Random selection (Page 2/5)

Applied probability Page 2 / 5

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 Y _n (hence of the X _n ), we rarely need to consider this possibility.

Independent selection from an iid incremental sequence

We assume throughout , unless specifically stated otherwise, that:

$X_{0} = Y_{0} = 0$
${Y_{k} : 1 \leq k}$ is iid
${N, Y_{k} : 0 \leq k}$ is an independent class

We utilize repeatedly two important propositions:

$E [h (D) | N = n] = E [h (X_{n})], n \geq 0$ .
$M_{D} (s) = g_{N} [M_{Y} (s)]$ . If the Y _n are nonnegative integer valued, then so is D and $g_{D} (s) = g_{N} [g_{Y} (s)]$

DERIVATION

We utilize properties of generating functions, moment generating functions, and conditional expectation.

$E [I_{{n}} (N) h (D)] = E [h (D) | N = n] P (N = n)$ by definition of conditional expectation, given an event. Now, $I_{{n}} (N) h (D) = I_{{n}} (N) h (X_{n})$ and $E [I_{{n}} (N) h (X_{n})] = P (N = n) E [h (X_{n})]$ . Hence $E [h (D) | N = n] P (N = n) = P (N = n) E [h (X_{n})]$ . Division by $P (N = n)$ gives the desired result.
By the law of total probability (CE1b), $M_{D} (s) = E [e^{s D}] = E {E [e^{s D} | N]}$ . By proposition 1 and the product rule for moment generating functions,
$E [e^{s D} | N = n] = E [e^{s X_{n}}] = \prod_{k = 1}^{n} E [e^{s Y_{k}}] = M_{Y}^{n} (s)$
Hence
$M_{D} (s) = \sum_{n = 0}^{\infty} M_{Y}^{n} (s) P (N = n) = g_{N} [M_{Y} (s)]$
A parallel argument holds for g _D in the integer-valued case.

— $□$

Remark . The result on M _D and g _D may be developed without use of conditional expectation.

M_{D} (s) = E [e^{s D}] = \sum_{k = 0}^{\infty} E [I_{{N = n}} e^{s X_{n}}] = \sum_{k = 0}^{\infty} P (N = n) E [e^{s X_{n}}]

= \sum_{k = 0}^{\infty} P (N = n) M_{Y}^{n} (s) = g_{N} [M_{Y} (s)]

— $□$

A service shop

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 inone 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, theindividual shop hour charges Y _k have the common distribution

Y = [0 1 2] with probabilities P Y = [1 / 4 1 / 2 1 / 4]

Make the basic assumptions of our model. Determine $P (D \leq 4)$ .

SOLUTION

g_{N} (s) = e^{8 (s - 1)} g_{Y} (s) = \frac{1}{4} (1 + 2 s + s^{2})

According to the formula developed above,

g_{D} (s) = g_{N} [g_{Y} (s)] = exp ((8 / 4) (1 + 2 s + s^{2}) - 8) = e^{4 s} e^{2 s^{2}} e^{- 6}

Expand the exponentials in power series about the origin, multiply out to get enough terms. The result of straightforward but somewhat tedious calculations is

g_{D} (s) = e^{- 6} (1 + 4 s + 10 s^{2} + \frac{56}{3} s^{3} + \frac{86}{3} s^{4} + \dots)

Taking the coefficients of the generating function, we get

P (D \leq 4) \approx e^{- 6} (1 + 4 + 10 + \frac{56}{3} + \frac{86}{3}) = e^{- 6} \frac{187}{3} \approx 0.1545

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A result on bernoulli trials

Suppose the counting random variable $N \sim$ binomial $(n, p)$ and $Y_{i} = I_{E_{i}}$ , with $P (E_{i}) = p_{0}$ . Then

g_{N} = {(q + p s)}^{n} and g_{Y} (s) = q_{0} + p_{0} s

By the basic result on random selection, we have

g_{D} (s) = g_{N} [g_{Y} (s)] = {[q + p (q_{0} + p_{0} s)]}^{n} = {[(1 - p p_{0}) + p p_{0} s]}^{n}

so that $D \sim$ binomial $(n, p p_{0})$ .

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In the next section we establish useful m-procedures for determining the generating function g _D and the moment generating function M _D for the compound demand for simple random variables, hence for determining the complete distribution. Obviously, these will not workfor all problems. It may helpful, if not entirely sufficient, in such cases to be able to determine the mean value $E [D]$ and variance $Var [D]$ . To this end, we establish the following expressions for the mean and variance.

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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