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The usual treatments deal with a single random variable or a fixed, finite number of random variables, considered jointly. However, there are many common applications in which we select at random a member of a class of random variables and observe its value, or select a random number of random variables and obtain some function of those selected. This is formulated with the aid of a counting or selecting random variable N, which is nonegative, integer valued. It may be independent of the class selected, or may be related in some sequential way to members of the class. We consider only the independent case. Many important problems require optional random variables, sometimes called Markov times. These involve more theory than we develop in this treatment.
Some common examples:
Total demand of N customers— N independent of the individual demands.
Total service time for N units— N independent of the individual service times.
Net gain in N plays of a game— N independent of the individual gains.
Extreme values of N random variables— N independent of the individual values.
Random sample of size N— N is usually determined by propereties of the sample observed.
Decide when to play on the basis of past results— N dependent on past.
As a basic model, we consider the sum of a random number of members of an iid class. In order to have a concrete interpretation to help visualize the formal patterns, we think of the demand of a random number of customers. We suppose the number of customers N is independent of the individual demands. We formulate a model to be used for a variety of applications.
A basic sequence [Demand of n customers]
An incremental sequence [Individual demands]
These are related as follows:
A counting random variable N. If N=n then n of the Yk are added to give the compound demand D (the random sum)
Note. In some applications the counting random variable may take on the idealized value ∞. For example, in a game that is played until some specified result occurs, this may never happen, so that no finite value can be assigned to N. In such a case, it is necessary to decide what value X∞ is to be assigned. For N independent of the Yn (hence of the Xn), we rarely need to consider this possibility.
Independent selection from an iid incremental sequence
We assume throughout, unless specifically stated otherwise, that:
X0=Y0=0
is iid
is an independent class
We utilize repeatedly two important propositions:
.
.
If the Yn are nonnegative integer valued,
then so is D and
DERIVATION
We utilize properties of generating functions, moment generating functions, and conditional expectation.
by definition of conditional expectation, given an event.
Now,
and
.
Hence
. Division by P(N=n) gives the desired result.
By the law of total probability (CE1b), . By proposition 1 and the product rule for moment generating functions,
Hence
A parallel argument holds for gD in the integer-valued case.
— □
Remark. The result on MD and gD may be developed without use of conditional expectation.
— □
Suppose the number N of jobs brought to a service shop in a day is Poisson (8). One fourth of these are items under warranty for which no charge is made. Others fall in one of two categories. One half of the arriving jobs are charged for one hour of shop time; the remaining one fourth are charged for two hours of shop time. Thus, the individual shop hour charges Yk have the common distribution
Make the basic assumptions of our model. Determine P(D≤4).
SOLUTION
According to the formula developed above,
Expand the exponentials in power series about the origin, multiply out to get enough terms. The result of straightforward but somewhat tedious calculations is
Taking the coefficients of the generating function, we get
Suppose the counting random variable N∼ binomial and Yi=IEi,
with
. Then
By the basic result on random selection, we have
so that D∼ binomial .
In the next section we establish useful m-procedures for determining the generating function
gD and the moment generating function MD for the compound demand for simple random
variables, hence for determining the complete distribution. Obviously, these will not work
for all problems. It may helpful, if not entirely sufficient, in such cases to be able to
determine the mean value E[D] and variance . To this end, we establish the
following expressions for the mean and variance.
DERIVATION
Hence
. By symmetry E[Y]=1.
. Hence,
We have m-procedures for performing the calculations necessary to determine the distribution for a composite demand D when the counting random variable N and the individual demands Yk are simple random variables with not too many values. In some cases, such as for a Poisson counting random variable, we are able to approximate by a simple random variable.
