A detector is tested over and over, and on each trial it either reports a signal or stays quiet. Two natural questions concern the same run of trials. How many signals appear among the first four trials, and on which trial does the first signal appear? This lesson models the run as a Bernoulli process, answers the first question with a binomial count and the second with a geometric waiting time, and shows how events about the waiting time translate into events about counts.
Key terms
Random process
Definition A random process is a collection of random variables , one for each value of an index in an index set , all defined on the same sample space with the same probability . The index is called the time index. In discrete time the index set is a set of integers such as , and in continuous time it is an interval such as . For each fixed the value is one random variable. For each fixed outcome in , the values , viewed as a function of , form a sample path of the process.
In plain words A random process is a whole family of random quantities studied together, so a question can involve several times at once, such as how a later value relates to earlier ones. The time index puts the variables in order. It can count trials, days or seconds.
Example A detector is tested on trials , and if trial reports a signal and otherwise. The sequence is a random process whose time index counts trials. The running totals of the same results form a second process on the same sample space. If the first five results are , the running totals after each of those trials are .
Bernoulli process
Definition Let . A Bernoulli process with success probability is a sequence of random variables that are mutually independent and each have the distribution, so and for every . Together these conditions say that for every and every choice of in ,
where is the number of ones. The outcome coded is called a success and the outcome coded a failure.
In plain words Every trial has the same chance of success, and no group of trials carries information about the other trials. The product formula needs independence of every finite group of trials, not only of pairs. Success is just the name of the outcome coded , and it need not be a good outcome.
Example For a detector with , the results no signal, signal, signal, no signal on the first four trials have probability . If a signal on one trial made a signal on the next trial more likely, or if the detector lost sensitivity as the trials went on, the results would not form a Bernoulli process, and the formulas on this page would not be justified for them.
Cumulative count
Definition In a Bernoulli process with success probability , the cumulative count after trials is
the number of successes among trials to , with . For each fixed the count has the binomial distribution , so for
with and . The sequence is itself a random process. It never decreases, and from one trial to the next it rises by or by .
In plain words The cumulative count is a running total of successes, and each count includes all earlier trials. The binomial coefficient counts the ways to place successes among ordered trials, and each placement has the same probability. The count and the relative frequency are different quantities on different scales.
Example With and four trials, there are placements of two signals, so
No signals has probability , so at least one signal has probability . The expected count is , which is not a possible count, and the variance is . The relative frequency has mean .
First success time
Definition In a Bernoulli process with success probability , the first success time is
the number of the first trial that succeeds. The event requires failures on trials to and a success on trial , so by independence
for . This is the geometric distribution on , and . The event that no trial ever succeeds is contained in for every , so its probability is at most for every . It is therefore , and is finite with probability . Some books record instead the number of failures before the first success. That variable is , it takes the values , and its mean is .
In plain words Count the trials up to and including the first success. For , a first success on trial and a first success by trial are different events. The first puts the earliest success exactly at trial , while the second allows it at any of the trials to .
Example With , the first signal comes on trial with probability . It comes by trial unless the first three trials all fail, so . The mean waiting time is trials, and the mean number of failures before the first signal is . The two conventions differ by exactly one trial, so state which one is in use before quoting a mean or a formula.
Count and waiting time events
Definition For every , the first success time and the cumulative count satisfy the event identities
so and . Also . For integers the conditioning event has probability , and since is contained in ,
This equals , which is the memoryless property of the geometric waiting time.
In plain words A count question fixes a number of trials and asks how many successes occur. A waiting question fixes a number of successes, here one, and asks when it occurs. Waiting longer than trials is the same event as having no success in the first trials. So every event about the first success time can be written with counts, while a count event such as is not determined by the first success time alone. Memorylessness says that after a run of failures the remaining wait has the same distribution as a fresh start, so in a Bernoulli process a success is never overdue. It rests on the independence of the trials and the constant .
Example With , the first signal comes after trial exactly when the first four trials have no signal, which has probability . Given that the first five trials all failed, the chance that the wait lasts more than three further trials is
the same as for a fresh start. If the chance of a signal grew after each quiet trial, the trials would not form a Bernoulli process, and the remaining wait would depend on how long the wait had already lasted.
Disjoint blocks of trials
Definition A block of trials is a set of trial numbers, and blocks are disjoint when no trial belongs to two of them. In a Bernoulli process, the counts of successes in disjoint blocks are mutually independent, because each count is a function of the trials in its own block and the trials are mutually independent. A block of trials has a count wherever it lies. Cumulative counts overlap instead. For ,
where and the count of trials to are independent. The cumulative counts and themselves are dependent, since
In plain words Counts that use different trials carry no information about each other. A cumulative count includes every earlier trial, so two cumulative counts share their common trials, and knowing the earlier count changes the probabilities for the later one. Independence of the trials does not carry over to every quantity built from them.
Example With , the counts in trials to and in trials to are independent, and each is . Both are zero with probability , which matches computed directly. The cumulative counts and are dependent. Four signals in the first four trials has probability , and given that event, the event is impossible, although . Their covariance is .
Check yourself
- A Bernoulli process has success probability . Find the probability of no successes in the first five trials, the probability that the first success comes on trial , and the probability that the first success comes by trial . Which of these events are stated with the count and which with the waiting time ?
Show the answer
No successes in five trials is the count event , with probability . A first success on trial needs four failures followed by a success, so . The first success comes by trial exactly when at least one of the first five trials succeeds, so . The first event is a count event. The other two are waiting time events, and the third can be rewritten as the count event .
- In the same process the first four trials have all failed. A classmate says a success on trial is now more likely than because it is overdue. Find and compare it with .
Show the answer
The conditioning event has probability , and is contained in , so
The chance is still , the same as for any single trial, so the classmate is wrong. The unconditional probability is smaller because it also requires the four failures, which in this question have already happened. In a Bernoulli process past failures do not make a success overdue.
- With , compute and compare it with the product . Are and independent? Are and independent?
Show the answer
The event says that trials and succeed and trials , and fail, so it equals . The two block counts are independent, so
Separately, and , whose product is . The joint probability differs from the product, so and are dependent. Knowing that the first two trials succeeded forces . The counts and use the disjoint blocks of trials to and to , so they are independent.