• suppose: $g(x)$ strictly monotonic, differentiable function of $x$
  • 👨‍🏫 i.e. inverse function exists
  • then r.v. $Y=g(X)$ has pdf given by $$f_Y(y)=\{\begin{array}{ll}{f}_X(g^{-1}(y))|\frac{\mathrm{d}}{\mathrm{d}y}{g}^{-1}(y)|& \text{if }y=g(x)\text{ for some x}\\ 0& \text{if }y\ne g(x)\text{ for all x}\end{array}$$
  • $g^{-1}(y)$: value of $x$ s.t. $g(x)=y$

• assume: $g(x)$ increasing
  • and same for $g^{-1}$
• $F_Y(y)=P(g(X)\le y)=P(X\le g^{-1}(y))=F_X(g^{-1}(y))$
  • and, from chain rule:
    • ⭐ $f_Y(y)=f_X(g^{-1}(y))\frac{\mathrm{d}}{\mathrm{d}y}{g}^{-1}(y)$
    • and, as $g^{-1}(x)$ non-decreasing, its derivative is non-negative
  • when $y\ne g(x)$, $F_Y(y)$ either $0$ or $1$: leading to $f_Y(y)=0$

1. $F\sim U(0,1)$
2. $X\sim F_X^{-1}(u),u\sim U(0,1)$
   • thus, we can perform inverse transformation method
   • generating cont. r.v. w/ distribution $F_X$ from $u\sim U(0,1)$
   • $Y=F_X^{-1}(U)$, then $$\begin{array}{rl}{F}_Y(y)& =P(Y\le y)=P(F_X^{-1}(U)\le y)\\ & =P(U\le F_X(y))=F_X(y)\end{array}$$

• if $y<0$: $F_Y(y)=0$
• if $y\ge 0$ $$\begin{array}{rl}{F}_Y(y)=P(Y\le y)=P(X^2\le y)& =P(-\sqrt{y}\le X\le \sqrt{y})\\ & ={\int }_{-\sqrt{y}}^{\sqrt{y}}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}\mathrm{d}x\\ & =2{\int }_0^{\sqrt{y}}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}\mathrm{d}x\\ f_Y(y)=F_Y^{\prime }(y)& =(2{\int }_0^{\sqrt{y}}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}\mathrm{d}x{)}^{\prime }\\ & =\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y}{2}}(\sqrt{y}{)}^{\prime }\\ & =\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y}{2}}{y}^{-1/2}\end{array}$$
• actually $Y=X^2\sim \Gamma (0.5,0.5)$
  • aka Chi-squared distribution

• define pdf $f_Y$
• $Y=e^X>0$
• for $y\le 0$, $F_Y(y)=0,f_Y(y)=0$
• else: $$\begin{array}{rl}{F}_Y(y)& =P(Y\le y)=P(e^X\le y)\\ & =P(X\le \ln y)={\int }_{-\infty }^{\ln y}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}\mathrm{d}x\\ f_Y(y)& =F_Y(y{)}^{\prime }=({\int }_{-\infty }^{\ln y}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}\mathrm{d}x{)}^{\prime }\\ & =\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(\ln y{)}^2}{2}}(\ln y{)}^{\prime }\\ & =\frac{1}{y\sqrt{2\pi }}{e}^{-\frac{(\ln y{)}^2}{2}}\end{array}$$
• also, as $g(Y)$ strictly monotonic
• $g^{-1}(y)=\ln y$ for $y>0$
  • also $(g^{-1}{)}^{\prime }=1/y$
• using formula, we obtain: $$\begin{array}{rl}{f}_Y& =f_X(g^{-1}(y))|\frac{\mathrm{d}}{\mathrm{d}y}{g}^{-1}(y)|\\ & =f_X(g^{-1}(y))|\frac{\mathrm{d}}{\mathrm{d}y}{g}^{-1}(y)|\\ & =\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(\ln y{)}^2}{2}}\cdot |\frac{1}{y}|\\ & =\frac{1}{y\sqrt{2\pi }}{e}^{-\frac{(\ln y{)}^2}{2}}\end{array}$$

• $Y=X^n$ and $n$ is odd
• find: $f_Y$
• 👨‍🏫 exercise $$\begin{array}{rl}{f}_Y(y)& =f_X(\sqrt[n]{y})\cdot \frac{\sqrt[n]{y}}{ny}\end{array}$$

