• and let joint pdf $X,Y,Z$ be cylinder: extending in Z axis from the circle
• then, can we say $(X,Y)\perp \perp Z$?
  • 👨‍🏫 yes!

• 👨‍🏫 its cdf is not well established

Examples

• joint density of $X,Y$: given by $$f(x,y)=\{\begin{array}{ll}{e}^{-(x+y)}& 0<x<\infty ,0<y<\infty \\ 0& \text{otherwise}\end{array}$$
  • find: density of $X/Y$
  • $F(a)=P(A\le a)=P(X/Y\le a)=P(X\le aY)$ $$\begin{array}{rl}{\int }_0^{\infty }{\int }_0^{ay}{e}^{-(x+y)}\mathrm{d}x\mathrm{d}y& ={\int }_0^{\infty }-e^{-(x+y)}{|}_0^{ay}\mathrm{d}y\\ & ={\int }_0^{\infty }-e^{-(a+1)y}+e^{-y}\mathrm{d}y\\ & =\frac{1}{a+1}{e}^{-(a+1)y}-e^{-y}{|}_0^{\infty }\\ & =\frac{1}{a+1}-1\end{array}$$
  • and $f(a)=F^{\prime }(a)=-\frac{1}{(a+1{)}^2}$

Independent r.v.

• two r.v. $X,Y$: independent if: $$P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$$
  • for any $A,B\subset \mathbb{R}$
  • i.e. value of $X$ not influencing probability of $Y$, vice versa
  • else: they are dependent
• theorem: following statements are equivalent for jointly discrete / cont. r.v.
  1. r.v. $X,Y$ are independent
  2. $\forall x,y\in \mathbb{R}$, we have: $$p_{X,Y}(x,y)=p_X(x)p_Y(y)$$
     • joint pmf = product of marginal pmf
  3. $\forall x,y\in \mathbb{R}$, we have: $$F_{X,Y}(x,y)=F_X(x)F_Y(y)$$
     • joint cdf = product of marginal cdf
• r.v. $X,Y$ independent: iff $\exists g,h:\mathbb{R}\to \mathbb{R}$ s.t. for all $x,y\in \mathbb{R}$
  • $f_{X,Y}(x,y)=g(x)h(y)$
  • ($g,h$) not necessarily being $pdf$
  • 👨‍🏫 joint pdf: factorizable
    • same for cdf
• proof:
  • for continuous case, independence $⟹$ factorizable
    • for its pdf
  • now: $f_{X,Y}=h(x)g(y)$ $$\begin{array}{rl}1& ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{f}_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y\\ & ={\int }_{-\infty }^{\infty }h(x)\mathrm{d}x{\int }_{-\infty }^{\infty }g(y)\mathrm{d}y\\ & =C_1{C}_2\end{array}$$
    • where $C_1={\int }_{-\infty }^{\infty }h(x)\mathrm{d}y,C_2={\int }_{-\infty }^{\infty }g(y)\mathrm{d}y$:
      • $f_X(x)={\int }_{-\infty }^{\infty }{f}_X(x,y)\mathrm{d}y=C_{2}h(x)$
      • $f_Y(y)={\int }_{-\infty }^{\infty }{f}_Y(x,y)\mathrm{d}x=C_{1}g(y)$
    • as $C_1{C}_2=1$, $f_{X,Y}(x,y)=f_X(x)f_Y(y)$

