Question

For an interarrival distribution \(F\) having mean \(\mu\), we defined the equilibrium distribution of \(F\), denoted \(F_{e}\), by $$ F_{e}(x)=\frac{1}{\mu} \int_{0}^{x}[1-F(y)] d y $$ (a) Show that if \(F\) is an exponential distribution, then \(F=F_{e}\). (b) If for some constant \(c\), $$ F(x)=\left\\{\begin{array}{ll} 0, & x

Short answer

In summary: (a) For the exponential distribution, we proved that \(F(x) = F_e(x) = 1 - e^{-\lambda x}\). (b) For a constant interarrival function with a cutoff at \(c\), we showed that the equilibrium distribution is the uniform distribution on the interval \((0, c)\), with \(F_e(x) = \frac{x}{c}\). (c) In the City of Berkeley parking scenario, the probability of getting a parking ticket if you return after 3 hours is 1, meaning you will definitely receive a ticket.

Step-by-step solution

(a) Proving F(x) = Fe(x) for exponential distribution

Given the equilibrium distribution of \(F_e\): \[ F_{e}(x)=\frac{1}{\mu} \int_{0}^{x}[1-F(y)] d y \] Let \(F(x) = 1 - e^{-\lambda x}\) be the exponential distribution function. Now, we need to find \(F_e(x)\) using the equilibrium distribution formula: \[ F_{e}(x)=\frac{1}{\mu} \int_{0}^{x}[1-(1 - e^{-\lambda y})] d y \] Since the mean of an exponential distribution is given by \(\mu = \frac{1}{\lambda}\), the equation becomes: \[ F_{e}(x)=\lambda \int_{0}^{x} e^{-\lambda y} d y \] Integrating the above equation, we have: \[ F_{e}(x)=-e^{-\lambda y}\bigg]_0^x = -e^{-\lambda x} + 1 \] Thus, we have shown that: \[ F(x)=F_{e}(x) = 1 - e^{-\lambda x} \]

(b) Proving Fe(x) is a uniform distribution on (0, c) for F(x) as a constant function

Given the constant interarrival distribution function \(F(x)\): \[ F(x)=\left\{\begin{array}{ll} 0, & x<c \\ 1, & x \geqslant c \end{array}\right. \] Now, we need to find \(F_e(x)\) using the equilibrium distribution formula. First, notice that the mean of \(F(x)\) is equal to the constant value \(c\), so we can write: \[ F_{e}(x)=\frac{1}{c} \int_{0}^{x}[1-F(y)] d y \] Now, we must break the integral into two parts. For the interval \(0 \leq y < c\), \(F(y) = 0\), and for the interval \(y \geq c\), \(F(y) = 1\). Thus: \[ F_{e}(x)=\frac{1}{c} \left[\int_{0}^{x}[1-0] d y + \int_{x}^{c}[1-1] d y\right] = \frac{1}{c} \left[\int_{0}^{x} 1 d y\right] \] Integrating the above equation, we get: \[ F_{e}(x)=\frac{1}{c} \left[y\bigg]_0^x\right] = \frac{x}{c} \] This is the equation for a uniform distribution on the interval \((0, c)\), thus proving the claim.

(c) Calculating the probability of getting a parking ticket in Berkeley

If you park your car in Berkeley and return after 3 hours, you can receive a ticket if and only if a parking official passed by after you parked your car but before 2 hours had passed. Assuming that the probability of an official passing by between you parking and exactly 2 hours later follows a uniform distribution on the interval \((0, 2)\), we can calculate the probability of getting a ticket: Let T be the time at which a parking official visits your car. The probability density function of T is given by: \[ f_T(t)=\left\{\begin{array}{ll} \frac{1}{2}, & 0 \leq t < 2 \\ 0, & t \geq 2 \end{array}\right. \] Now, we need to calculate the probability that a parking official visited your car before the 2 hours mark, i.e., we need to find \(\text{P}(T<2)\): \[ \text{P}(T<2) = \int_{0}^{2} f_T(t) dt = \int_{0}^{2} \frac{1}{2} dt = \left[\frac{t}{2}\bigg]_0^2 = 1 \] Since the probability of an official passing by before 2 hours is 1, you will definitely get a ticket if you return after 3 hours.

Key concepts

Exponential Distribution

The exponential distribution is a probability distribution that describes time between events in a Poisson process. This is a situation where events occur continuously and independently at a constant average rate. It's characterized by its rate parameter \( \lambda \), which is the average number of events per unit time. The exponential distribution has the unique property of being memoryless, meaning that the probability of an event occurring in the next time period is independent of how long it's been since the last event.

The exponential distribution is defined by the probability density function \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \) and \( f(x) = 0 \) otherwise. The mean or expected value of an exponential distribution is given by \( \mu = \frac{1}{\lambda} \). What makes the exponential distribution particularly interesting in the study of equilibrium distributions is that it is equal to its own equilibrium distribution, as shown in the exercise. This self-equilibrium property is a reflection of the memoryless feature mentioned earlier.

An Example in Real-Life Scenarios

Imagine waiting for a bus that arrives on average every 10 minutes; the time you'll wait for the next bus is most likely to follow an exponential distribution.

Uniform Distribution

A uniform distribution, in the context of probability models, represents a situation where all outcomes are equally likely. For a continuous uniform distribution, any value within the lower bound \( a \) and upper bound \( b \) has equal probability of occurring. It is represented by the probability density function \( f(x) = \frac{1}{b - a} \) for \( a \leq x \leq b \) and \( f(x) = 0 \) elsewhere.

The uniform distribution is often used to model situations where there is no preference for any particular outcome or when information is lacking to give a more precise distribution. An interesting aspect of the uniform distribution is its relation to other distributions in certain contexts, like when a distribution with a constant delay leads to a uniform equilibrium distribution, as in the textbook exercise.

Illustrating the Uniform Distribution

A practical example could be a lottery draw, where every ticket has an equally likely chance to be drawn. Similarly, the problem in our exercise models the interarrival times as identically equal to a constant, resulting in an equilibrium distribution that is uniform.

Probability Models

Probability models are mathematical representations of complex stochastic, or random, processes. They describe the likely outcomes of these processes and the probabilities associated with these outcomes. Models help statisticians and researchers understand the underlying dynamics of real-world phenomena by making assumptions and using known probability distributions, such as the exponential or uniform distributions described earlier.

These models are built upon a foundation of axioms and principles that govern probability theory, such as the rules of probabilities being between 0 and 1, the total probability summing to 1, and the treatment of independent and mutually exclusive events. When reality fits neatly into a model's assumptions, the model can be used to make accurate predictions, calculate risk, and inform decision-making.

Applying Probability Models

The exercise involving the Berkeley parking ticket scenario leverages probability models to predict the likelihood of receiving a parking ticket. Such models are not only academic exercises but are also applied in various fields, including finance, insurance, logistics, and even public policy to manage resources, evaluate risk, and optimize strategies.