Functions of a Random Variable

Summary: 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)). Suppose we have the distribution for X. How can we determine P(Z∈M), the probability Z takes a value in the set M? Mapping approach. Simply find the amount of probability mass mapped into the set M by the random variable X. In the absolutely continuous case, integrate the density function for X over the set M. In the discrete case, as an alternative, select those possible values for X which are in the set M and add their probabilities. For a Borel function g and set M, determine the set N of all those t which are mapped into M, then determine the probability X is in N as in the previous case.

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 gis 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 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 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

g(X)=200+18I_M1(X)(X−10)+(16−18)I_M2(X)(X−20)(1)

+(15−16)I_M3(X)(X−30)+(13−15)I_M4(X)(X−50)(2)

whereM1=[10,∞),M2=[20,∞),M3=[30,∞),M4=[50,∞)(3)

The function rule is more complicated than in Example 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 ∫_Mf_X.
  - 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, thenX(ω)∈N. Hence
  {ω:g(X(ω))∈M{={ω:X(ω)∈N{(4)
  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 g(t_i) 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 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

P(g(X)>0)=P(X=−2)+P(X=6)=0.2+0.2=0.4(5)

Second solution. The discrete alternative

TABLE 1

X=	-2	0	1	3	6
PX=	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

P(Z>0)=0.2+0.2=0.4(6)

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 4: An absolutely continuous example

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

Z=g(X)=(X+1)(X−4)(7)

Determine P(Z>0).

SOLUTION

First we determine N={t:g(t)>0{. As in Example 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 [−3,7] is 0.1 times the length of that subinterval. Thus, the desired probability is

P(g(X)>0)=0.1[(−1−(−3))+(7−4)]=0.5(8)

We consider, next, some important examples.

EXAMPLE 5: The normal distribution and standardized normal distribution

To show that if X∼N(μ,σ²) then

Z=g(X)=

X−μ

∼N(0,1)(9)

VERIFICATION

We wish to show the denity function for Z is

φ(t)=

\2π

e^−t²/2(10)

Now

g(t)=

t−μ

≤vifft≤σv+μ(11)

Hence, for given M=(−∞,v] the inverse image is N=(−∞,σv+μ], so that

F_Z(v)=P(Z≤v)=P(Z∈M)=P(X∈N)=P(X≤σv+μ)=F_X(σv+μ)(12)

Since the density is the derivative of the distribution function,

f_Z(v)=F_Z^'(v)=F_X^'(σv+μ)σ=σf_X(σv+μ)(13)

Thus

f_Z(v)=

σ\2π

exp[−

(

σv+μ−μ

)²=

\2π

e^−v²/2=φ(v)(14)

We conclude that Z∼N(0,1)

EXAMPLE 6: Affine functions

Suppose X has distribution function F_X. If it is absolutely continuous, the corresponding density is f_X. Consider Z=aX+b(a≠0). Here g(t)=at+b, an affine function (linear plus a constant). Determine the distribution function for Z (and the density in the absolutely continuous case).

SOLUTION

F_Z(v)=P(Z≤v)=P(aX+b≤v)(15)

There are two cases

a>0:
F_Z(v)=P(X,≤,

v−b

a

)=F_X(

v−b

a

)(16)
a<0
F_Z(v)=P(X,≥,

v−b

a

)=P(X,>,

v−b

a

)+P(X,=,

v−b

a

)(17)
So that
F_Z(v)=1−F_X(

v−b

a

)+P(X,=,

v−b

a

)(18)

For the absolutely continuous case, P(X,=,

v−b

)=0, and by differentiation

for a>0f_Z(v)=

1

a

f_X(

v−b

a

)
for a<0f_Z(v)=−

1

a

f_X(

v−b

a

)

Since for a<0, −a=|a|, the two cases may be combined into one formula.

f_Z(v)=

|a|

f_X(

v−b

)(19)

EXAMPLE 7: Completion of normal and standardized normal relationship

Suppose Z∼N(0,1). Show that X=σZ+μ(σ>0) is N(μ,σ²).

VERIFICATION

Use of the result of Example 6 on affine functions shows that

f_X(t)=

φ(

t−μ

σ\2π

exp[−,

t−μ

)²](20)

EXAMPLE 8: Fractional power of a nonnegative random variable

Suppose X≥0 and Z=g(X)=X^1/a for a>1. Since for t≥0, t^1/a is increasing, we have 0≤t^1/a≤v iff 0≤t≤v^a. Thus

F_Z(v)=P(Z≤v)=P(X≤v^a)=F_X(v^a)(21)

In the absolutely continuous case

f_Z(v)=F_Z^'(v)=f_X(v^a)av^a−1(22)

EXAMPLE 9: Fractional power of an exponentially distributed random variable

Suppose X∼ exponential (λ). Then Z=X^1/a∼ Weibull (a,λ,0).

According to the result of Example 8,

F_Z(t)=F_X(t^a)=1−e^{−λt^a}(23)

which is the distribution function for Z∼ Weibull (a,λ,0).

