Applied Probability by Paul E Pfeiffer - HTML preview

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Chapter 10. Functions of Random Variables

10.1. Functions of a Random Variable^*

Introduction

Frequently, we observe a value of some random variable, but are really interested in a value derived from this by a function rule. If X is a random variable and g is a reasonable function (technically, a Borel function), then Z=g(X) is a new random variable which has the value g(t) for any ω such that X(ω)=t. Thus Z(ω)=g(X(ω)).

The problem; an approach

We consider, first, functions of a single random variable. A wide variety of functions are utilized in practice.

Example 10.1. A quality control problem

In a quality control check on a production line for ball bearings it may be easier to weigh the balls than measure the diameters. If we can assume true spherical shape and w is the weight, then diameter is kw^1/3, where k is a factor depending upon the formula for the volume of a sphere, the units of measurement, and the density of the steel. Thus, if X is the weight of the sampled ball, the desired random variable is D=kX^1/3.

Example 10.2. Price breaks

The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase ten tickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule:

11-20, $18 each
21-30, $16 each
31-50, $15 each
51-100, $13 each

If the number of purchasers is a random variable X, the total cost (in dollars) is a random quantity Z=g(X) described by

(10.1) g ( X ) = 200 + 18 I_{M
1} ( X ) ( X – 10 ) + ( 16 – 18 ) I_{M
2} ( X ) ( X – 20 )

(10.2) + ( 15 – 16 ) I_{M
3} ( X ) ( X – 30 ) + ( 13 – 15 ) I_{M
4} ( X ) ( X – 50 )

(10.3)

The function rule is more complicated than in Example 10.1, but the essential problem is the same.

The problem

If X is a random variable, then Z=g(X) is a new random variable. Suppose we have the distribution for X. How can we determine P(Z∈M), the probability Z takes a value in the set M?

An approach to a solution

We consider two equivalent approaches

To find P(X∈M).
1. Mapping approach. Simply find the amount of probability mass mapped into the set M by the random variable X.
  - In the absolutely continuous case, calculate .
  - In the discrete case, identify those values t_i of X which are in the set M and add the associated probabilities.
2. Discrete alternative. Consider each value t_i of X. Select those which meet the defining conditions for M and add the associated probabilities. This is the approach we use in the MATLAB calculations. Note that it is not necessary to describe geometrically the set M; merely use the defining conditions.
To find P(g(X)∈M).
1. Mapping approach. Determine the set N of all those t which are mapped into M by the function g. Now if X(ω)∈N, then g(X(ω))∈M, and if g(X(ω))∈M, then X(ω)∈N. Hence
  (10.4){ω:g(X(ω))∈M}={ω:X(ω)∈N}
  Since these are the same event, they must have the same probability. Once N is identified, determine P(X∈N) in the usual manner (see part a, above).
2. Discrete alternative. For each possible value t_i of X, determine whether meets the defining condition for M. Select those t_i which do and add the associated probabilities.

— □

Remark. The set N in the mapping approach is called the inverse image N=g^–1(M).

Example 10.3. A discrete example

Suppose X has values -2, 0, 1, 3, 6, with respective probabilities 0.2, 0.1, 0.2, 0.3 0.2.

Consider Z=g(X)=(X+1)(X–4). Determine P(Z>0).

SOLUTION

First solution. The mapping approach

g(t)=(t+1)(t–4). N={t:g(t)>0} is the set of points to the left of –1 or to the right of 4. The X-values –2 and 6 lie in this set. Hence

(10.5) P ( g ( X ) > 0 ) = P ( X = – 2 ) + P ( X = 6 ) = 0 . 2 + 0 . 2 = 0 . 4

Second solution. The discrete alternative

Table 10.1.

X =	-2	0	1	3	6
P X =	0.2	0.1	0.2	0.3	0.2
Z =	6	-4	-6	-4	14
Z > 0	1	0	0	0	1

Picking out and adding the indicated probabilities, we have

(10.6) P ( Z > 0 ) = 0 . 2 + 0 . 2 = 0 . 4

In this case (and often for “hand calculations”) the mapping approach requires less calculation. However, for MATLAB calculations (as we show below), the discrete alternative is more readily implemented.

Example 10.4. An absolutely continuous example

Suppose X∼ uniform [–3,7]. Then f_X(t)=0.1,–3≤t≤7 (and zero elsewhere). Let

(10.7) Z = g ( X ) = ( X + 1 ) ( X – 4 )

Determine P(Z>0).

SOLUTION

First we determine N={t:g(t)>0}. As in Example 10.3, g(t)=(t+1)(t–4)>0 for t<–1 or t>4. Because of the uniform distribution, the integral of the density over any subinterval of is 0.1 times the length of that subinterval. Thus, the desired probability is