Gamma and Exponential Distributions

• Exponential is a special case of the gamma distribution.
• Play an important role in queuing theory and reliability problems.
• Time between arrivals at service facilities, time to failure of component parts and electrical systems.
• Definition 6.2:
  
• also
  $$\Gamma(n+1)=n\Gamma(n)=n!$$
  $$\Gamma(1/2)=\sqrt{\pi}~exception$$

• Gamma Distribution: The continuous random variable $X$ has a gamma distribution, with parameters $\alpha$ and $\beta$,
• If its density function is given by
  $$f(x)=\left\lbrace \begin{array}{lr} \frac{1}{\beta^\alpha\... ...}e^{-x/\beta}, & x>0 \\ 0,& elsewhere\\ \end{array}\right.$$
  where $\alpha>0$ and $\beta>0$
• The mean and variance of the gamma distribution are (Proof is in Appendix A.28)
  $$\mu=\alpha \beta and \sigma^2=\alpha \beta^2$$

Figure 6.26: Gamma Distributions.

• Exponential Distribution ($\alpha=1$, special gamma distribution): The continuous random variable $X$ has an exponential distribution, with parameters $\beta$,
• In real life, we observe the lifetime of certain products decreased as time goes.
• To model life-lengths, especially the exponential curve seemed be good to fit these data rather well.
• If its density function is given by
  $$f(x)=\left\lbrace \begin{array}{lr} \frac{1}{\beta}e^{-x/\beta}, & x>0 \\ 0,& elsewhere\\ \end{array}\right.$$
  where $\beta>0$
• The mean and variance of the exponential distribution are
  $$\mu=\beta~and~\sigma^2= \beta^2$$

• The exponential distribution has a single tail. The single parameter $\beta$ determines the shape of the distribution.

• Relationship to the Poisson Process: The most important applications of the exponential distribution are situations where the Poisson process applies.
• An industrial engineer may be interested in modeling the time $T$ between arrivals at a congested interaction during rush hour in a large city. An arrival represents the Poisson event.
• Using Poisson distribution, the probability of no events occurring in the span up to time $t$
  $$p(0,\lambda t)=\frac{e^{-\lambda t}(\lambda t)^0}{0!}=e^{-\lambda t}$$

• Let $X$ be the time to the first Poisson event.
• The probability that the length of time until the first event will exceed $x$ is the same as the probability that no Poisson events will occur in $x$.
  $$P(X\geq x)=e^{-\lambda x} \Rightarrow P(0 \leq X \leq x)=1-e^{-\lambda x}$$

• Differentiate the cumulative distribution function for the exponential distribution
  $$f(x)=\lambda e^{-\lambda x} ~with~ \lambda=1/\beta$$

• Applications of Gamma and Exponential Distributions
• The mean of the exponential distribution is the parameter $\beta$, the reciprocal of the parameter in the Poisson distribution.
• Poisson distribution has no memory, implying that occurrences in successive time periods are independent. They immediately ``forget'' their past behavior.
• The important parameter $\beta$ is the mean time between events.
• The equipment failure often conforms to this Poisson process, $\beta$ is called mean time between failures.
• Many equipment breakdowns do follow the Poisson process, and thus the exponential distribution does apply.
• Other applications include survival times in bio-medical experiments and computer response time.

• Example 6.17: Suppose that a system contains a certain type of component whose time in years to failure is given by $T$.
• The random variable $T$ is modeled nicely by the exponential distribution with mean time to failure $\beta = 5$.
• Solution:
  $$P(T>8)=\frac{1}{5}\int_8^{\infty} e^{-t/5}dt=e^{-8/5} \approx 0.2$$

  Let $X$ represent the number of components functioning after 8 years.

• If 5 of these components are installed in different systems, what is the probability that at least 2 are still functioning at the end of 8 years.
• Solution:
  $$P(X \geq 2)=\sum_{x=2}^{5} b(x;5,0.2)=1-\sum_{x=0}^{1} b(x;5,0.2)$$
  $$=1-0.7373=0.2627$$

• Example 6.18: Suppose that telephone calls arriving at a switchboard follow a Poisson process with an average of 5 calls coming per minute.
• What is the probability that up to a minute will elapse until 2 calls have come in to the switchboard?
• Solution:

  The Poisson process applies with time until 2 Poisson events following a gamma distribution with $\beta=1/5$ and $\alpha=2$.
  Let represent the time in minutes that transpires before 2 calls come.

  $$P(X \leq x)=\int_0^{x} \frac{1}{\beta^2} x e^{-x/\beta}dx$$
  $$P(X \leq 1)=25\int_0^{1} x e^{-5x}dx$$
  $$=1-e^{-5*1}(1+5)=0.96$$

• Example 6.19: In a biomedical study with rats a dose-response investigation is used to determine the effect of the dose of a toxicant on their survival time.
• For a certain dose of the toxicant the study determines that the survival time, in weeks, has a gamma distribution with $\alpha= 5$ and $\beta=10$.
• What is the probability that a rat survives no longer than 60 weeks?
• Solution:

  Let $X$ be the survival time

  $$P(X \leq x)=\int_0^{x} \frac{1}{\beta^\alpha \Gamma(\alpha)} x^{\alpha-1} e^{-x/\beta}dx$$
  $$P(X \leq 60)=\frac{1}{\beta^5} \int_0^{60} \frac{x^{\alpha-1} e^{-x/\beta}}{\Gamma(5)} dx$$

  Using incomplete gamma function $F(x;\alpha)=\int_0^{x} \frac{y^{\alpha-1}e^{-y}}{\Gamma(\alpha)} dy$
  Let $y=x/\beta$, and $x=\beta y$

  $$\Rightarrow P(X \leq 60)=\int_0^{6} \frac{y^{4} e^{-y}}{\Gamma(5)} dy$$
  $$=F(6;5)=0.715,~see~Appendix~A.24$$