]>
The strong renewal assumption means that the Poisson process must probabilistically restart at a fixed time . In particular, if the first arrival has not occurred by time , then the time remaining until the arrival occurs must have the same distribution as the first arrival time itself. This is known as the memoryless property and can be stated in terms of a generic interarrival time as follows: the conditional distribution of given is the same as the distribution of . Equivalently,
Let denote the right-tail distribution function of , so that for .
Show that the memoryless property is equivalent to the law of exponents:
Show that the only solutions of the functional equation in Exercise 1, which are continuous from the right, are exponential functions. Let . Successively show that
In the context of Exercise 2, let . Then (since ) so
Hence has a continuous distribution with cumulative distribution function given by
Show that the probability density function of is
A random variable with this probability density function is said to have the exponential distribution with rate parameter . The reciprocal is known as the scale parameter.
Show directly that the exponential probability density function is a valid probability density function.
In the gamma experiment, set so that the simulated random variable has an exponential distribution. Vary with the scroll bar and watch how the shape of the probability density function changes. For selected values of , run the experiment 1000 times with an update frequency of 10, and watch the apparent convergence of the empirical density function to the probability density function.
Show that the quantile function of is
In particular, show that
Recall that if a nonnegative random variable with a continuous distribution is interpreted as the lifetime of a device, then the failure rate function is
where, as usual, denotes the probability density function and the cumulative distribution function.
Show that the exponential distribution with rate parameter has constant failure rate , and is the only such distribution.
The following exercises give the mean, variance, and moment generating function of the exponential distribution.
Show that .
Show that .
Show that for .
In the context of the Poisson process, the parameter is known as the rate of the process. On average, there are time units between arrivals, so the arrivals come at an average rate of per unit time.
Note also that the mean and standard deviation are equal for an exponential distribution, and that the median is always smaller than the mean.
In the gamma experiment, set so that the simulated random variable has an exponential distribution. Vary with the scroll bar and watch how the mean/standard deviation bar changes. Now set , run the experiment 1000 times with an update frequency of 10, and watch the apparent convergence of the empirical mean and standard deviation to the distribution mean and standard deviation, respectively.
Show that for where is the gamma function. In particular, if .
The exponential distribution has an amazing number of interesting mathematical properties; some of these properties are satisfied only by the exponential distribution, and thus serve as characterizations.
Suppose that has the exponential distribution with rate parameter . Show that the following random variables have geometric distributions on and on , respectively, each with parameter .
In many respects, the geometric distribution is a discrete version of the exponential distribution.
Suppose that and have exponential distributions with parameters and , respectively, and are independent. Show that
Suppose that is a sequence of independent random variables, and that has the exponential distribution with rate parameter for each .
Note that the minimum in part (a) has the exponential distribution with parameter . In the context of reliability, if a series system has independent components, each with an exponentially distributed lifetime, then the lifetime of the system is also exponentially distributed, and the failure rate of the system is the sum of the component failure rates. In the context of random processes, if we have independent Poisson process, then the new process obtained by combining the random times is also Poisson, and the rate of the new process is the sum of the rates of the individual processes (we will return to this point latter).
In the order statistic experiment, select the exponential distribution.
We can now generalize Exercise 15:
In the setting of Exercise 16, show that for ,
The results in Exercise 16 and Exercise 18 are very important in the theory of continuous-time Markov chains. Suppose that
is the time until an event of interest occurs (the arrival of a customer, the failure of a device, etc.) for each
; these times are independent and exponentially distributed. Then the first time
that one of the events occurs is also exponentially distributed (Exercise 16 (a)), and the probability that the first event to occur is event
is proportional to the rate
.
The next exercise gives a randomized
version of the memoryless property:
Suppose that
Consider again the setting of Exercise 16. Show that
Of course, the probabilities of other orderings can be computed by permuting the parameters appropriately in the formula on the right.
Suppose that the length of a telephone call (in minutes) is exponentially distributed with rate parameter
Suppose that the lifetime of a certain electronic component (in hours) is exponentially distributed with rate parameter
Suppose that the time between requests to a web server (in seconds) is exponentially distributed with rate parameter 2.
Suppose that the lifetime
The position