STA256 Lecture 18

• Test Prep
• 1
  • Let $X$ have a pdf of: $f(x)=6x(1-x)$ for $0<x<1$
  • a: find pdf of transformation $y=\frac{1-x}{x}$
    • support of y
      • $x=1$ then $y=\frac{1-1}{1}=0$
      • $x=0$ then $y=\frac{1-0}{0}$ undefined
        • $\underset{x\to 0}{lim}\frac{1-x}{x}=\mathrm{\infty }$
      • So the support of $y$ is $0<y<\mathrm{\infty }$
    • inverse map
      • $y=\frac{1-x}{x}$
      • $yx=1-x$
      • $\dots$
      • $x=\frac{1}{1+y}$
      • $x=(1+y{)}^{-1}$
      • #tk
    • jacobian
      • $\left|J\right|=\frac{dX}{dY}=\frac{d}{dY}(1+y{)}^{-1}$
      • $\left|J\right|=-\frac{1}{(1+y{)}^2}$
    • transformation theorem
      • $f(x)=6x(1-x)$
      • $g^{-1}(y)=\frac{1}{1+y}$
      • $\left|J\right|=-\frac{1}{(1+y{)}^2}$
      • The density of $y$
      • $f_Y(y)=f_X(g^{-1}(y))\left|J\right|$
      • $f_Y(y)=f_X\left(\frac{1}{1+y}\right)(-\frac{1}{(1+y{)}^2})$
      • $f_Y(y)=\frac{6}{1+y}(1-\frac{6}{1+y})(-\frac{1}{(1+y{)}^2})$
      • $f_Y(y)=(\frac{6}{1+y}-\frac{36}{(1+y{)}^2})(-\frac{1}{(1+y{)}^2})$
      • $\dots$
      • $f_Y(y)=\frac{6y}{(1+y{)}^4}$ for $0<y<\mathrm{\infty }$
  • b: find ${\mu }_Y$
    • method 1:
      • $E[Y]=\int yf(y)dy$
      • $E[Y]=\int \frac{y(6y)}{(1+y{)}^4}dy$
      • $E[Y]=\int \frac{6y^2}{(1+y{)}^4}dy$
      • $E[Y]={\int }_0^{\mathrm{\infty }}\frac{6y^2}{(1+y{)}^4}dy$
      • $E[Y]={\int }_0^{\mathrm{\infty }}6y^2(1+y{)}^{-4}dy$
      • $u=1+y$
      • $y=u-1$
      • $du=dy$
      • $y=0⟹u=1$
        $y=\mathrm{\infty }⟹u=\mathrm{\infty }$
      • $E[Y]={\int }_{u=1}^{u=\mathrm{\infty }}6(u-1{)}^2(u{)}^{-4}du$
      • $E[Y]=6{\int }_{u=1}^{u=\mathrm{\infty }}(u-1{)}^2(u{)}^{-4}du$
      • $E[Y]=6{\int }_{u=1}^{u=\mathrm{\infty }}\frac{u^2-2u+1}{u^4}du$
      • $E[Y]=6{\int }_{u=1}^{u=\mathrm{\infty }}{u}^2-\frac{2}{u^3}+u^{-4}du$
      • $E[Y]=6[\frac{1}{3}{u}^3-2(\dots )]{}_1^{\mathrm{\infty }}$
      • $\dots$
      • $=2$
      • #tk
    • method 2
      • $E(Y)=E\left(\frac{1-x}{x}\right)={\int }_{x=0}^{x=1}\frac{1-x}{x}f(x)dx$
      • $E(Y)=E\left(\frac{1-x}{x}\right)={\int }_{x=0}^{x=1}\frac{1-x}{x}(6x(1-x))dx$
      • $\dots$
• 2
  • Let $X$ have a pdf of: $f(x)=\frac{2x}{3}$ for $1<x<3$
