S^2{]}^2$
    • $\mathrm{Var}(S^2)=\mathrm{E}[S^4]-{\sigma }^4$
    • $\mathrm{E}[S^4]$
    • $S^2=\frac{1}{n-1}\sum _{i=1}^n[X_i^2-2X_i\overline{X}+{\overline{X}}^2]$
    • $S^4={(\frac{1}{n-1}\sum _{i=1}^n[X_i^2-2X_i\overline{X}+{\overline{X}}^2])}^2$
    • $S^4=(\frac{1}{n-1}\sum _{i=1}^n[X_i^2-2X_i\overline{X}+{\overline{X}}^2])(\frac{1}{n-1}\sum _{i=1}^n[X_i^2-2X_i\overline{X}+{\overline{X}}^2])$
    • $S^2=\frac{1}{n-1}[\sum _{i=1}^n{X}_i^2-2\overline{X}\sum _{i=1}^n{X}_i+n{\overline{X}}^2]$
    • $\mathrm{Var}(S^2)$
    • $\mathrm{Var}\left(\frac{(n-1)S^2}{{\sigma }^2}\right)=\frac{(n-1{)}^2}{{\sigma }^4}\mathrm{Var}(S^2)$
    • Variance of any Chi-squared Distribution is $2\nu$
    • $\mathrm{Var}({\chi }_{(n-1)}^2)=\frac{(n-1{)}^2}{{\sigma }^4}\mathrm{Var}(S^2)$
    • $2(n-1)=\frac{(n-1{)}^2}{{\sigma }^4}\mathrm{Var}(S^2)$
    • $\frac{2{\sigma }^4(n-1)}{(n-1{)}^2}=\mathrm{Var}(S^2)$
    • $\frac{2{\sigma }^4}{(n-1)}=\mathrm{Var}(S^2)$

• 6.3.10

  • Show how to use the Chi-squared Distribution to calculate $P(a<\frac{S^2}{{\sigma }^2}<b)$
  • We know that $\frac{(n-1)S^2}{{\sigma }^2}\sim {\chi }_{(n-1)}^2$
  • So introduce this
  • $P[(n-1)a<\frac{(n-1)S^2}{{\sigma }^2}<(n-1)b]$
  • Let $a^{\prime }=(n-1)a$ and $b^{\prime }=(n-1)b$
  • $P(a^{\prime }<{\chi }_{(n-1)}^2<b^{\prime })$
  • We can then grab values from the chi-squared table.

• 6.3.11

  • Let $X_1,\dots ,X_n$ be a sample from an $N({\mu }_X,{\sigma }^2)$ distribution.
  • Let $Y_1,\dots ,Y_m$ be an independent sample from $N({\mu }_Y,{\sigma }^2)$
  • Show how to use the F Distribution to find $P(\frac{S_X^2}{S_Y^2}>c)$
  • $F_{1,2}=\frac{\frac{{\chi }_{(1)}^2}{1}}{\frac{{\chi }_{(2)}^2}{2}}$
  • We start with: $P[\frac{S_X^2}{S_Y^2}>c]$. They're independent as given.
  • So the variances are independent
  • $P[\frac{(n-1)S_X^2}{(m-1)S_Y^2}>\frac{n-1}{m-1}c]$
  • $P[\frac{\frac{(n-1)S_X^2}{{\sigma }^2}}{\frac{(n-1)S_Y^2}{{\sigma }^2}}>\frac{(n-1)c}{(m-1){\sigma }^2}]$
  • $U=\frac{(n-1)S_X^2}{{\sigma }^2}$
  • $V=\frac{(m-1)S_Y^2}{{\sigma }^2}$
  • $F=\frac{\frac{U}{n-1}}{\frac{V}{m-1}}$
  • $F={\displaystyle \frac{\frac{\frac{(n-1)S_X^2}{{\sigma }^2}}{n-1}}{\frac{V}{m-1}}}$
    • $\frac{\frac{(n-1)S_X^2}{{\sigma }^2}}{n-1}$
    • $\frac{(n-1)S_X^2}{{\sigma }^2}÷(n-1)$
    • $\frac{(n-1)S_X^2}{{\sigma }^2}\cdot \frac{1}{n-1}$
    • $\frac{S_X^2}{{\sigma }^2}$
  • $F=\frac{\frac{S_X^2}{{\sigma }^2}}{\frac{S_Y^2}{{\sigma }^2}}$
  • $F=\frac{S_X^2}{{\sigma }^2}\cdot \frac{{\sigma }^2}{S_Y^2}$
  • $F=\frac{S_X^2}{S_Y^2}$
  • This means that $P(F_{n-1,m-1}>c)=P(\frac{S_X^2}{S_Y^2}>c)$
  • So we can just use the F table to evaluate this.