Examples

• given 3 balls randomly selected from urn
  • containing 3 red, 4 white, 5 blue
  • let $R,W$: no. of red / white chosen
  • are $R,W$: independent?
  • $P(R=3,W=3)=0\ne P(R=3)P(W=3)$
• w/ $n+m$ independent trials, having common success probability $p$
  • $X$: no. of success in first $n$ trials
  • $Y$: no. of success in last $m$ trials
  • $X$ doesn't influence $Y$, and vice versa $$\begin{array}{rl}P(X=x,Y=y)& =\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x(1-p{)}^{n-x}(\genfrac{}{}{0ex}{}{m}{y})p^y(1-p{)}^{m-y}\\ =P(X=x)P(Y=y)& \end{array}$$
  • $X,Y$: independent
  • similarly $$P(Z=z)=\left(\genfrac{}{}{0ex}{}{m+n}{z}\right)p^z(1-p{)}^{m+n-z}$$ $$\begin{array}{rl}P(X=x,Z=z)& =P(X=x,X+Y=z)\\ & =P(X=x,Y=z-x)\\ & =\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x(1-p{)}^{n-x}(\genfrac{}{}{0ex}{}{m}{z-x})p^{z-x}(1-p{)}^{m-z+x}\end{array}$$
  • and $P(X=x,Z=z)\ne P(X=x)P(Z,z)$
• man an woman: decided to meet at a location
  • each person: independently arrive at u.a.r. between 12 noon ro 1 pm
  • find probability: first to arrive waiting longer than 10 minutes
  • $X$: time past 12 noon, in minutes, man arrives
  • $Y$: time past 12 noon, in minutes, woman arrives
  • $f_X(x)=\frac{1}{60}$ if $0\le x\le 60$
  • $f_Y(y)=\frac{1}{60}$ if $0\le y\le 60$
  • $f(x,y)={\frac{1}{60}}^2$ if $0\le x,y\le 60$
  • finding: $P(X\ge Y+10)+P(X+10\le Y)=2P(X\ge Y+10)$ $$\begin{array}{rl}P(X\ge Y+10)& ={\int }_0^{50}{\int }_{y+10}^{60}(\frac{1}{60}{)}^2\mathrm{d}x\mathrm{d}y\\ & ={\int }_0^{50}(\frac{1}{60}{)}^{2}x{|}_{y+10}^{60}\mathrm{d}y\\ & ={\int }_0^{50}\frac{1}{60}-\frac{y+10}{60^2}\mathrm{d}y\\ & ={\int }_0^{50}\frac{1}{60}-\frac{y}{60^2}-\frac{10}{60^2}\mathrm{d}y\\ & =\frac{y}{60}-\frac{y^2/2}{60^2}-\frac{10y}{60^2}{|}_0^{50}\\ & =\frac{3000-1250-500}{3600}\\ & =\frac{1250}{3600}=\frac{25}{72}\end{array}$$
  • thus $\frac{25}{36}$..?
• Buffon's needle problem: table ruled w/ equidistant parallel lines a $D$ apart
  • needle: length $L\le D$
    • randomly thrown on the table
  • what is the probability that needle: intersect one of the lines?
• if $f(x,y)=24xy$, $0<x<1,0<1,0<x+y<1$
  • are r.v. independent?
  • solution 1
    • find marginal pdf, and compare $$\begin{array}{rl}{f}_X(x)& ={\int }_0^{1-x}24xy\mathrm{d}y\\ & =12x{y}^2{|}_0^{1-x}\\ & =12x(1-x{)}^2\\ f_Y(y)& =12y(1-y{)}^2\\ f(x,y)& \ne f_X(x)f_Y(y)\end{array}$$
  • solution 2
    • let $$I(x,y)=\{\begin{array}{ll}1& \text{if }0<x<1,0<y<1,0<x+y<1\\ 0& \text{otherwise}\end{array}$$
  • $f(x,y)=24xyI(x,y)$ for all $x,y\in \mathbb{R}$
    • and $f(x,y)$: not factorizable
      • and thus: $X,Y$ are not independent
    • i.e. factorizable: we can define $I(x,y)=I(x)I(y)$ for all $x,y\in \mathbb{R}$
  • basically: $I$ is factorizable if its region is shown as square, cube, etc.
    • which edges are all parallel to the axis
    • otherwise: impossible
• $N$ traffic accidents occur per day w/ $N\sim Poisson(\lambda )$
  • each accident: major & minor
    • and it is a major accident w/ $p\in (0,1)$
  • let $X,Y$: denote no. of major & minor accidents, respectively
  • $X+Y=N$, if $N=n$, then $X\sim Bin(n,p)$
  1. find joint pmf of $X,Y$
     • for $i,j\ge 0$ $$\begin{array}{rlrl}P(X=i,Y=j)& =P(X=i,X+Y=i+j)& & \\ & =P(X=i,N=i+j)& & \\ & =P(X=i|N=i+j)P(N=i+j)& & \\ & =\left(\genfrac{}{}{0ex}{}{i+j}{i}\right)p^i(1-p{)}^{j}P(N=i+j)& & \text{as }X\sim Bin(n,p)\\ & =\left(\genfrac{}{}{0ex}{}{i+j}{i}\right)p^i(1-p{)}^j{e}^{-\lambda }\frac{{\lambda }^{i+j}}{(i+j)!}& & \\ & =\frac{(i+j)!}{i!j!}{p}^i(1-p{)}^j{e}^{-\lambda }\frac{{\lambda }^{i+j}}{(i+j)!}& & \\ & =p^i(1-p{)}^j{e}^{-\lambda }\frac{{\lambda }^{i+j}}{i!j!}& & \\ & =e^{-\lambda }\frac{(\lambda p{)}^i}{i!}\frac{(\lambda (1-p){)}^j}{j!}& & \end{array}$$
  2. are $X,Y$ independent?
     • yes, as it's factorizable by:
       • $g(i)=e^{-\lambda }\frac{(\lambda p{)}^i}{i!}$ within range
       • $h(i)=\frac{(\lambda (1-p){)}^j}{j!}$ within range
  3. can you identify: distribution of $X,Y$?
     • 👨‍🏫 exercise!
     • $X\sim Poisson(\lambda p)$
     • $Y\sim Poisson(\lambda )$