  • a: find the mean and variance
    • mean is $\mu$
      • $E[X]=\int xf(x)dx$
      • $E[X]=\int \frac{2x^2}{3}dx$
      • $E[X]=\frac{1}{3}{\int }_1^{3}2x^{2}dx$
      • $E[X]=\frac{2}{3}{\int }_1^3{x}^{2}dx$
      • $E[X]=\frac{2}{3}\left[\frac{1}{3}{x}^3\right]{}_1^3$
      • $E[X]=\frac{2}{3}(\left[\frac{27}{3}\right]-\left[\frac{1}{3}\right])$
      • $E[X]=\frac{2}{3}\left[\frac{26}{3}\right]$
      • $E[X]=\frac{52}{9}$
      • $\dots$
      • #tk
    • variance
      • $E[X^2]=\int x^{2}f(x)dx$
      • $={\int }_1^3{x}^2\left(\frac{2x}{3}\right)dx$
      • $\dots$
      • $\frac{40}{3}$
      • ${\sigma }_x^2=var(x)=E[x^2]-(E[x]{)}^2$
      • ${\sigma }_x^2=var(x)=\left(\frac{40}{3}\right)-{\left(\frac{52}{9}\right)}^2$
      • ${\sigma }_x^2=var(x)=$
• 3
  • Let $X$ have a pdf of $f(x)=e^{-x}$ for $x>0$
  • a: find the MGF
    • $E[e^{tx}]=\int e^{tx}f(x)dx$
    • $E[e^{tx}]=\int e^{tx}{e}^{-x}dx$
    • $E[e^{tx}]={\int }_0^{\mathrm{\infty }}{e}^{tx-x}dx$
    • $E[e^{tx}]={\int }_0^{\mathrm{\infty }}{e}^{t(x-1)}dx$
    • $E[e^{tx}]={\int }_0^{\mathrm{\infty }}{e}^{(t-1)x}dx$
    • $E[e^{tx}]=\left[\frac{1}{t-1}{e}^{tx-x}\right]{}_{x=0}^{x=\mathrm{\infty }}$
    • $E[e^{tx}]=\underset{x\to \mathrm{\infty }}{lim}\left[\frac{1}{t-1}{e}^{tx-x}\right]-\left[\frac{1}{t-1}\right]$
    • $\dots$
    • $E[e^{tx}]=\frac{1}{1-t}$
    • Neighbourhood of $t$
      • Examine $\frac{1}{t-1}{e}^{(t-1)x}$
      • $t-1\ne 0$
      • $(t-1)<0$
      • $t<1$
      • That's when this converges
    • $M_X(t)=-\frac{1}{t-1}$ for $t<1$
  • b: find $E[X]$ with MGF
    • $M_X(t)=(1-t{)}^{-1}$
    • $M_X^{\prime }(t)=-1(1-t{)}^{-2}$
    • $M_X^{\prime }(t)=\frac{1}{(1-t{)}^2}$
    • $M_X^{\prime }(0)=\frac{1}{1}=1$
    • So the $\mu$ is $1$
  • c: find the variance
    • #tk
• 4
  • Let $(X_1,X_2)$ have Joint Probability Density Function of: $f(x_1,x_2)=\frac{1}{2}(x_1+x_2)$ for $0<x_1<1$ and $0<x_2<1$
  • a: find the marginal pdfs
    • $f_{X_1}(x_1)={\int }_0^1{f}_{X_1,X_2}(x_1,x_2)d{x}_2$
      • $f_{X_1}(x_1)={\int }_0^1\frac{1}{2}(x_1+x_2)d{x}_2$
      • $f_{X_1}(x_1)=\frac{1}{2}{\int }_0^1{x}_1+x_{2}d{x}_2$
      • $f_{X_1}(x_1)=\frac{1}{2}[x_1{x}_2+\frac{1}{2}{x}_2^2]{}_0^1$
