An Example, Treated From the Point of View of Numerical n-Tuple Valued Random Phenomena

In the next two sections we discuss an example that illustrates the need to introduce various concepts concerning random variables, which will, in turn, be presented in the course of the discussion. We begin in this section by discussing the example in terms of the notion of a numerical valued random phenomenon in order to show the similarities and differences between this notion and that of a random variable.

Let us consider a commuter who is in the habit of taking a train to the city; the time of departure from the station is given in the railroad timetable as 7:55 A.M. However, the commuter notices that the actual time of departure is a random phenomenon, varying between 7:55 and 8 A.M. Let us assume that the probability law of the random phenomenon is specified by a probability density function ; further, let us assume

\begin{align} f_{1}\left(x_{1}\right) &= \begin{cases} \frac{2}{25}\left(5-x_{1}\right), & \text{for } 0 \leq x_{1} \leq 5 \tag{3.1}\\ 0, & \text{otherwise.} \end{cases} \end{align} 

in which represents the number of minutes after 7:55 A.M. that the train departs.

Let us suppose next that the time it takes the commuter to travel from his home to the station is a numerical valued random phenomenon, varying between 25 and 30 minutes. Then, if the commuter leaves his home at 7:30 A.M. every day, his time of arrival at the station is a random phenomenon, varying between 7:55 and 8 A.M. Let us suppose that the probability law of this random phenomenon is specified by a probability density function ; further, let us assume that is of the same functional form as , so that \begin{align} f_{2}\left(x_{2}\right) &= \begin{cases} \frac{2}{25}\left(5-x_{2}\right), & \text{for } 0 \leq x_{2} \leq 5 \tag{3.2}\\ 0, & \text{otherwise,} \end{cases} \end{align} in which represents the number of minutes after 7:55 A.M. that the commuter arrives at the station.

The question now naturally arises: will the commuter catch the 7:55 A.M. train? Of course, this question cannot be answered by us; but perhaps we can answer the question: what is the probability that fhe commuter will catch the 7:55 A.M. train?

Before any attempt can be made to answer this question, we must express mathematically as a set on a sample description space the random event described verbally as the event that the commuter catches the train. Further, to compute the probability of the event, a probability function on the sample description space must be defined.

As our sample description space , we take the space of 2-tuples of real numbers, where represents the time (in minutes after 7:55 A.M.) at which the train departs from the station, and denotes the time (in minutes after 7:55 A.M.) at which the commuter arrives at the station. The event that the man catches the train is then given as a set of sample descriptions by , since to catch the train his arrival time must be less than the train’s departure time . The event is diagrammed in Fig. 3A .

We define next a probability function on the events in . To do this, we use the considerations of section 7 , Chapter 4, concerning numerical 2-tuple valued random phenomena. In particular, let us suppose that the probability function is specified by a 2-dimensional probability density function . From a knowledge of we may compute the probability that the commuter will catch his train by the formula

in which the second and third equations follow by the usual rules of calculus for evaluating double integrals (or integrals over the plane) by means of iterated (or repeated) single integrals.

We next determine whether the function is specified by our having specified the probability density functions and by (3.1) and (3.2) .

More generally, we consider the question: what relationship exists between the individual probability density functions and and the joint probability density function ? We show first that from a knowledge of , one may obtain a knowledge of and by the formulas, for all real numbers and , \begin{align} & f_{1}\left(x_{1}\right)=\int_{-\infty}^{\infty} f\left(x_{1}, x_{2}\right) d x_{2} \\ & f_{2}\left(x_{2}\right)=\int_{-\infty}^{\infty} f\left(x_{1}, x_{2}\right) d x_{1}. \end{aligned} \tag{3.4}\] 

Conversely, we show by a general example that from a knowledge of and one cannot obtain a knowledge of , since is not uniquely determined by and ; more precisely, we show that to given probability density functions and there exists an infinity of functions that satisfy (3.4) with respect to and .

To prove (3.4), let and be the distribution functions of the first and second random phenomena under consideration; in the example discussed, is the distribution function of the departure time of the train from the station, and is the distribution function of the arrival time of the man at the station. We may obtain expressions for and in terms of , for is equal to the probability, according to the probability function , of the set , and similarly . Consequently,

We next use the fact that

By differentiation of (3.5), in view of (3.6), we obtain (3.4).

Conversely, given any two probability density functions and , let us show how one may find many probability density functions to satisfy (3.4). Let be a positive number. Choose a finite nonempty interval to such that for . Similarly, choose a finite nonempty interval to such that for . Define a function of two variables by

Clearly, by construction, for all and

Define the function for any real numbers and by

It may be verified, in view of (3.8), that is a probability density function satisfying (3.4).

We now return to the question of how to determine . There is one (and, in general, only one) circuinstance in which the individual probability density functions and determine the joint probability density function , namely, when the respective random phenomena, whose probability density functions are and , are independent. 

We define two random phenomena as independent, letting and denote their respective probability functions and their joint probability function, if it holds that for all real numbers , and

Equivalently, two random phenomena are independent, letting and denote their respective distribution functions and their joint distribution function, if it holds that for all real numbers and

Equivalently, two continuous random phenomena are independent, letting and denote their respective probability density functions and their joint probability density, if it holds that for all real numbers and

Equivalently, two discrete random phenomena are independent, letting and denote their respective probability mass functions and their joint probability mass function if it holds that for all real numbers and

The equivalence of the foregoing statements concerning independence may be shown more or less with ease by using the relationships developed in Chapter 4; indications of the proofs are contained in section 6.

Independence may also be defined in terms of the notion of an event depending on a phenomenon , which is analogous to the notion of an event depending on a trial developed in section 2 of Chapter 3. An event is said to depend on a random phenomenon if a knowledge of the outcome of the phenomenon suffices to determine whether or not the event has occurred . We then define two random phenomena as independent if, for any two events and , depending, respectively, on the first and second phenomenon, the probability of the intersection of and is equal to the product of their probabilities:

As shown in section 2 of Chapter 3, two random phenomena are independent if and only if a knowledge of the outcome of one of the phenomena does not affect the probability of any event depending upon the other phenomenon.

Let us now return to the problem of the commuter catching his train, and let us assume that the commuter’s arrival time and the train’s departure time are independent random phenomena. Then (3.12) holds, and from (3.3) 

Since and are specified by (5.1) and (5.2) , respectively, the probability that the commuter will catch his train can now be computed by evaluating the integrals in (3.15) . However, in the present example there is a very special feature present that makes it possible to evaluate without any laborious calculation.

The reader may have noticed that the probability density functions and have the same functional form. If we define by or 0, depending on whether or otherwise, we find that for all real numbers . In terms of , we may write (3.15), making the change of variable and in the second integral,

By adding the two integrals in (3.16), it follows that

We conclude that the probability that the man will catch his train is equal to .

Exercises