28  Exponential Distributions

• Exponential distributions are often used to model the waiting times between events in a random process that occurs continuously over time.

Example 28.1 Suppose that we model the time, measured continuously in hours, from now until the next earthquake (of any magnitude) occurs in southern CA as a continuous random variable \(X\) with pdf

\[ f_X(x) = 2 e^{-2x}, \; x \ge0 \]

1. Sketch the pdf of \(X\). What does this tell you about waiting times?
   
   
   
2. Without doing any integration, approximate the probability that \(X\) rounded to the nearest minute is 0.5 hours.
   
   
   
3. Compute and interpret \(\text{P}(X \le 3)\).
   
   
   
4. Find the cdf of \(X\).
   
   
   
5. Find the median time between earthquakes.
   
   
   
6. Set up the integral you would solve to find \(\text{E}(X)\). Interpret \(\text{E}(X)=1/2\). How does the median compare to the mean? Why?
   
   
   
7. Set up the integral you would solve to find \(\text{E}(X^2)\).
   
   
   
8. Find \(\text{Var}(X)\) and \(\text{SD}(X)\).
   
   
   

• A continuous random variable \(X\) has an Exponential distribution with rate parameter \(\lambda>0\) if its pdf is

\[ f_X(x) = \begin{cases}\lambda e^{-\lambda x}, & x \ge 0,\\ 0, & \text{otherwise} \end{cases} \]

• If \(X\) has an Exponential(\(\lambda\)) distribution then \[\begin{align*} \text{cdf:} \quad F_X(x) & = 1-e^{-\lambda x}, \quad x\ge 0\\ \text{P}(X>x) & = e^{-\lambda x}, \quad x\ge 0\\ \text{E}(X) & = \frac{1}{\lambda}\\ \text{Var}(X) & = \frac{1}{\lambda^2}\\ \text{SD}(X) & = \frac{1}{\lambda} \end{align*}\]
• Exponential distributions are often used to model the waiting time in a random process until some event occurs.
  • \(\lambda\) is the average rate at which events occur over time (e.g., 2 per hour)
  • \(1/\lambda\) is the mean time between events (e.g., 1/2 hour)
• If \(X\) has an Exponential(\(\lambda)\) distribution and \(a>0\) is a constant, then \(aX\) has an Exponential(\(\lambda/a\)) distribution.
  • If \(X\) is measured in hours with rate \(\lambda = 2\) per hour and mean 1/2 hour
  • Then \(60X\) is measured in minutes with rate \(2/60\) per minute and mean \(60(1/2)=30\) minutes.
• If \(X\) has an Exponential(1) distribution and \(\lambda>0\) is a constant then \(X/\lambda\) has an Exponential(\(\lambda\)) distribution.
• If \(U\) has a Uniform(0, 1) distribution then \(-\log(U)\) has Exponential(1) distribution. (\(-\log(1-U)\) also has an Exponential(1) distribution.)

Example 28.2 \(X\) and \(Y\) are continuous random variables with joint pdf

\[ f_{X, Y}(x, y) = \frac{1}{3x^2}\exp\left(-\left(\frac{y}{x^2} + \frac{x}{3}\right)\right), \qquad x > 0, y>0 \]

1. Identify by name the marginal distribution and one-way conditional distributions that you can obtain from the joint pdf without doing any calculus.
   
   
   
   
   
2. How could you use an an Exponential(1) spinner to simulate \((X, Y)\) pairs with this joint distribution?
   
   
   
   
   
3. Sketch a plot of the joint pdf.
   
   
   
   
   
4. Find \(\text{E}(Y)\) without doing any calculus.
   
   
   
   
   
5. Find \(\text{Cov}(X, Y)\) without doing any calculus. (Well, without doing any multivariable calculus.)
   
   
   
   
   
6. Use simulation to approximate the distribution of \(Y\). Does \(Y\) have an Exponential distribution?
   
   
   
   
   

28.1 Memoryless property

Example 28.3

Let \(X\) be the waiting time (hours) until the next earthquake and assume \(X\) has an Exponential(2) distribution.

1. Find the probability that the waiting time is greater than 1 hour.
   
   
   
   
   
2. Suppose that no earthquakes occur in the next 3 hours. Find the conditional probability that the waiting time (from now) is greater than 4 hours. Be sure to write a valid probability statement involving \(X\) before computing. What do you notice?
   
   
   
   
   

• Memoryless property. If \(W\) has an Exponential(\(\lambda\)) distribution then for any \(w,h>0\)

\[ \text{P}(W>w+h\,\vert\, W>w) = \text{P}(W>h) \]

• Given that we have already waited \(w\) units of time, the conditional probability that we wait at least an additional \(h\) units of time is the same as the unconditional probability that we wait at least \(h\) units of time.
• A continuous random variable \(W\) has the memoryless property if and only if \(W\) has an Exponential distribution. That is, Exponential distributions are the only continuous¹ distributions with the memoryless property.
• If \(W\) has an Exponential(\(\lambda\)) distribution then the conditional distribution of the excess waiting time, \(W - w\), given \(\{W>w\}\) is Exponential(\(\lambda\)).

28.2 Exponential race

Example 28.4

Xiomara and Rogelio each leave work at noon to meet the other for lunch. The amount of time, \(X\), that it takes Xiomara to arrive is a random variable with an Exponential distribution with mean 10 minutes. The amount of time, \(Y\), that it takes Rogelio to arrive is a random variable with an Exponential distribution with mean 20 minutes. Assume that \(X\) and \(Y\) are independent. Let \(W\) be the time, in minutes after noon, at which the first person arrives.

