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  • STA256 Lecture 07

Probability Density Function

• Note for Continuous Random Variable
  • $P(a\le X\le b)=P(a<X<b)$
  • This is because the area under a line is $0$
  • ${\int }_1^{1}f(x)dx=0$
• Review from calculus
  • $n\ne -1$
  • $\int x^{n}dx$
  • $\frac{1}{n+1}{x}^{n+1}+C$
• Example:
  • Let Continuous Random Variable $X$ have a Probability Density Function
  • $f(x)=(\begin{array}{cc}-x+kx& 0\le x\le 1\\ 0& \text{otherwise}\end{array})$
  • Solve for $k$
    • To be a valid Probability Density Function
    • ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx=1$
    • ${\int }_0^1-x^2+kxdx=1$
    • Antiderive:
      • $(-\frac{x^3}{3}+\frac{k{x}^2}{2}){}_0^1=1$
      • $(-\frac{1}{3}+\frac{k}{2})-(-\frac{0^3}{3}+\frac{k{0}^2}{2})=1$
      • $(-\frac{1}{3}+\frac{k}{2})=1$
      • $k=\frac{8}{3}$
  • Therefore the Probability Density Function is $f(x)=\begin{array}{cc}-x^2+\frac{8x}{3}& 0\le x\le 1\\ 0& \text{otherwise}\end{array}$
  • Find $P(0\le x\le \frac{1}{2})$

        left=-1; right=1;
        top=2; bottom=-1;
        ---
        y=-x^2+\frac{8x}{3}|0<=x<= 1/2
    

  • $P(0\le x\le \frac{1}{2})$
    • ${\int }_0^{\frac{1}{2}}f(x)dx$
    • ${\int }_0^{\frac{1}{2}}-x^2+\frac{8x}{3}dx$
    • Antiderive:
      • $(-\frac{1}{3}{x}^3+\frac{1}{6}(8x^2)){}_0^{\frac{1}{2}}$
      • $\frac{7}{24}$
• Exercise:
  • Suppose we were provided with the Cumulative Distribution Function, but not the Probability Density Function.
  • #tk Review
  1. Verify $F(x)$ is a valid Cumulative Distribution Function
  2. Use the Cumulative Distribution Function to derive the Probability Density Function
     • Take derivative
• Question
  • Let $X$ be a Continuous Random Variable with a Cumulative Distribution Function, $F(x)=\begin{array}{cc}1-e^{-2x}& x\ge 0\\ 0& \text{otherwise}\end{array}$
  • Verify $F(x)$ is a valid Cumulative Distribution Function.
  • Show $x\to \mathrm{\infty }$ goes to $1$ and $x\to -\mathrm{\infty }$ goes to $0$
  • To show non-decreasing
    • $x=0,F(0)1-e^{-2(0)}=1-1=0$
      • $0\ge 0$
      • True
    • $\underset{x\to \mathrm{\infty }}{lim}1-e^{-2x}=1$
    • $\underset{x\to -\mathrm{\infty }}{lim}F(x)=0$
    • To show it's non-negative
    • $0<e^{-2x}\le 1$
    • So $1>\underset{F(x)}{\underbrace{1-e^{-2x}}}\ge 0$
    • Non-decreasing between $0$ and $1$
  • Find the Probability Density Function
    • Cumulative Distribution Function is $F(x)$
    • We want $f(x)$ the Probability Density Function
    • $f(x)=\frac{d}{dx}F(x)dx$
    • $\frac{d}{dx}(1-e^{-2x})$
    • $-2(-e^{-2x})$
    • $2e^{-2x}$
    • Probability Density Function is
    • $f(x)=\begin{array}{cc}2e^{-2x}& x\ge 0\\ 0& \text{otherwise}\end{array}$