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MANE 3332.05

Lecture 13

Agenda

  • Continue Chapter 4 lectures - Exponential and Weibull Distributions
  • Normal Practice Problems (assigned 10/9/2025, due 10/14/2025)
  • Normal Quiz (assigned 10/14/2025, due 10/16/2026)
  • Exponential Practice Problems (assigned 10/14/2025, due 10/16/2025)
  • Schedule

Handouts

Tuesday Date and Topic(s) Thursday Date and Topic(s)
10/14: Exponential and Weibull distributions 10/16: Chapter 5 (not on midterm)
**10/21: ** Midterm Review 10/23: Midterm Exam

Exponential Distribution

  • The exponential distribution is widely used in the area of reliability and life-test data.

  • Ostle, et. al. (1996) list the following applications of the exponential distribution

    • the number of feet between two consecutive erroneous records on a computer tape,

    • the lifetime of a component of a particular device,

    • the length of a life of a radioactive material and

    • the time to the next customer service call at a service desk

Exponential Distribution

  • The PDF for an exponential distribution with parameter \(\lambda >0\) is
\[ f(x)=\lambda e^{-\lambda x},\;\;\mbox{ for }0\leq x < \infty \]
  • The mean of \(X\) is
\[ \mu=E(X)=\frac{1}{\lambda} \]
  • The variance of \(X\) is
\[ \sigma^2=V(X)=\frac{1}{\lambda^2} \]

Note that other authors define \(f(x)=\frac{1}{\theta}e^{-x/\theta}\). Either definition is acceptable. However one must be aware of which definition is being used.

The Exponential CDF

The CDF for the exponential distribution is easy to derive

\[ \begin{aligned} F(x)&=&P(X\leq x)=\int_{-\infty}^x\lambda e^{-\lambda y}\,dy\\ &=&\int_{0}^x\lambda e^{-\lambda y}\,dy\\ &=&\left.\left(-e^{-\lambda y}\right)\right|_{y=0}^x\\ &=& -e^{-\lambda x}-(-e^0)\\ &=&-e^{-\lambda x}+1\\ &=&1-e^{-\lambda x}\end{aligned} \]

Problem 4-79

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Lack of Memory Property

  • The mathematical definition is
\[ P(X<t_1+t_2|X>t_1)=P(X<t_2) \]

  • That is "the probability of a failure time that is less than \(t_1+t_2\) given the failure time is greater than \(t_1\) is the probability that the item's failure time is less than \(t_2\)

  • This property is unique to the exponential distribution

  • Often used to model the reliability of electronic components.

Problem 4--80

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Relationship to the Poisson Distribution

  • Let \(Y\) be a Poisson random variable with parameter \(\lambda\). Note: \(Y\) represents the number of occurrences per unit

  • Let \(X\) be a random variable that records the time between occurrences for the same process as \(Y\)

  • \(X\) has an exponential distribution with parameter \(\lambda\)

Lognormal Distribution

  • Let \(W\) have a normal distribution with mean \(\theta\) and variance \(\omega^2\); then \(X=\exp(W)\) is a lognormal random variable with pdf
\[ f(x)=\frac{1}{x\omega\sqrt{2\pi}}\exp\left[-\frac{\left(\ln(x)-\theta\right)^2}{2\omega^2}\right]\;\;0<x<\infty \]
  • The mean of \(X\) is
\[ E(X)=e^{\theta+\omega^2/2} \]
  • The variance of \(X\) is
\[ V(X)=e^{2\theta+\omega^2}\left(e^{\omega^2}-1\right) \]

Example Problem

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Gamma Distribution

  • The random variable \(X\) with pdf
\[ f(x)=\frac{\lambda^rx^{r-1}e^{-\lambda x}}{\Gamma(r)},\;\;\mbox{for }x>0 \]

is a gamma random variable with parameters \(\lambda>0\) and \(r>0\).

  • The gamma function is
\[ \Gamma(r)=\int_0^\infty x^{r-1}e^{-x}\,dx\;\;\mbox{for }r>0 \]

with special properties:

  • \(\Gamma(r)\) is finite

  • \(\Gamma(r)=(r-1)\Gamma(r-1)\)

  • For any positive integer \(r\), \(\Gamma(r)=(r-1)!\)

  • \(\Gamma(1/2)=\pi^{1/2}\)

Gamma Distribution

  • The mean and variance are
\[ \mu=E(X)=r/\lambda\mbox{ and }\sigma^2=V(X)=r/\lambda^2 \]
  • We will not work any probability problems using the gamma distribution

Gamma Tables

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Weibull Distribution

  • The random variable \(X\) with pdf

$$ f(x)=\frac{\beta}{\delta}\left(\frac{x}{\delta}\right)^{\beta-1}\exp\left[-\left(\frac{x}{\delta}\right)^\beta\right],\;\; \mbox{ for }x>0 $$ is a Weibull random variable with scale parameter \(\delta>0\) and shape parameter \(\beta>0\)

  • The CDF for the Weibull distribution is
\[ F(x)=1-\exp\left[-\left(\frac{x}{\delta}\right)^\beta\right] \]
  • The mean of the Weibull distribution is
\[ \mu=E(X)=\delta\Gamma\left(1+\frac{1}{\beta}\right) \]
  • The variance of the Weibull distribution is
\[ \sigma^2=V(X)=\delta^2\Gamma\left(1+\frac{2}{\beta}\right)-\delta^2\left[\Gamma\left(1+\frac{1}{\beta}\right)\right]^2 \]

Weibull Problem

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