1. What is the relationship between \(W\) and \(X, Y\)?
   
   
   
   
   
2. Find and interpret \(\text{P}(W>40)\).
   
   
   
   
   
3. Find \(\text{P}(W > w)\) and identify by name the distribution of \(W\).
   
   
   
   
   
4. Find \(\text{E}(W)\). Is it equal to \(\min(\text{E}(X), \text{E}(Y))\)?
   
   
   
   
   
5. Is \(\text{P}(Y>X)\) greater than or less than 0.5? Make an educated guess for \(\text{P}(Y > X)\).
   
   
   
   
   
6. Find \(\text{P}(Y>X|X=10)\).
   
   
   
   
   
7. Find \(\text{P}(Y>X|X=x)\) for general \(x\).
   
   
   
   
   
8. Find \(\text{P}(Y > X)\). Hint: Use a continuous version of the law of total probability.
   
   
   
   
   
9. Find \(\text{P}(W>40, Y>X)\).
   
   
   
   
   
10. Find \(\text{P}(W>w, Y>X)\) for general \(w\).
    
    
    
    
    
11. Let \(\text{I}_{\{Y > X\}}\) be the indicator that Xiomara arrives first. What can you say about \(W\) and \(\text{I}_{\{Y > X\}}\)?
    
    
    
    
    

• Exponential race (a.k.a., competing risks.) Let \(W_1, W_2, \ldots, W_n\) be independent random variables. Suppose \(W_i\) has an Exponential distribution with rate parameter \(\lambda_i\). Let \(W = \min(W_1, \ldots, W_n)\) and let \(I\) be the discrete random variable which takes value \(i\) when \(W=W_i\), for \(i=1, \ldots, n\). Then
  • \(W\) has an Exponential distribution with rate \(\lambda = \lambda_1 + \cdots+\lambda_n\)
  • \(\text{P}(I=i) = \text{P}(W=W_i) = \frac{\lambda_i}{\lambda_1+\cdots+\lambda_n}, i = 1, \ldots, n\)
  • \(W\) and \(I\) are independent
• Imagine there are \(n\) contestants in a race, labeled \(1, \ldots, n\), racing independently, and \(W_i\) is the time it takes for the \(i\)th contestant to finish the race. Then \(W = \min(W_1, \ldots, W_n)\) is the winning time of the race, and \(W\) has an Exponential distribution with rate parameter equal to sum of the individual contestant rate parameters.
• The discrete random variable \(I\) is the label of which contestant is the winner. The probability that a particular contestant is the winner is the contestant’s rate divided by the total rate. That is, the probability that contestant \(i\) is the winner is proportional to the contestant’s rate \(\lambda_i\).
• Furthermore, \(W\) and \(I\) are independent. Information about the winning time does not influence the probability that a particular contestant won. Information about which contestant won does not influence the distribution of the winning time.

28.3 Gamma distributions

Example 28.5

Suppose that elapsed times (hours) between successive earthquakes are independent, each having an Exponential(2) distribution. Let \(T\) be the time elapsed from now until the third earthquake occurs.

1. Compute \(\text{E}(T)\).
   
   
   
   
   
2. Compute \(\text{SD}(T)\).
   
   
   
   
   
3. Does \(T\) have an Exponential distribution? Explain.
   
   
   
   
   
4. Use simulation to approximate the distribution of \(T\).
   
   
   
   
   

• A continuous random variable \(X\) has a Gamma distribution with shape parameter \(\alpha\), a positive integer², and rate parameter³ \(\lambda>0\) if its pdf is \[\begin{align*} f_X(x) & = \frac{\lambda^\alpha}{(\alpha-1)!} x^{\alpha-1}e^{-\lambda x} , & x \ge 0,\\ & \propto x^{\alpha-1}e^{-\lambda x} , & x \ge 0 \end{align*}\] If \(X\) has a Gamma(\(\alpha\), \(\lambda\)) distribution then \[\begin{align*} \text{E}(X) & = \frac{\alpha}{\lambda}\\ \text{Var}(X) & = \frac{\alpha}{\lambda^2}\\ \text{SD}(X) & = \frac{\sqrt{\alpha}}{\lambda} \end{align*}\]
• If \(W_1, \ldots, W_n\) are independent and each \(W_i\) has an Exponential(\(\lambda\)) distribution then \((W_1+\cdots+W_n)\) has a Gamma(\(n\),\(\lambda\)) distribution.
• Each \(W_i\) represents the waiting time between two occurrences of some event, so \(W_1 + \cdots+ W_n\) represents the total waiting time until a total of \(n\) occurrences.
• Exponential distributions are continuous analogs of Geometric distributions, and Gamma distributions are continuous analogs of Negative Binomial distributions.
• For a positive integer \(d\), the Gamma(\(d/2, 1/2\)) distribution is also known as the chi-square distribution with \(d\) degrees of freedom.

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1. Geometric distributions are the only discrete distributions with the discrete analog of the memoryless property.

2. There is a more general expression of the pdf which replaces \((\alpha-1)!\) with the Gamma function \(\Gamma(\alpha)=\int_0^\infty u^{\alpha-1}e^{-u} du\), that can be used to define a Gamma pdf for any \(\alpha>0\). When \(\alpha\) is a positive integer, \(\Gamma(\alpha)=(\alpha-1)!\).

3. Like Exponential distributions, Gamma distributions are sometimes parametrized directly by their mean \(1/\lambda\), instead of the rate parameter \(\lambda\). The mean \(1/\lambda\) is called the scale parameter.