      • $f_{X_1}(x_1)=\frac{1}{2}\left([x_1+\frac{1}{2}]\right)$
      • $f_{X_1}(x_1)=\frac{1}{2}\left([x_1+\frac{1}{2}]\right)$ for $0<x_1<1$
    • $f_{X_2}(x_2)={\int }_0^1{f}_{X_1,X_2}(x_1,x_2)d{x}_1$
      • $f_{X_2}(x_2)=\frac{1}{2}{\int }_0^1{x}_1+x_{2}d{x}_1$
      • $f_{X_2}(x_2)=\frac{1}{2}[x_1{x}_2+\frac{1}{2}{x}_1^2]{}_{x_1=0}^{x_1=1}$
      • $f_{X_2}(x_2)=\frac{1}{2}[x_2+\frac{1}{2}]$ for $0<x_2<1$
  • b: Are $X_1$ and $X_2$ independent
    • only if $f_{X_1}(x_1)f_{X_2}(x_2)=f_{X_1,X_2}(x_1,x_2)$
    • $(\frac{1}{2}{x}_1+\frac{1}{4})(\frac{1}{2}{x}_2+\frac{1}{4})=\frac{1}{2}(x_1+x_2)$
    • $\frac{1}{4}{x}_1{x}_2+\frac{1}{8}{x}_1+\frac{1}{8}{x}_2+\frac{1}{16}$
    • See that they're not independent because $LHS\ne RHS$
  • c: find the conditional distribution given $x_2$
    • $f(x_1|x_2)$
    • $f(x_1|x_2)=\frac{f(x_1,x_2)}{f(x_2)}$
    • $f(x_1|x_2)=\frac{\frac{1}{2}(x_1+x_2)}{\frac{1}{2}{x}_2+\frac{1}{4}}$
    • $f(x_1|x_2)=\frac{\frac{1}{2}(x_1+x_2)}{\frac{1}{2}(x_2+\frac{1}{2})}$
    • $f(x_1|x_2)=\frac{x_1+x_2}{x_2+\frac{1}{2}}$
    • $\dots$
    • $f(x_1|x_2)=\frac{2x_1+2x_2}{2x_2+1}$ for $0<x_1<1$
    • Treat $x_1|x_2=\mathcal{X}$ some new random variable, so it's the support of $x_1$
  • d: find the expected value of $X_1|X_2=1$
    • Need $f(x_1|x_2=1)$
    • $f(x_1|x_2=1)=\frac{2x_1+2(1)}{2(1)+1}=\frac{2x_1+2}{3}$ for $0<x_1<1$
    • $E[X_1|X_2=1]=\int x_{1}f(x_1|x_2=1)d{x}_1$
    • $E[X_1|X_2=1]=\int \frac{x_1(2x_1+2)}{3}d{x}_1$
    • $E[X_1|X_2=1]=\frac{1}{3}\int x_1(2x_1+2)d{x}_1$
    • $E[X_1|X_2=1]=\frac{1}{3}\int 2x_1^2+2x_{1}d{x}_1$
    • $E[X_1|X_2=1]=\frac{1}{3}{\int }_0^{1}2x_1^2+2x_{1}d{x}_1$
    • $E[X_1|X_2=1]=\frac{1}{3}[2x_1^2+2x_1]{}_0^1$
    • $\dots$
    • $=\frac{5}{9}$
    • #tk
• 5
  • Let $X$ have a pmf of $p(x)=\frac{e-1}{e^x}$ for $y=1,2,3,\dots$
  • a: find $P(x>2)$
    • $P(x>2)=1-P(x\le 2)$
    • $P(x\le 2)=\sum _{i=1}^{2}p(X=x_i)$
    • $P(x\le 2)=P(X=1)+P(X=2)$
    • $P(x\le 2)=\frac{e-1}{e}+\frac{e-1}{e^2}$
    • $P(x\le 2)=\frac{e^2-e}{e^2}+\frac{e-1}{e^2}$
    • $P(x\le 2)=\frac{e^2-e+e-1}{e^2}$
    • $P(x\le 2)=\frac{e^2-1}{e^2}=0.8646647168$
    • $1-0.8646647168=0.1353352832$