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  • STA256 Lecture 06

Transformations

• If we have a random variable $X$ and distribution of $X$ and have another variable $Y=g(X)$, then we need to find the distribution of $Y$.
• Example:
  • Random variable $X$ with known Probability Mass Function.
    • $P_X(x)$
  • Random variable $Y$ with relationship $Y=g(X)$ where $g$ is one-to-one, injective.
  • Find the Probability Mass Function of $Y$
    • $P_Y(Y)=P_X(g^{-1}(g))$

Discrete Random Variable

• Example (Slide 76):
  • We have random variable $X$ it has a Probability Mass Function
  • $P_X(x)=\frac{1}{3},x=1,2,3$
  • $\begin{array}{cccc}X=x& 1& 2& 3\\ P(X=x)& \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\end{array}$
  • Find Probability Mass Function of $Y=2x+1$
  • $g(x)=2x+1$
  • $y=2x+1$
  • Inverse
    • $x=\frac{y-1}{2}$
    • So $y^{-1}(x)=\frac{x-1}{2}$
  • By the transformation theorem for one-to-one mappings. We have $P_Y(y)=P_X(g^{-1}(y))$
  • $P_X\left(\frac{y-1}{2}\right),y=3,5,7$
  • $\begin{array}{cc}y=3& P_Y(3)=P_X\left(\frac{3-1}{2}\right)=P_X(1)=\frac{1}{3}\\ y=5& P_Y(5)=P_X\left(\frac{5-1}{2}\right)=P_X(2)=\frac{1}{3}\\ y=7& P_Y(7)=P_X\left(\frac{7-1}{2}\right)=P_X(3)=\frac{1}{3}\end{array}$

  ---

  • $\begin{array}{cccc}X=x& 1& 2& 3\\ P(X=x)& \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \downarrow y=2x+1\\ Y=y& 3& 5& 7\\ P(Y=y)& \frac{1}{3}& \frac{1}{3}& \frac{1}{3}\end{array}$
• Example:
  • $P_X(x)=\frac{x}{10},x=1,2,3,4$
  • Find the Probability Mass Function of $Y=5x+2$
  • $\begin{array}{ccccc}X=x& 1& 2& 3& 4\\ P(X=x)& \frac{1}{10}& \frac{2}{10}& \frac{3}{10}& \frac{4}{10}\end{array}$
  • $g(x)=5x+2$
  • Inverse
    • $y=5x+2$
    • $y-2=5x$
    • $\frac{y-2}{5}=x$
    • $g^{-1}(x)=\frac{x-2}{5}$
    • Support:
      • Using $g$
      • $g^{-1}$: support is $x=7,12,17,22$
  • Theorem for one-to-one transformations
    • $P_Y(y)=P_X(g^{-1}(y))$
    • $P_X(x)=\frac{x}{10}$
    • $g^{-1}(y)=\frac{y-2}{5}$
    • $P_Y(y)=P_X\left(\frac{g^{-1}(y)}{10}\right)$
    • $P_Y(y)=P_X\left(\frac{\frac{y-2}{5}}{10}\right)$
    • $P_Y(y)=P_X\left(\frac{y-2}{50}\right)$
    • for $y=,7,12,17,22$
• Example: Ref
  • 2
    • Let $X$ have the Probability Mass Function $P_X(x)=\frac{1}{3},x=-1,0,1$
    • Find the Probability Mass Function of $Y=X^2$
    • $g(x)=x^2$
    • inverse
      • $y=x^2$
      • $\sqrt{y}=x$
      • $g^{-1}(x)=\sqrt{y}$
      • $g^{-1}(y)=\sqrt{x}$
    • Theorem for one to one transformations doesn't apply.
    • So we do it manually
    • $P_Y(1)$
      • $P(Y=1)$
      • $Y=1$ only if $X=1,-1$
      • So $P(X=1)+P(X=-1)=\frac{2}{3}$
    • $P_Y(0)$
      • $P(Y=0)$
      • $=P(X=0)=\frac{1}{3}$

Continuous Random Variable

• When would you use either?:
  • Cumulative Distribution Function is $F(x)=\left\{\begin{array}{c}\mathrm{\_}\mathrm{\_}\\ \mathrm{\_}\mathrm{\_}\\ \mathrm{\_}\mathrm{\_}\end{array}\right\}$
  • Provided $Y=g(x)$ find the Probability Density Function of $Y$
  • To use the Jacobian
  • $f_X(x)=\frac{d}{dx}F(x)$
  • Now we have the Probability Density Function of $X$. Then continue with the jacobian.