• assume $F$: strictly increasing function
• define the r.b. by $X=F(Y)$
• possible values of $X\in [0,1]$
• for $x\in [0,1]$ $$\begin{array}{rl}{F}_X(x)& =P(X\le x)=P(F(Y)\le x)\\ & =P(F^{-1}(F(Y))\le F^{-1}(x))\\ & =P(Y\le F^{-1}(x))=F(F^{-1}(x))=x\end{array}$$
• PDF of $x$: $f_X(x)=(F_X(x){)}^{\prime }=x^{\prime }=1$ for $x\in [0,1]$
• thus, $X\sim U(0,1)$

• $F(x)=1-e^{-\lambda x}$ for $x\ge 0$
• then $F^{-1}(u)$: value $x$ s.t.
  • $u=F(x)=1-e^{-\lambda x}$
  • $x=-\frac{1}{\lambda }\log (1-u)$
• $X=-\frac{1}{\lambda }\log (1-U)\sim Exp(\lambda )$
• and $1-U$: also follows $\sim U(0,1)$
  • thus $X=-\frac{1}{\lambda }\log (U)\sim Exp(\lambda )$

• e.g. a student's age, gender, major, year of study
• on particular day, number of vehicle accidents, deaths, and major injuries

• joint distribution function

• generalization of $F_X(x)=P(X\le x)$
• often abbreviated: joint d.f.

• then $F_X$: marginal distribution function of $X$ (= cdf)

• 👨‍🏫 all derived from:
  • $X\le \infty =S$
  • $X\le -\infty =\varnothing$
• $F_{X,Y}(-\infty ,y)=0$ for any $y\in \mathbb{R}$
• $F_{X,Y}(\infty ,y)=F_Y(y)$ for any $y\in \mathbb{R}$
• $F_{X,Y}(x,-\infty )=0$ for any $x\in \mathbb{R}$
• $F_{X,Y}(x,\infty )=F_X(x)$ for any $x\in \mathbb{R}$
• $F_{X,Y}(\infty ,\infty )=1$
• $F_{X,Y}(-\infty ,-\infty )=0$
• $F_{X,Y}(\infty ,-\infty )=F_{X,Y}(-\infty ,\infty )=0$

• $P(X>a,Y>b)=1-F_X(a)-F_Y(b)+F_{X,Y}(a,b)$
  • think about it in geometry! $$\begin{array}{rl}P(a_1<X\le a_2,b_1<Y<b_2)& =F_{X,Y}(a_2,b_2)-F_{X,Y}(a_1,b_2)\\ & -F_{X,Y}(a_2,b_1)+F_{X,Y}(a_1,b_1)\end{array}$$
• inclusion matters here!

• $A=\{X\le a\},B=\{Y\le b\}$
• then $\{X>a,Y>b\}=A^c{B}^c=(A\cup B{)}^c$
• thus: $$\begin{array}{rl}P(X>a,Y>b)& =P((A\cup B{)}^c)=1-P(A\cup B)\\ & =1-P(A)-P(B)+P(AB)\\ & =1-F_X(a)-F_Y(b)+F_{X,Y}(a,b)\end{array}$$

• $p_X(x)=P(X=x)={\sum }_{y\in \mathbb{R}}{p}_{X,Y}(x,y)$
• $p_Y(y)=P(Y=y)={\sum }_{x\in \mathbb{R}}{p}_{X,Y}(x,y)$

• $p_{X,Y}(x,y)\ge 0,\forall x,y\in \mathbb{R}$
• ${\sum }_x{\sum }_y{p}_{X,Y}(x,y)=1$

1. $P(a_1<X\le a_2,b_1<Y\le b_2)={\sum }_{a_1<x\le a_2}{\sum }_{b_1<y\le b_2}{p}_{X,Y}(x,y)$
2. $F_{X,Y}(a,b)=P(X\le a,Y\le b)={\sum }_{x\le a}{\sum }_{y\le b}{p}_{X,Y}(x,y)$
3. $P(X>a,Y>b)={\sum }_{x>a}{\sum }_{y>b}{p}_{X,Y}(x,y)$

suppose: $3$ balls randomly selected from urn

• w/ $3,4,5$ red, white, and blue balls respectively
• $R,W$: rn. of white balls chosen
• pmf of $X,Y$ is:

• last row & column: marginal pmf

suppose: 15% of family in community w/ no children

• 20% w/ 1
• 35% w/ 2
• 30% w/ 3
• each child: equally likely to be a boy or girl
• $B$: no. of boys, $G$: no. of girls

• $f_{X,Y}(x,y)\ge 0,\forall x,y\in \mathbb{R}$
• ${\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{f}_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y=1$

1. for $A,B\subset R$, take $C=A\times B$ $$\begin{array}{r}P(X\in A,Y\in B)={\int }_A{\int }_B{f}_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x\end{array}$$
2. for $a_1,a_2,b_1,b_2\in \mathbb{R}$ where $a_1<a_2,b_1<b_2$ $$\begin{array}{r}P(a_1<X\le a_2,b_1<Y\le b_2)={\int }_{a_1}^{a_2}{\int }_{b_1}^{b_2}{f}_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x\end{array}$$
3. ⭐ for $a,b\in \mathbb{R}$ $$F_{X,Y}(a,b)=P(X\le a,Y\le b)={\int }_{-\infty }^a{\int }_{-\infty }^b{f}_{X,Y}(x,y)\mathrm{d}y\mathrm{d}x$$

• thus: $$f_{X,Y}(x,y)=\frac{{\partial }^2}{\partial x\partial y}{F}_{X,Y}(x,y)$$

1. $P(X>1,Y<1)$ $$\begin{array}{rl}P(X>1,Y<1)& ={\int }_1^{\infty }{\int }_0^{1}2e^{-x}{e}^{-2y}\mathrm{d}y\mathrm{d}x\\ & ={\int }_1^{\infty }(-e^{-x}{e}^{-2y}){|}_0^1\mathrm{d}x\\ & ={\int }_1^{\infty }-e^{-x}(e^{-2}-1)\mathrm{d}x\\ & =e^{-x}(e^{-2}-1){|}_1^{\infty }\\ & =-e^{-1}(e^{-2}-1)=e^{-1}-e^{-3}\end{array}$$
2. $P(X<Y)$ $$\begin{array}{rl}P(X<Y)& ={\int }_0^{\infty }{\int }_x^{\infty }2e^{-x}{e}^{-2y}\mathrm{d}y\mathrm{d}x\\ & ={\int }_0^{\infty }-e^{-x}{e}^{-2y}{|}_x^{\infty }\mathrm{d}x\\ & ={\int }_0^{\infty }{e}^{-3x}\mathrm{d}x\\ & =-\frac{1}{3}{e}^{-3x}{|}_0^{\infty }=\frac{1}{3}\end{array}$$
3. $P(X\le x)$ $$\begin{array}{rl}P(X\le a)& ={\int }_0^a{\int }_0^{\infty }2e^{-x}{e}^{-2y}\mathrm{d}y\mathrm{d}x\\ & ={\int }_0^a-e^{-x}{e}^{-2y}{|}_0^{\infty }\mathrm{d}x\\ & ={\int }_0^a{e}^{-x}\mathrm{d}x\\ & =-e^{-x}{|}_0^a\\ & =1-e^{-a}\\ P(X\le x)& =1-e^{-x}\end{array}$$
4. $p_X(x)$ $$\begin{array}{rl}{p}_X(x)& ={\int }_0^{\infty }2e^{-x}{e}^{-2y}\mathrm{d}y\\ & =-e^{-x}{e}^{-2y}{|}_0^{\infty }\\ & =e^{-x}\end{array}$$ = $X\sim Exp(1)$
5. $F_Y(y)$ $$\begin{array}{rl}{F}_Y(a)& ={\int }_0^{\infty }{\int }_0^{a}2e^{-x}{e}^{-2y}\mathrm{d}y\mathrm{d}x\\ & ={\int }_0^{\infty }-e^{-x}{e}^{-2y}{|}_0^a\mathrm{d}x\\ & ={\int }_0^{\infty }-e^{-x}(e^{-2a}-1)\mathrm{d}x\\ & =e^{-x}(e^{-2a}-1){|}_0^{\infty }\\ & =-(e^{-2a}-1){|}_0^{\infty }=1-e^{-2a}\\ F_Y(y)& =1-e^{-2y}\end{array}$$