STA256 Tutorial W12

• 3.4.10
  • Suppose $e^{3t+8t^2}$ is the MGF of $X$
  • What distribution is this?
  • Normal Distribution
  • $X\sim N(\mu ,\sigma )$
  • $M_X(t)=e^{\mu t+\frac{t^2{\sigma }^2}{2}}$
  • $3t+8t^2=\mu t+\frac{t^2{\sigma }^2}{2}$
  • $3t+8t^2=\mu t+\frac{t^2{\sigma }^2}{2}$
  • $\mu =3$
  • $8t^2=\frac{t^2{\sigma }^2}{2}$
  • $16t^2=t^2{\sigma }^2$
  • $16={\sigma }^2$
  • $\sigma =4$
  • What is $P(-1<x<9)$
  • Want to standardize it:
    • $X\sim N(\mu ,\sigma )$
    • $\frac{X-\mu }{\sigma }\sim N(0,1)$
    • $P(\frac{-1-3}{4}<\frac{x-\mu }{3}<\frac{9-3}{4})$
    • $P(-1<z<1.5)$
    • $F_Z(z)=\mathrm{\Phi }(z)=P(Z\le z)$
    • $P(-1<z<1.5)=\mathrm{\Phi }(1.5)-\mathrm{\Phi }(-1)$
    • $P(-1<z<1.5)=0.9332-\mathrm{\Phi }(-1)=0.7745$
    • #tk
• 3.4.30
  • $P(X_1+2X_2-2X_3>7)$
  • Compute the above
  • Where $X_1,X_2,X_3\stackrel{\text{iid}}{\sim }N(1,{\sigma }^2=4)$
    • With normal distributions, if you add them together, it's still normal. But we have new mean and variance
  • $X_1+2X_2-2X_3\sim N({\mu }_s,{\sigma }_s^2)$
  • $\mathrm{E}[X_1+2X_2-2X_3]={\mu }_s$
  • $\mathrm{E}[X_1]+2\mathrm{E}[X_2]-2\mathrm{E}[X_3]={\mu }_s$
  • $1+2-2={\mu }_s$
  • ${\mu }_s=1$
  • $\mathrm{E}[X_1]=\mathrm{E}[X_2]=\mathrm{E}[X_3]=1$
  • $\mathrm{Var}(X_1)=\mathrm{Var}(X_2)=\mathrm{Var}(X_3)=4$
  • $\mathrm{Var}(X_1+2X_2-2X_3)={\sigma }_s^2$
  • $\mathrm{Var}(X_1)+4\mathrm{Var}(X_2)+4\mathrm{Var}(X_3)+4\mathrm{Cov}(X_1,X_2)-4\mathrm{Cov}(X_1,X_3)-8\mathrm{Cov}(X_2,X_3)={\sigma }_s^2$
  • $4+(4)(4)+(4)(4)={\sigma }_s^2$
    • No Covariance because it's independent
  • $4+(4)(4)+(4)(4)=36$
  • $\mathrm{SD}=6=\sigma$
  • To compute
  • $P(X_1+2X_2-2X_3>7)$
  • $P(Z>\frac{7-1}{6})=1-\mathrm{\Phi }(1)$
• 5.1.7
  • Let $X_1,\dots ,X_n\stackrel{\text{iid}}{\sim }f(x)=\{\begin{array}{ll}{e}^{-(x-\mathrm{\Theta })}& x>\mathrm{\Theta },\mathrm{\Theta }\in \mathbb{R}\\ 0& \text{otherwise}\end{array}$
  • Let $Y_n=min(\{X_1,\dots ,X_n\})$
  • Show that $Y_n\stackrel{P}{\to }\mathrm{\Theta }$ by finding it's CDF
  • First find CDF
    • ${\int }_{\mathrm{\Theta }}^x{e}^{-x+\mathrm{\Theta }}\text{differential}x$
    • $[e^{-x+\mathrm{\Theta }}]{}_{\mathrm{\Theta }}^x$
    • $=1-e^{\mathrm{\Theta }-x}$
  • Solution
    • $F_{Y_n}(y)=P(Y_n\le y)$
    • $=P(min(\{X_1,\dots ,X_n\}\le y))$
    • $=1-P(min(\{X_1,\dots ,X_n\})>y)$
    • So if the minimum is $>y$ then all $X_1,\dots ,X_n>y$
    • $=1-P(X_1>y,\dots ,X_n>y)$
    • $=1-P(X_1>y)P(X_2>y)\dots P(X_n>y)$
    • Since these are iid they're all the same
    • $=1-P(X_1>y{)}^n$
    • $=1-[1-\underset{\text{CDF}}{\underbrace{P(X_1<y)}}{]}^n$
    • $=1-[1-F_{X_1}(y){]}^n$
    • Find the cdf
    • $F_{X_1}(y)=1-e^{\mathrm{\Theta }-y}$
    • $=1-[1-(1-e^{\mathrm{\Theta }-y}){]}^n\text{ for }y>\mathrm{\Theta }$
    • $=1-[e^{\mathrm{\Theta }-y}{]}^n$
    • $=1-[e^{n\mathrm{\Theta }-ny}]$
    • $=1-e^{n\mathrm{\Theta }-ny}$
    • $X_n\stackrel{P}{\to }\mathrm{\Theta }⟺P(|X_n\to \mathrm{\Theta }|\ge ϵ)\to 0$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=P(|Y_n-\mathrm{\Theta }|>ϵ)+P(Y_n-\mathrm{\Theta }<-ϵ)$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=P(Y_n>\mathrm{\Theta }+ϵ)+\underset{\text{Probability of }0}{\underbrace{P(Y_n<-ϵ+\mathrm{\Theta })}}$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=1-1+e^{-n(\mathrm{\Theta }+ϵ-\mathrm{\Theta })}$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=e^{-nϵ}$
    • To show this goes to $0$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=\underset{n\to \mathrm{\infty }}{lim}{e}^{-nϵ}$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=\underset{n\to \mathrm{\infty }}{lim}\frac{e^{ϵ}}{e^n}$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=\underset{n\to \mathrm{\infty }}{lim}\frac{e^{ϵ}}{e^n}=0$
    • $P(|Y_n-\mathrm{\Theta }|>ϵ)=0$
    • So $e^{n\mathrm{\Theta }}\to 0$