Jacobian Technique

#tk Theorem 33

• Example:
  • If Continuous Random Variable $X$ has a Probability Density Function $f_X(x),y=g(x)$ $x=g^{-1}(y)$
  • The Probability Density Function of $Y$ is:
    • $f_Y(y)=f_X(g^{-1}(y))\cdot \underset{\text{Jacobian (J)}}{\underbrace{\left|\frac{dx}{dy}\right|}}$
      • $\frac{dx}{dy}=\frac{d}{dy}{g}^{-1}(y)$
      • Jacobian accounts for stretches and modifications to the equations
  • #tk slide 90
• Example 2:
  • $f_X(x)=2(1-x)=2-2x$
    • support of $x$ is $0<x<1$
  • $y=-\ln (1-x)$
  • Probability Density Function of $Y$
  • $f_Y(y)=?$
  • Support of $Y$
    • $0<x<1$
    • $x=0$, $y=-\ln (1-0)=0$
    • $x=1,y=-\ln (1-1)=-\ln (0)\to \mathrm{\infty }$
    • support of $y:0<y<\mathrm{\infty }$
  • Find the inverse
    • $Y=-\ln (1-x)$
    • $g(x)=-\ln (1-x)$
    • $y=-\ln (1-x)$
    • $-y=\ln (1-x)$
    • $e^{-y}=1-x$
    • $e^{-y}-1=-x$
    • $1-e^{-y}=x$
    • $g^{-1}(x)=1-e^{-x}$
    • $g^{-1}(y)=1-e^{-y}$
    • Jacobian
      • $\left|\frac{dx}{dy}\right|=|\frac{d}{dy}1-e^{-y}|=0+1e^{-y}$
      • $0+e^{-y}=e^{-y}$
    • By the theorem:
      • $f_X(x)=2-2x$
      • $g^{-1}(y)=1-e^{-y}$
      • $f_Y(y)=f_x(g^{-1}(y))\cdot \left|\frac{dx}{dy}\right|$
      • $=(2(1-(1-e^{-y})))\cdot e^{-y}$
      • $=(2e^{-y})\cdot e^{-y}$
      • $=(2e^{-2y})$
      • We need the support
      • $0<y<\mathrm{\infty }$
• Example 3:
  • Let a Continuous Random Variable $X$ have a Probability Density Function $f_X(x)$. Let $y=g(x)$ the Probability Density Function of $Y$ be $f_Y(y)=f_X(g^{-1}(y))\left|\frac{dx}{dy}\right|$
  • $X$ has a Probability Density Function $f_X(x)=\frac{1}{2}(1+x),-1\le x{\le }_1$
  • Find the PDF of the transformation $y=e^x$
  • Find if it's one-to-one and support of $y$
    • Support of $Y$
      • $y=e^x,-1\le x\le 1$
      • $e^x$ is one-to-one
      • $\begin{array}{cc}x=-1,& y=e^{-1}\\ x=1,& y=e^1\end{array}$
    • Inverse Map
      • $y=e^x$ means $g(x)=e^x$
      • $x=\ln y$ means $g^{-1}(x)=\ln x$
      • Jacobian
      • $\left|J\right|=\left|\frac{dx}{dy}\right|$
      • $|\frac{d}{dy}\ln y|$
      • $\left|\frac{1}{y}\right|$
      • Since $e^x$ is always positive, we can just say $\frac{1}{y}$
    • By the transformation theorem
      • $f_Y(y)=f_X(g^{-1}(y))$
      • $f_X(x)=\frac{1}{2}(1+x)$
      • $g^{-1}(y)=\ln y$
      • $f_X(x)=\frac{1}{2}(1+\ln y)\left|J\right|$
      • $f_X(x)=\frac{1}{2}(1+\ln y)\frac{1}{y}$
      • $f_X(x)=\frac{1}{2y}(1+\ln y)$
      • Support
        • $e^{-1}\le y\le e$
• Example 4:
  • $f(x)=\frac{3}{4}(1-x^2),-1\le x\le 1$
  • Find PDF of $y=x^2$
  • Support of $Y$
    • Not one-to-one
    • $\begin{array}{cc}x& y\\ (-1,0)& (0,1)\\ (0,1)& \underset{\begin{array}{c}\text{Support of }y\\ 0\le y\le 1\end{array}}{\underbrace{(0,1)}}\end{array}$
  • Inverse map
    • $y=x^2$
    • $g(x)=x^2$
    • $x=\pm \sqrt{y}$
    • $g^{-1}(y)=\pm \sqrt{y}$
    • Jacobian
    • $\left|\frac{dx}{dy}\right|$
    • $|\frac{d}{dy}(\pm \sqrt{y})|$
    • $|\pm \frac{1}{2\sqrt{y}}|$
    • $\frac{1}{2\sqrt{y}}$
  • By the transformation theorem
    • $f_Y(y)=f_X(g^{-1}(y))$
    • $f_X(x)=\frac{3}{4}(1-x^2),-1\le x\le 1$
    • $(f_X(-\sqrt{y})+f_X(\sqrt{y}))\left|J\right|$
    • $f_Y(y)=\frac{3}{4}(1-{(\pm \sqrt{y})}^2)$
    • $f_Y(y)=(\left(\frac{3}{4}(1-{\left(\sqrt{y}\right)}^2)\right)+\left(\frac{3}{4}(1-{(-\sqrt{y})}^2)\right))\frac{1}{2\sqrt{y}}$
    • $f_Y(y)=\frac{3(1-y)}{4\sqrt{y}},0\le y\le 1$