EXAMPLE 10: A simple approximation as a function of X

If X is a random variable, a simple function approximation may be constructed (see Distribution Approximations). We limit our discussion to the bounded case, in which the range of X is limited to a bounded interval I=[a,b]. Suppose I is partitioned into n subintervals by points t_i, 1≤i≤n−1, with a=t₀ and b=t_n. Let M_i=[t_i−1,t_i) be the ith subinterval, 1≤i≤n−1 and M_n=[t_n−1,t_n]. Let E_i=X⁻¹(M_i) be the set of points mapped into M_i by X. Then the E_i form a partition of the basic space Ω. For the given subdivision, we form a simple random variable X_s as follows. In each subinterval, pick a point s_i,t_i−1≤s_i<t_i. The simple random variable

X_s=

∑

i=1

s_iI_{E_i}(24)

approximates X to within the length of the largest subinterval M_i. Now I_{E_i}=I_{M_i}(X), since I_{E_i}(ω)=1 iff X(ω)∈M_i iff I_{M_i}(X(ω))=1. We may thus write

X_s=

∑

i=1

s_iI_{M_i}(X),afunctionofX(25)

Use of MATLAB on simple random variables

For simple random variables, we use the discrete alternative approach, since this may be implemented easily with MATLAB. Suppose the distribution for X is expressed in the row vectors X andPX.

We perform array operations on vector X to obtain
G=[g(t₁)g(t₂)⋯g(t_n)](26)
We use relational and logical operations on G to obtain a matrix M which has ones for those t_i (values of X) such that g(t_i) satisfies the desired condition (and zeros elsewhere).
The zero-one matrix M is used to select the the corresponding p_i=P(X=t_i) and sum them by the taking the dot product of M and PX.

EXAMPLE 11: Basic calculations for a function of a simple random variable

X = -5:10;                     % Values of X
PX = ibinom(15,0.6,0:15);      % Probabilities for X
G = (X + 6).*(X - 1).*(X - 8); % Array operations on X matrix to get G = g(X)
M = (G > - 100)&(G < 130);     % Relational and logical operations on G
PM = M*PX'                     % Sum of probabilities for selected values
PM =  0.4800
disp([X;G;M;PX]')              % Display of various matrices (as columns)
   -5.0000   78.0000    1.0000    0.0000
   -4.0000  120.0000    1.0000    0.0000
   -3.0000  132.0000         0    0.0003
   -2.0000  120.0000    1.0000    0.0016
   -1.0000   90.0000    1.0000    0.0074
         0   48.0000    1.0000    0.0245
    1.0000         0    1.0000    0.0612
    2.0000  -48.0000    1.0000    0.1181
    3.0000  -90.0000    1.0000    0.1771
    4.0000 -120.0000         0    0.2066
    5.0000 -132.0000         0    0.1859
    6.0000 -120.0000         0    0.1268
    7.0000  -78.0000    1.0000    0.0634
    8.0000         0    1.0000    0.0219
    9.0000  120.0000    1.0000    0.0047
   10.0000  288.0000         0    0.0005
[Z,PZ] = csort(G,PX);          % Sorting and consolidating to obtain
disp([Z;PZ]')                  % the distribution for Z = g(X)
 -132.0000    0.1859
 -120.0000    0.3334
  -90.0000    0.1771
  -78.0000    0.0634
  -48.0000    0.1181
         0    0.0832
   48.0000    0.0245
   78.0000    0.0000
   90.0000    0.0074
  120.0000    0.0064
  132.0000    0.0003
  288.0000    0.0005
P1 = (G<-120)*PX '           % Further calculation using G, PX
P1 =  0.1859
p1 = (Z<-120)*PZ'            % Alternate using Z, PZ
p1 =  0.1859

EXAMPLE 12

X=10I_A+18I_B+10I_C with {A,B,C{ independent and P=[0.60.30.5].

We calculate the distribution for X, then determine the distribution for

Z=X^1/2−X+50(27)

c = [10 18 10 0];
pm = minprob(0.1*[6 3 5]);
canonic
 Enter row vector of coefficients  c
 Enter row vector of minterm probabilities  pm
Use row matrices X and PX for calculations
Call for XDBN to view the distribution
disp(XDBN)
         0    0.1400
   10.0000    0.3500
   18.0000    0.0600
   20.0000    0.2100
   28.0000    0.1500
   38.0000    0.0900
G = sqrt(X) - X + 50;       % Formation of G matrix
[Z,PZ] = csort(G,PX);       % Sorts distinct values of g(X)
disp([Z;PZ]')               % consolidates probabilities
   18.1644    0.0900
   27.2915    0.1500
   34.4721    0.2100
   36.2426    0.0600
   43.1623    0.3500
   50.0000    0.1400
M = (Z < 20)|(Z >= 40)      % Direct use of Z distribution
M =    1     0     0     0     1     1
PZM = M*PZ'
PZM =  0.5800

Remark. Note that with the m-function csort, we may name the output as desired.

EXAMPLE 13: Continuation of Example 12, above.

H = 2*X.^2 - 3*X + 1;
[W,PW] = csort(H,PX)
W  =     1      171     595     741    1485    2775
PW =  0.1400  0.3500  0.0600  0.2100  0.1500  0.0900

EXAMPLE 14: A discrete approximation

Suppose X has density function f_X(t)=

(3t²+2t) for 0≤t≤1. Then F_X(t)=

(t³+t²). Let Z=X^1/2. We may use the approximation m-procedure tappr to obtain an approximate discrete distribution. Then we work with the approximating random variable as a simple random variable. Suppose we want P(Z≤0.8). Now Z≤0.8 iff X≤0.8²=0.64. The desired probability may be calculated to be

P(Z≤0.8)=F_X(0.64)=(0.64³+0.64²)/2=0.3359(28)

Using the approximation procedure, we have

tappr
Enter matrix [a b] of x-range endpoints  [0 1]
Enter number of x approximation points  200
Enter density as a function of t  (3*t.^2 + 2*t)/2
Use row matrices X and PX as in the simple case
G = X.^(1/2);
M = G <= 0.8;
PM = M*PX'
PM =   0.3359       % Agrees quite closely with the theoretical