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  • STA256 Lecture 13

Transformations - Bivariate

Discrete Random Vector

• Let $P_{x_1,x_2}(x_1,x_2)$ be a Joint Probability Mass Function of $X_1$ and $X_2$
• Let $\{\begin{array}{l}{Y}_1=g_1(x_1,x_2)\\ Y_2=g_2(x_1,x_2)\end{array}$
• The Joint Probability Mass Function of $Y_1,Y_2$ is:
• $P_{y_1,y_2}(y_1,y_2)=P_{x_1,x_2}[w_1(g_1,g_2),w_2(g_1,g_2)]$
• Where $w_1,w_2$ are the inverse maps
• Example:
  • Non-one-to-one function
  • Let $(x_1,x_2)$ have a Joint Probability Mass Function
  • $$\begin{array}{cccc}& & & x_1\\ & & 0& 1\\ & 0& 0.2& 0.3\\ x_2& 1& 0.1& 0.4\end{array}$$
  • Let $\{\begin{array}{ll}{Y}_1=x_1+x_2& =g_1(x_1,x_2)\\ Y_2=x_1{x}_2& =g_2(x_1,x_2)\end{array}$
  • Find the Probability Mass Function of $(Y_1,Y_2)$
  • $\begin{array}{ccccc}& & & & (Y_1,Y_2)\\ x_1=0& x_2=0⟹& Y_1=0+0=0& Y_2=(0)(0)=0& (0,0)\\ x_1=0& x_2=1⟹& Y_1=0+1=1& Y_2=(0)(1)=0& (1,0)\\ x_1=1& x_2=0⟹& Y_1=1+0=1& Y_2=(1)(0)=0& (1,0)\\ x_1=1& x_2=1⟹& Y_1=1+1=2& Y_2=(1)(1)=1& (2,1)\end{array}$
  • See that it's not injective as there are two output for the same input.
  • $f(x_1,x_2)=f(c_1,c_2)\text{⧸}⟹(x_1,x_2)=(c_1,c_2)$
  • $P(Y_1=0,Y_2=0)=P(X_1=0,X_2=0)=0.2$
  • $P(Y_1=1,Y_2=0)=P(X_1=0,X_2=1)+P(X_1=1,X_2=0)=0.1+0.3=0.4$
  • $P(Y_1=2,Y_2=1)=P(X_1=1,X_2=1)=0.4$
  • $\begin{array}{cccc}& & & Y_1\\ & \times & 0& 1& 2\\ & 0& 0.2& 0.4& 0\\ Y_2& 1& 0& \underset{\begin{array}{c}\text{We don’t get the}\\ \text{chance for these}\\ \text{two probabilities}\\ \text{so we get }0\end{array}}{\underbrace{0}}& 0.4\end{array}$
  • $f_{y_1,y_2}(y_1,y_2)=\{\begin{array}{ll}0.2& y_1=0,y_2=0\\ 0.4& y_1=1,y_2=0\\ 0.4& y_1=2,y_2=1\\ 0& \text{otherwise}\end{array}$
  • #tk slides 28

