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  • Special Distributions

Exponential Distribution

• Commonly used to model time between events, survival, failure.
• Probability Density Function
  • $f(x)=\frac{1}{\beta }{e}^{-\frac{x}{\beta }}$ for $x\ge 0,\beta >0$
  • $\beta =1$
  • $f(x)=e^{-x}$
  • Exponential decay

        left=-5; right=5;
        top=5; bottom=-0.5;
        ---
        y=e^{-x}
        y=1/2 e^{- x/2}
        y=1/3 e^{- x/3}
        y=1/0.5 e^{- x/0.5}
    

• Cumulative Distribution Function
  • #tk
  • $F(x)=P(X\le x)=1-e^{-\frac{x}{\beta }}$
  • Obtain $F_X(x)$ from $f_X(x)$
• Properties:
  • Mean: $\mu =E[X]=\beta$
  • Variance: ${\sigma }^2=var(x)={\beta }^2$
  • MGF: $M_X(t)=\frac{1}{1-\beta t}$ for $t<\frac{1}{\beta }$
  • Memoryless
• Exercise:
  • #tk
  • Try to obtain the Exponential Distribution from the Gamma Distribution
  • Set $\alpha =1$ in the Gamma Distribution
• Memoryless Property

  	left=-0.50; right=5;
  	top=2; bottom=-0.5;
  	---
  	f(x)= e^{- x}
  	x=1
  	(1,0)|label:A
  	x=2
  	(2,0)|label:B
  	x=3
  	(3,0)|label:A+B
  

  • $P(X>a+b|X>a)=P(X>b)$
  • $=\frac{P(X>a+b\cap X>a)}{P(X>a)}$
  • $=\frac{P(X>a+b)}{P(X>a)}$
  • Complement
  • $=\frac{1-P(X\le a+b)}{1-P(X\le a)}$
  • $=\frac{1-F(a+b)}{1-F(a)}$
  • $=\frac{1-(1-e^{-\frac{a+b}{\beta }})}{1-(1-e^{-\frac{a}{\beta }})}$
  • $=\frac{e^{-\frac{a+b}{\beta }}}{e^{-\frac{a}{\beta }}}$
  • $=e^{-\frac{a+b}{\beta }-(-\frac{a}{\beta })}$
  • $=e^{-\frac{a+b}{\beta }+\frac{a}{\beta }}$
  • $=e^{-\frac{b}{\beta }}$
  • $P(X>x)=e^{-\frac{x}{\beta }}$
• Example:
  • Slide 39
  • Arrivals for a clinic is $\approx$ exponential
  • $E[X]=4$ minutes
  • $X\sim \exp (\beta =?)$
  • The mean will help us recover $\beta$
  • Since $E[X]=\beta$
  • $\beta =4$
  • $f_X(x)=\frac{1}{4}{e}^{-\frac{x}{4}}$
  • What's the time that arrival time is $<1$ minute

    	left=-0.50; right=5;
    	top=1; bottom=-0.5;
    	---
    	f(x)=1/4 e^{- x/4}
    	x<1|0<=y<=f(x)
    

    • $P(X<1)=\frac{1}{4}{\int }_0^1{e}^{-\frac{x}{4}}\text{differential}x$
    • $\frac{1}{4}{\int }_0^1{e}^{-\frac{x}{4}}\text{differential}x=1-e^{-\frac{1}{4}}$
    • ${[-\frac{\cancel{\frac{1}{4}}{e}^{-\frac{1}{4}x}}{\cancel{\frac{1}{4}}}]}_0^1$
    • $[-e^{-\frac{1}{4}x}]{}_0^1$
    • $-e^{-\frac{1}{4}}+1\approx 0.2212$
    • $F_X(x)=1-e^{-\frac{x}{4}}$
    • $P(X<1)=F(1)$
  • What's the probability that arrival times exceed 10 minutes

    	left=-0.50; right=15;
    	top=0.25; bottom=-0.1;
    	---
    	f(x)=1/4 e^{- x/4}
    	x>10|0<=y<=f(x)
    

    • Directly
      • $P(X>10)={\int }_{10}^{\mathrm{\infty }}f(x)\text{differential}x$
      • $P(X>10)={\int }_{10}^{\mathrm{\infty }}\frac{1}{4}{e}^{-\frac{x}{4}}\text{differential}x$
      • ${\int }_{10}^{\mathrm{\infty }}\frac{1}{4}{e}^{-\frac{x}{4}}\text{differential}x=e^{-\frac{5}{2}}$
      • ${[-e^{-\frac{x}{4}}]}_{10}^{\mathrm{\infty }}$
      • $\underset{x\to \mathrm{\infty }}{lim}[-e^{-\frac{x}{4}}]+e^{-2.5}$
      • $0+e^{-2.5}$
      • $e^{-2.5}$
    • Indirectly
      • Use CDF
      • $P(X>10)=1-P(X\le 10)$
      • $P(X>10)=1-F(10)$
  • Find the ${90}^{\text{th}}$ percentile $X_{90}$ of the arrival times

    	left=-0.50; right=15;
    	top=0.25; bottom=-0.1;
    	---
    	f(x)=1/4 e^{- x/4}
    	x<4\ln(10)|0<=y<=f(x)
    

    • $X_{90}$
    • $P(X\le X_{90})=F(X_{90})=0.9$
    • $F_X(X_{90})=0.9$
    • ${\int }_0^{x_{90}}\frac{1}{4}{e}^{-\frac{x}{4}}\text{differential}x=0.9$
    • ${\int }_0^{x_{90}}\frac{1}{4}{e}^{-\frac{x}{4}}\text{differential}x=1-e^{-\frac{x_{90}}{4}}$
    • $1-e^{-\frac{x_{90}}{4}}=0.9$
    • #tk exercise
    • Answer is $x_{90}=4\ln (10)$
• Example:
  • Wait times are exponential with $\beta =12$
  • Find the probability that someone has to wait at least $15$ minutes given they have waited at least $10$ minutes.
  • $X\ge 15|X\ge 10$
  • Probability Density Function
    • $f(x)=\frac{1}{12}{e}^{-\frac{1}{12}x}$
  • $P(X\ge 15|X\ge 10)$
    • Directly
      • $\frac{P(X\ge 15\cap X\ge 10)}{P(X\ge 10)}$
      • $\frac{P(X\ge 15)}{P(X\ge 10)}$
    • Memoryless property
      • $P(X\ge a+b|X>a)=P(X>b)$
      • $P(X\ge 15|X\ge 10)$
        • $15=a+b$
        • $10=a$
        • $15=10+b$
        • $5=b$
      • $P(X>5)={\int }_0^5\frac{1}{12}{e}^{-\frac{1}{12}x}\text{differential}x$
      • $=\frac{1}{12}{\int }_0^5{e}^{-\frac{1}{12}x}\text{differential}x=1-e^{-\frac{5}{12}}$