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  • Discrete Random Vector

Joint Cumulative Distribution Function

• $F_{X,Y}(x,y)=P(X\le x,Y\le y)$

Discrete Random Variable

• $F_{X,Y}(x,y)=\sum _x^{}\sum _{y}P(X\le x,Y\le y)$
• Example:
  • $\begin{array}{cccc}& & & X_1\\ & & 0& 1& 2\\ X_2& 0& 0.10& 0.15& 0.05& 0.3\\ & 1& 0.05& 0.30& 0.35& 0.7\\ & & 0.15& 0.45& 0.4& 1\end{array}$
  • $F(\underset{X_1}{\underbrace{1}},\underset{X_2}{\underbrace{1}})=P(X_1\le 1,X_2\le 1)$
  • $P(X_1=0,X_2=0)+P(X_1=1,X_2=0)+\dots$

Continuous Random Vector

• For univariate we did integration: Joint Cumulative Distribution Function
• This is just a level surface
• Now we get $P(a<x_1<b,c<x_2<d)$
• ${\int }_{x_1=c}^{x_1=d}{\int }_{x_2=a}^{x_2=b}dx$
• Example:
  • Bivariate Probability Density Function
  • $f(x_1,x_2)=\frac{24-3(x_1^2+x_2^2)}{256},\begin{array}{c}-2\le x_1\le 2\\ -2\le x_2\le 2\end{array}$
  • Find the probability that $P(\underset{\text{Region of integration}}{\underbrace{0.2\le x_1\le 1.8,0.2\le x_2\le 1.8}})$
    • This is a square that we're integrating

      	left=-0.5; right=2;
      	top=2; bottom=-0.5;
      	---
      	0.2<=x<=1.8|0.2<=y<=1.8
      	0.2<=y<=1.8|0.2<=x<=1.8
      

    • ${\int }_{x_1=1.8}^{x_2=1.8}{\int }_{x_1=0.2}^{x_1=1.8}f(x_1,x_2)d{x}_{1}d{x}_2$
    • ${\int }_{x_1=1.8}^{x_2=1.8}{\int }_{x_1=0.2}^{x_1=1.8}\frac{24-3(x_1^2+x_2^2)}{256}d{x}_{1}d{x}_2$
    • $\frac{1}{256}{\int }_{x_1=1.8}^{x_2=1.8}({\int }_{x_1=0.2}^{x_1=1.8}24-3x_1^2+3x_2^{2}d{x}_1)d{x}_2$
    • $\frac{1}{256}{\int }_{x_1=1.8}^{x_2=1.8}\left([24x_1-\frac{3x_1^3}{3}+3x_2^2{x}_1]{}_{x_1=0.2}^{x_1=1.8}\right)d{x}_2$
    • $\frac{1}{256}{\int }_{x_1=1.8}^{x_2=1.8}([24x_1-\frac{3x_1^3}{3}+3x_2^2{x}_1]-[24x_1-\frac{3x_1^3}{3}+3x_2^2{x}_1])d{x}_2$
    • $\frac{1}{256}{\int }_{x_2=0.2}^{x_2=1.8}32.576-\frac{4.8x_2^3}{3}dx$
    • $\frac{42.8032}{256}$
• Example:
  • Suppose we want to find $P(0.2\le x_1\le 1.8,0.2\le x_2\le 1.8,x_2<x_1)$
  • $x_2<x_1$ is another restriction that we have.

    	left=-0.25; right=2;
    	top=2; bottom=-0.25;
    	---
    	0.2<=x<=1.8|0.2<=y<=1.8
    	0.2<=y<=1.8|0.2<=x<=1.8
    	y=x
    	(1.8,0.2)|label:(X1=1.8,X2=0.2)
    

  • Consider the boundary, where $x_2=x_1$
  • Find out which triangle satisfies the original condition that $x_2<x_1$
  • Find the point at the end
  • See that on the bottom right, we get that $x_1>x_2$
  • So we need the bottom triangle
  • $f(x_1,x_2)=\frac{24-3(x_1^2+x_2^2)}{256}$
  • ${\int }_{x_1=0.2}^{x_1=1.8}{\int }_{x_2=0.2}^{x_2=x_1}\frac{24-3(x_1^2+x_2^2)}{256}d{x}_{1}d{x}_2$
• Example:
  • Slide 16
• Example:
  • Slide 18
  • Joint Probability Density Function
  • $f(x_1,x_2)=3x_1$ for $0\le x_2\le x_1\le 1$
  • Find $P(0\le x_1\le 0.5,0.25\le x_2)$

    	left=-0.25; right=1.25;
    	top=1.25; bottom=-0.25;
    	---
    	y=x
    	0<=x<=1|0<=y<=1
    	0<=x<=0.5|0.25<=y
    	
    

  • The area shaded, that's under the line $x_2=x_1$
  • ${\int }_0^1{\int }_0^{1}f(x_1,x_2)dxdx$
  • ${\int }_{x_2=0.25}^{x_2=0.5}{\int }_{x_1=x_2}^{x_1=0.5}3x_{1}d{x}_{1}d{x}_2$
  • ${\int }_{x_2=0.25}^{x_2=0.5}\left[\frac{3x_1^2}{2}\right]{}_{x_1=x_2}^{x_1=0.50}d{x}_2$
  • ${\int }_{x_2=0.25}^{x_2=0.5}\frac{3}{8}-\frac{3x_2^2}{2}d{x}_2$
  • $[\frac{3x_2}{8}-\frac{3x_2^3}{6}]{}_{x_2=0.25}^{x_2=0.5}$
  • $\frac{5}{128}$
  • #tk Try integral the other way $d{x}_2,d{x}_1$ instead of the way we did
• Example:
  • Let $f(x_1,x_2)=\frac{1}{9}{x}_2$ for $0\le x_2\le x_1\le 3$
  • Find $P(x_1+x_2\le 3)$

    	left=-0.25; right=3.25;
    	top=3.25; bottom=-0.25;
    	---
    	y=x
    	y=3-x
    	y=3|0<=x<=3
    	x=3|0<=y<=3
    	y=1.5|0<=x<=1.5|dashed
    	x=1.5|0<=y<=1.5|dashed
    

  • Consider $x_1+x_2=3$
  • $x_2=3-x_1$
  • Bottom right triangle satisfies $x_2=x_1$
  • Bottom triangle satisfies $x_1+x_2\le 3$ and $x_2=x_1$
  • By symmetry, finding the bottom triangle bisected by the line $x=1.5$
  • If we find the left half, $\times 2$ gets us the right half two.
  • $2{\int }_{x_1=0}^{x_1=1.5}{\int }_{x_2=0}^{x_2=x_1}\frac{1}{9}{x}_{2}d{x}_{2}d{x}_1$
  • Or we get ${\int }_{x_1=0}^{x_1=1.5}{\int }_{x_2=0}^{x_2=x_1}\frac{1}{9}{x}_{2}d{x}_{2}d{x}_1+{\int }_{x_1=1.5}^{x_1=3}{\int }_{x_2=0}^{x_2=3-x_1}\frac{1}{9}{x}_{2}d{x}_{2}d{x}_1$
    • #tk Hint u-sub
  • $\frac{1}{8}$ as our result