Continuous Random Vector

Jacobian

• Suppose we have a Joint Probability Density Function of Continuous Random Variables $X_1,X_2$ and transformations:
  • $\{\begin{array}{l}{Y}_1=g_1(x_1,x_2)\\ Y_2=g_2(x_1,x_2)\end{array}$
• The Joint Probability Density Function of $(Y_1,Y_2)$ is $f_{y_1,y_1}(y_1,y_1)=f_{x_1,x_2}(h_1(x_1,x_2),h_2(x_1,x_2))\cdot \left|J\right|$ where $h_1,h_2$ are the inverse maps.
• Here $\begin{array}{cccc}& & y_1& y_2\\ \left|J\right|=& x_1& \frac{\partial x_1}{\partial y_1}& \frac{\partial x_1}{\partial y_2}\\ & x_2& \frac{\partial x_2}{\partial y_1}& \frac{\partial x_2}{\partial y_2}\end{array}$
• Determinant of the above
• If we don't have a one-to-one transformation and only one function $g_1(x_1,x_2)$, we introduce a dummy variable
• Example:
  • Let $(x_1,x_2)$ be continuous, with a Joint Probability Density Function given by:
    • $f_{x_1,x_2}(x_1,x_2)=8x_1{x}_2$ for $0<x_1<x_2<1$
  • Let $\{\begin{array}{ll}{Y}_1=\frac{x_1}{x_2}& =g_1(x_1,x_2)\\ Y_2=x_2& =g_2(x_1,x_2)\end{array}$
  • Find the Joint Probability Density Function of $(Y_1,Y_2)$
  • Consider the support of $(Y_1,Y_2)$
    • We want things in terms of $x_1$ and $x_2$
    • $y_2=x_2⟹x_2=y_2=h_1$
      • Sub this into the next
    • $y_1=\frac{x_1}{x_2}⟹x_1=y_1{x}_2=y_1{y}_2=h_2$
    • $y_1=\frac{x_1}{x_2}$ for $0<x_1<x_2<1$
    • Since $x_1<x_2$, $y_1=\frac{x_1}{x_2}$ we always have a smaller value over a bigger value
    • So we get that $0<y_1<1$
    • $y_2=x_2$
    • if $0<x_1<x_2<1$ then $0<y_2<1$
  • $x_1=y_1{y}_2$
  • $x_2=y_2$
  • $\left|J\right|=\left|\begin{array}{ccc}& y_1& y_2\\ x_1& \frac{\partial x_1}{\partial y_1}& \frac{\partial x_1}{\partial y_2}\\ x_2& \frac{\partial x_2}{\partial y_1}& \frac{\partial x_2}{\partial y_2}\end{array}\right|$
  • $\left|J\right|=\left|\begin{array}{ccc}& y_1& y_2\\ x_1& y_2& y_1\\ x_2& 0& 1\end{array}\right|$
  • $\left|J\right|=(y_2)(1)-(y_1)(0)$
  • $\left|J\right|=y_2$
  • By the transformation theorem
  • $f(x_1,x_2)=8x_1{x}_2$
  • $f_{y_1,y_2}(y_1,y_2)=f_{x_1,x_2}(h_1(y_1,y_2),h_2(y_1,y_2))$
  • We know above that we get $h_1=y_1{y}_2$ and $h_2=y_2$
  • $f_{y_1,y_2}(y_1,y_2)=8h_1{h}_2$
  • $f_{y_1,y_2}(y_1,y_2)=8(y_1{y}_2)(y_2)\left|J\right|$
  • $f_{y_1,y_2}(y_1,y_2)=8(y_1{y}_2)(y_2)(y_2)$
  • $f_{y_1,y_2}(y_1,y_2)=8(y_1{y}_2)(y_2{)}^2$
  • $f_{y_1,y_2}(y_1,y_2)=8y_1{y}_2^3$
• Example:
  • Slide 33 #tk
  • Introduce a dummy variable
  • Let $y_2=x_2$
• Example:
  • $(x_1,x_2)$ have joint pdf
  • $f(x_1,x_2)=4x_1{x}_2$ for $0<x_1<1$ and $0<x_2<1$
  • Let $Y_1=x_1+x_2$
  • $Y_2=x_1-x_2$
    • Since we have one-to-one, we don't need a dummy variable
  • Find the Joint Probability Density Function of $(Y_1,Y_2)$
  • Inverse map, $y_1=x_1+x_2$ is $x_1=\frac{y_1+y_2}{2}=h_1$
  • and $x_2=\frac{y_1-y_2}{2}=h_2$
  • Support is $0<x_1<1$ and $0<x_2<1$
  • we get that $0<\frac{y_1+y_2}{2}<1$ and $0<\frac{y_1-y_2}{2}<1$
  • So we get $0<y_1+y_2<2$ and $0<y_1-y_2<2$
  • Find the jacobian
  • $\left|\begin{array}{cc}\frac{\partial x_1}{\partial y_1}& \frac{\partial x_1}{\partial y_2}\\ \frac{\partial x_2}{\partial y_1}& \frac{\partial x_2}{\partial y_2}\end{array}\right|=\left|\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right|=-\frac{1}{2}$
  • Since we get $\left|J\right|$ we turn $-\frac{1}{2}\to \frac{1}{2}$
  • Joint Probability Density Function.
  • By the transformation theorem
  • $f(x_1,x_2)=4x_1{x}_2$ for $0<x_1<1$ and $0<x_2<1$
  • $f_{y,y}(y_1,y_2)=f_{x_1,x_2}(\underset{X_1}{\underbrace{h_1(y_1,y_2)}},\underset{X_2}{\underbrace{h_2(y_1,y_2)}})\cdot \left|J\right|$
  • Recall $h_1,h_2$ and $f_{x_1,x_2}$
  • $f_{y,y}(y_1,y_2)=4\left(\frac{y_1+y_2}{2}\right)\left(\frac{y_1-y_2}{2}\right)\frac{1}{2}$
  • $f_{y,y}(y_1,y_2)=\left(\frac{(y_1+y_2)(y_1-y_2)}{2}\right)$
    • for $0<y_1+y_2<2$
    • and $0<y_1-y_2<2$
• Example:
  • Let $(X_1,X_2)$ have a Joint Probability Density Function
  • $f(x_1,x_2)=x_1$ for $x_2<x_1<x_1+1$ and $0<x_2<1$