Cumulative Distribution Function Technique

• Let $X$ be a Continuous Random Variable with a Cumulative Distribution Function $F(x)$
• Let $Y=g(x)$ be a transformation
• The Cumulative Distribution Function of $Y$:
  • $F_Y(y)=P(Y\le y)$
  • $F_Y(y)=P(X\le g^{-1}(y))$
  • $F_Y(y)=F_X(g^{-1}(y))$ if $g$ is non-decreasing.
  • $F_Y(y)=1-F_X(g^{-1}(y))$ if $g$ is non-increasing.
• The Probability Density Function of $Y$:
  • The derivative
  • $f_Y(y)=\frac{d}{dy}{F}_Y(y)$
• Example 1:
  • Let $X$ have a Continuous Random Variable given by $F(x)=\left\{\begin{array}{cc}3x^2-2x^3& 0\le x<1\\ 0& \text{otherwise}\end{array}\right\}$
  • Find the Probability Density Function of $y=x^2$
  • Support of $y$
    • $y=x^2$ over $0\le x<1$

      	left=-1.5; right=1.5;
      	top=2; bottom=-0.5;
      	---
      	y=x^{2}|0<=x<1
      	y=x^{2}|dashed|x<0
      

    • In $0\le x\le 1$ this is one-to-one
    • $\begin{array}{cc}x=0& y=0^2=0\\ x=1& y=1^2=1\end{array}$
    • So $0\le y<1$
  • Inverse Map:
    • $y=x^2$ this is $g(x)$
    • $x=\sqrt{y}$ this is $g^{-1}(x)=\sqrt{x}$
    • Over $0\le x<1$
  • Cumulative Distribution Function of $Y$:
    • $F_Y(y)=F_X(g^{-1}(y))$
    • $F(x)=\left\{\begin{array}{cc}3x^2-2x^3& 0\le x<1\\ 0& \text{otherwise}\end{array}\right\}$
    • $g^{-1}(y)=\sqrt{y}$
    • $F_Y(y)=F_X(g^{-1}(y))$
    • $F_Y(y)=F_X(3{\sqrt{y}}^2-2{\sqrt{y}}^3)$
    • $F_Y(y)=F_X(3y-2{\sqrt{y}}^3)$
    • $F_Y(y)=3y-2\sqrt{y},0\le y<1$
  • Probability Density Function of $Y$
    • $f_Y(y)=\frac{d}{dy}{F}_Y(y)$
    • $f_Y(y)=\frac{d}{dy}3y-2\sqrt{y^3}$
    • $f_Y(y)=3-3\sqrt{y},0\le y<1$
• Example 2:
  • Let $X$ have a Cumulative Distribution Function $F(x)=\left\{\begin{array}{cc}0& x<-1\\ \frac{(x+1{)}^2}{4}& -1\le x<1\\ 1& x\ge 1\end{array}\right\}$
  • Support of $Y$
    • $Y=x^2,-1\le x\le 1$
    • Not one-to-one here
    • $0\le y<1$
  • Inverse Map
    • $y=x^2$ this is $g(x)=x^2$
    • $x=\pm \sqrt{y}$ this is $g^{-1}(x)=\pm \sqrt{x}$
  • Cumulative Distribution Function of $Y$
    • $F_Y(y)=P(Y\le y)$
    • $F_Y(y)=P(x^2\le y)$
    • $F_Y(y)=P(-\sqrt{y}\le x\le \sqrt{y})$
    • For a Continuous Random Variable:
      • $P(a\le X<b)=F(b)-F(a)$
      • This is because the area under a line is $0$.
      • $F(x)=\left\{\begin{array}{cc}0& x<-1\\ \frac{(x+1{)}^2}{4}& -1\le x<1\\ 1& x\ge 1\end{array}\right\}$
    • $F_Y(y)=F(\sqrt{y})-F(-\sqrt{y})$
    • $g^{-1}(y)=\pm \sqrt{y}$
    • $F_Y(y)=\left(\frac{(1+\sqrt{y}{)}^2}{4}\right)-\left(\frac{(1-\sqrt{y}{)}^2}{4}\right)$
    • $F_Y(y)=\dots$
    • $F_Y(y)=\sqrt{y},0<y<1$
  • Probability Density Function of $Y$
    • $f_Y(y)=\frac{d}{dy}{F}_Y(y)=\frac{d}{dy}\sqrt{y}=\frac{1}{2\sqrt{y}},0<y<1$
• Exercise #tk
  • $F(x)=\left\{\begin{array}{cc}0& x<0\\ x^5& 0\le x<1\\ 1& x\ge 1\end{array}\right\}$
  • Find the Probability Density Function of $y=1-x^3$
    • This is decreasing
  • Answer: $f_Y(y)=\frac{5}{3}(1-y{)}^{\frac{2}{3}},0<y<1$