The procedure gend
If the Yi are nonnegative, integer valued, then so is D, and there is a generating function. We examine a strategy for computation which is implemented in the m-procedure gend. Suppose
The coefficients of gN and gY are the probabilities of the values of N and Y, respectively. We enter these and calculate the coefficients for powers of gY:
We wish to generate a matrix P whose rows contain the joint probabilities. The probabilities in the ith row consist of the coefficients for the appropriate power of gY multiplied by the probability N has that value. To achieve this, we need a matrix, each of whose n+1 rows has nm+1 elements, the length of yn. We begin by “preallocating” zeros to the rows. That is, we set P=zeros(n+1,n*m+1). We then replace the appropriate elements of the successive rows. The replacement probabilities for the ith row are obtained by the convolution of gY and the power of gY for the previous row. When the matrix P is completed, we remove zero rows and columns, corresponding to missing values of N and D (i.e., values with zero probability). To orient the joint probabilities as on the plane, we rotate P ninety degrees counterclockwise. With the joint distribution, we may then calculate any desired quantities.
The number of customers in a major appliance store is equally likely to be 1, 2, or 3. Each customer buys 0, 1, or 2 items with respective probabilities 0.5, 0.4, 0.1. Customers buy independently, regardless of the number of customers. First we determine the matrices representing gN and gY. The coefficients are the probabilities that each integer value is observed. Note that the zero coefficients for any missing powers must be included.
gN = (1/3)*[0 1 1 1]; % Note zero coefficient for missing zero power gY = 0.1*[5 4 1]; % All powers 0 thru 2 have positive coefficients gend Do not forget zero coefficients for missing powers Enter the gen fn COEFFICIENTS for gN gN % Coefficient matrix named gN Enter the gen fn COEFFICIENTS for gY gY % Coefficient matrix named gY Results are in N, PN, Y, PY, D, PD, P May use jcalc or jcalcf on N, D, P To view distribution for D, call for gD disp(gD) % Optional display of complete distribution 0 0.2917 1.0000 0.3667 2.0000 0.2250 3.0000 0.0880 4.0000 0.0243 5.0000 0.0040 6.0000 0.0003 EN = N*PN' EN = 2 EY = Y*PY' EY = 0.6000 ED = D*PD' ED = 1.2000 % Agrees with theoretical EN*EY P3 = (D>=3)*PD' P3 = 0.1167 [N,D,t,u,PN,PD,PL] = jcalcf(N,D,P); EDn = sum(u.*P)./sum(P); disp([N;EDn]') 1.0000 0.6000 % Agrees with theoretical E[D|N=n] = n*EY 2.0000 1.2000 3.0000 1.8000 VD = (D.^2)*PD' - ED^2 VD = 1.1200 % Agrees with theoretical EN*VY + VN*EY^2
Note that the zero power is missing from gY, corresponding to the fact that P(Y=0)=0.
gN = 0.2*[1 1 1 1 1]; gY = 0.1*[0 5 3 2]; % Note the zero coefficient in the zero position gend Do not forget zero coefficients for missing powers Enter the gen fn COEFFICIENTS for gN gN Enter the gen fn COEFFICIENTS for gY gY Results are in N, PN, Y, PY, D, PD, P May use jcalc or jcalcf on N, D, P To view distribution for D, call for gD disp(gD) % Optional display of complete distribution 0 0.2000 1.0000 0.1000 2.0000 0.1100 3.0000 0.1250 4.0000 0.1155 5.0000 0.1110 6.0000 0.0964 7.0000 0.0696 8.0000 0.0424 9.0000 0.0203 10.0000 0.0075 11.0000 0.0019 12.0000 0.0003
p3 = (D == 3)*PD' % P(D=3) P3 = 0.1250 P4_12 = ((D >= 4)&(D <= 12))*PD' P4_12 = 0.4650 % P(4 <= D <= 12)
We are interested in the number of successes in N trials for a general counting random variable. This is a generalization of the Bernoulli case in Example 15.2. Suppose, as in Example 15.5, the number of customers in a major appliance store is equally likely to be 1, 2, or 3, and each buys at least one item with probability p=0.6. Determine the distribution for the number D of buying customers.
SOLUTION
We use