    	left=-0.50; right=6;
    	top=10; bottom=-10;
    	---
    	y=x-5|0<=x<=5
    	y=x|0<=x<=5
    	x=5|0<=y<=5
    	x=0|-5<=y<=0
    

  • That's how it looks if we graph it
  • Let $y=x_1-x_2$ and find the Probability Density Function of $Y$
  • Since this is not one-to-one, we introduce a dummy variable
  • Let $w=x_2$, where $w$ is our dummy variable.
    • Used to get the joint pdf, but we integrate it out at the end
  • Inverse Map
  • $Y$
    • $w=x_2$ so $x_2=w=h_2$
    • Finding something in terms of $x_1$
    • $y=x_1-x_2$
    • $x_1=y+\underset{\begin{array}{c}\text{We subbed in}\\ \text{from the dummy}\\ \text{variable}\end{array}}{\underbrace{x_2}}$
    • $x_1=y+w$
  • Support
    • $0<x_2<1$ we get that $0<w<1$
    • Since $x_2<x_1<x_2+1$ sub in what we had
    • $x_2<y+w<x_2+1$
    • $w<y+w<x_2+1$
    • $w<y+w<w+1$
    • $0<y<1$
  • Jacobian
    • $\left|J\right|=\left|\left|\begin{array}{cc}\frac{\partial x_1}{\partial y_1}& \frac{\partial x_1}{\partial y_2}\\ \frac{\partial x_2}{\partial y_1}& \frac{\partial x_2}{\partial y_2}\end{array}\right|\right|=\left|\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|\right|=1$
  • By the transformation theorem
    • $f_{y,w}(y,w)=f_{x_1,x_2}(h_1(y,w),h_2(y,w))\left|J\right|$
    • $f_{y,w}(y,w)=f_{x_1,x_2}(y+w)$
      • We don't have $h_2$ because it's just a dummy var
    • $y+w$ for $0<w<1$ and $0<y<1$
  • Integrate $w$ out
    • $f_y(y)=\int f_{y,w}(y,w)dw$
    • $f_y(y)=\int y+wdw$
      • Integrate from the support of $w$
    • $f_y(y)=\underset{0}{\overset{1}{\int }}y+wdw$
    • $f_y(y)=(yw+\frac{1}{2}{w}^2){}_0^1$
    • $f_y(y)=(y(1)+\frac{1}{2}(1{)}^2)$
    • $f_y(y)=\frac{1}{2}+y$ for $0<y<1$
  • Another example of slide 33 #tk