• 6.2.3

  • Let $\overline{X}$ be the average of a sample of $16$ independent normal random variables with $\mu =0$ and ${\sigma }^2=1$
  • Determine $c$ such that $P[\left|\overline{X}\right|<c]=0.5$
  • $X_i\sim N(0,1)$
  • $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
  • $\overline{X}\sim N(0,\frac{1}{16})$
  • $P[-c<\overline{X}<c]=0.5$
  • $P[-\frac{c-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{c-\mu }{\frac{\sigma }{\sqrt{n}}}]=0.5$
  • $P[-\frac{c}{4^{-1}}<\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{c}{4^{-1}}]=0.5$
  • $P[-4c<Z<4c]=0.5$
  • $a=-4c$
  • $b=4c$
  • $P[a<Z<b]=0.5$
  • $P(0<Z<b)=0.25$
  • $b\approx 0.675$
  • $a=-b$
  • $P(-0.675<Z<0.675)=0.5$
    • Verify
    • $1-(2)(0.250)=0.5$
  • $0.675=4c$
  • $c=0.16875$

• 6.2.4

  • If $T$ follows a $t_7$ distribution. Find $t_0$ such that:
    • a: $P(\left|T\right|<t_0)=0.9$
      • $T=\frac{N}{\sqrt{\frac{C}{n}}}$
      • $N\sim N(0,1)$
      • $C\sim {\chi }_{(7)}^2$
      • $P(-t_0<\frac{N}{\sqrt{\frac{C}{n}}}<t_0)=0.9$
      • $P(-t_0<T<t_0)=0.9$
      • $1-0.9=0.1$
      • $\frac{0.1}{2}=0.05$
      • Because a T Distribution is symmetric, we can find $t_{0.05}$ on the table to help us.
      • $t_0=1.895$
      • So $P(\left|T\right|<1.895)=0.9$
    • b: $P(T>t_0)=0.05$
      • To have the shaded area on the T table be over a certain value.
      • I'm given right tails.
      • It should just be $t_0$ from above then. Where $t_0=1.895$.

• 6.2.5

  • If $X\sim F_{n,m}$ then $X^{-1}\sim F_{m,n}$
  • $X\sim F_{n,m}$
  • So $X\approx {\displaystyle \frac{\frac{{\chi }_{(n)}^2}{n}}{\frac{{\chi }_{(m)}^2}{m}}}$
  • $X\approx \frac{{\chi }_{(n)}^2}{n}\frac{m}{{\chi }_{(m)}^2}$
  • By Continuous Mapping Theorem
  • $X^{-1}\sim F_{m,n}$
    • $X\approx \frac{{\chi }_{(n)}^2}{n}\frac{m}{{\chi }_{(m)}^2}$
    • $X^{-1}\approx \frac{n}{{\chi }_{(n)}^2}\frac{{\chi }_{(m)}^2}{m}$
    • $X^{-1}\approx \frac{\frac{{\chi }_{(m)}^2}{m}}{\frac{{\chi }_{(n)}^2}{n}}\sim F_{m,n}$

• 6.2.8

  • If $X$ and $Y$ are independent exp random variables with $\lambda =1$
  • Then $\frac{X}{Y}\sim F$ distribution
  • Exponential Distribution
  • $X\sim \mathrm{Exp}(1)$
  • $Y\sim \mathrm{Exp}(1)$
  • $f_X(x)=e^{-x}$
  • $f_Y(y)=e^{-y}$
  • $\frac{X}{Y}$
    • $=\frac{e^{-x}}{e^{-y}}$
    • How does exp relate to chi-squared?
      • Chi-squared Distribution is Gamma Distribution with $\alpha =\frac{\nu }{2},\beta =2$
      • So we have $\frac{1}{\mathrm{\Gamma }\left(\frac{\nu }{2}\right)2^{\nu }}{x}^{\frac{\nu }{2}-1}{e}^{-\frac{x}{2}}$
      • $\mathrm{\Gamma }\left(\frac{\nu }{2}\right)={\int }_0^{\mathrm{\infty }}{t}^{\frac{\nu }{2}-1}{e}^{-\frac{\nu }{2}}\text{differential}x$
      • We also know that we can get a chi-squared random variable by squaring and summing $\nu$ standard normal variables.
    • Since $\lambda =1⟹\alpha ,\beta =1$
    • See that $G\sim \mathrm{\Gamma }(1,1)$
      • $=e^{-x}$
    • $H=2G$
      • $M_X(t)=(1-\beta t{)}^{-\alpha }$
      • If we let $Y=cX$, the MGF of $Y$ is:
      • $M_Y(t)=E[e^{tY}]=E[e^{t(cX)}]=E[e^{(ct)X}]=M_X(ct)$
      • Substituting $ct$ into the original MGF:
      • $M_Y(t)=(1-\beta (ct){)}^{-\alpha }=(1-(c\beta )t{)}^{-\alpha }$
      • Notice that the structure is identical, but the $\beta$ has been replaced by $c\beta$. The $\alpha$ remains unchanged.
    • $H\sim \mathrm{\Gamma }(1,2)⟹\mathrm{\Gamma }(\frac{2}{2},2)⟹\nu =2$
    • $H\sim {\chi }_{(4)}^2$
    • Meaning $X\approx H\approx Y$
  • $F_{2,2}=\frac{\frac{{\chi }_{(2)}^2}{2}}{\frac{{\chi }_{(2)}^2}{2}}$
  • #tk