CDF Technique

• Joint Probability Density Function of $(X_1,X_2)$
• $F_{x_1,x_2}(x_1,x_2)=P(X_1\le x_1,X_2\le x_2)$
• Transformations - Bivariate
  • $Y_1=g_1(x_1,x_2)$
  • $Y_2=g_2(x_1,x_2)$
• The Cumulative Distribution Function:
  • $F_{y_1,y_2}(y_1,y_2)=P(Y_1\le y_1,Y_2\le y_2)$
  • $={\int }_{}^{}{\int }_{\begin{array}{c}{g}_1<y_1\\ g_2<y_2\end{array}}^{}{f}_{x_1,x_2}(x_1,x_2)d{x}_{1}d{x}_2$
• To recover the Probability Density Function
  • $f_{Y_1,Y_2}(y_1,y_2)=\frac{{\partial }^2}{\partial y_1\partial y_2}{F}_{Y_1,Y_2}(y_1,y_2)$
• The CDF technique is useful for 2-to-1 transformations without dummy variables
• Example:
  • $z=g(x_1,x_2)$
  • CDF
    • $F_z(z)=P(Z\le z)$
    • $={\int }_{x_2}^{}{\int }_{x_1}^{}{f}_{x_1,x_2}d{x}_{1}d{x}_2$
      • These limits on the integrals are $x_i$ in terms of $z$
      • $x_i(z)$
    • $={\int }_z^{}{f}_{x_1,x_2}dz$ if it's setup correctly
  • PDF
    • $f_Z(z)=\frac{d}{dz}{F}_Z(z)$
• Example:
  • Find the Probability Density Function of $z=x_1+x_2$
  • $f(x_1,x_2)=1$ for $\begin{array}{c}0<x_1<1\\ 0<x_2<1\end{array}$
  • We know this is two-to-one because we have one output $z$ not $(z_1,z_2)$.

    	left=-0.5; right=2;
    	top=2; bottom=-0.5;
    	---
    	y=1|0<=x<=1
    	x=1|0<=y<=1
    

  • This is a two-to-one function, so the
  • Support
    • Since $\begin{array}{c}0<x_1<1\\ 0<x_2<1\end{array}$
    • $\begin{array}{c}0<x_1+x_2<2\\ 0<z<2\end{array}$
    • $x_1+x_2=z$
    • So we rearrange for $x_2$
    • $x_2=z-x_1$
    • Line with slope $-1$ and variable height $z$

      	left=-0.5; right=2;
      	top=2; bottom=-0.5;
      	---
      	y=1|0<=x<=1
      	x=1|0<=y<=1
      	y=-x+0.5|0<=x<=0.5
      	y=-x+1|0<=x<=1
      	y=-x+1.5|0<=x<=1.5
      	y=-x+2|0<=x<=2
      

    • All of these are possible
    • We're looking for the region below however.
    • As the height of the line increases we have more and more area under the line inside the square we have to find
    • We want that $F(Z)=P(Z\le z)=P(X_1+X_2\le Z)$
    • Two cases:
      • Area to find is a triangle

        	left=-0.5; right=2;
        	top=2; bottom=-0.5;
        	---
        	y=1|0<=x<=1
        	x=1|0<=y<=1
        	y=-x+0.5|0<=x<=0.5
        	y=-x+1|0<=x<=1
        

        • $0<z<1$
        • $F(Z)=P(Z\le z)$
        • $={\int }_{?}^{?}{\int }_{?}^{?}f(x_1,x_2)d{x}_{2}d{x}_1$
        • $={\int }_{x_1=0}^{x_1=z}{\int }_{x_2=0}^{x_2=-x_1+z}1d{x}_{2}d{x}_1$
        • #tk
        • $=z^2-\frac{z^2}{2}$
        • $=\frac{z^2}{2}$ for $0<z<1$
      • Area to find is a square with a triangle cutout of the top right

        	left=-0.5; right=2;
        	top=2; bottom=-0.5;
        	---
        	y=1|0<=x<=1
        	x=1|0<=y<=1
        	y=-x+1.5|0<=x<=1.5
        	y=-x+2|0<=x<=2
        

        • $1<z<2$
        • It's easier to calculate the area of the top right triangle subtracted from the square
        • $F(Z)=P(Z\le z)$
        • $=1-P(Z\ge z)$
        • $=1-{\int }_{?}^{?}{\int }_{?}^{?}1d{x}_{2}d{x}_1$
        • $=1-{\int }_{x_1=z-1}^{x_1=1}{\int }_{x_2=z-x_1}^{x_2=1}1d{x}_{2}d{x}_1$
        • #tk
        • $1-\frac{(2-z{)}^2}{2}$ for $1\le z\le 2$
    • For both cases the CDF is case 1 + case 2
    • CDF
      • $F_X(x)=\{\begin{array}{ll}0& z<0\\ \frac{z^2}{2}& 0<z<1\\ 1-\frac{(2-z{)}^2}{2}& 1<z<2\\ 1& z>2\end{array}$
    • PDF
      • $\frac{d}{dx}{F}_X(x)$