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 kw1/3kw1/3, 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=kX1/3D=kX1/3.

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:

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

The function rule is more complicated than in Example 1, but the essential problem is the same.

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

SOLUTION

First solution. The mapping approach

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

Second solution. The discrete alternative

Picking out and adding the indicated probabilities, we have

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.

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

Determine P(Z>0)P(Z>0).

SOLUTION

First we determine N={t:g(t)>0}N={t:g(t)>0}. As in Example 3, g(t)=(t+1)(t-4)>0g(t)=(t+1)(t-4)>0 for t<-1t<-1 or t>4t>4. Because of the uniform distribution, the integral of the density over any subinterval of [-3,7][-3,7] is 0.1 times the length of that subinterval. Thus, the desired probability is

To show that if X∼N(μ,σ2)X∼N(μ,σ2) then

VERIFICATION

We wish to show the denity function for Z is

Now

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

Since the density is the derivative of the distribution function,

Thus

We conclude that Z∼N(0,1)Z∼N(0,1)

Suppose X has distribution function F_X. If it is absolutely continuous, the corresponding density is f_X. Consider Z=aX+b(a≠0)Z=aX+b(a≠0). Here g(t)=at+bg(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

There are two cases

For the absolutely continuous case, PX=v-ba=0PX=v-ba=0, and by differentiation

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

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

VERIFICATION

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

Suppose X≥0X≥0 and Z=g(X)=X1/aZ=g(X)=X1/a for a>1a>1. Since for t≥0t≥0, t1/at1/a is increasing, we have 0≤t1/a≤v0≤t1/a≤v iff 0≤t≤va0≤t≤va. Thus

In the absolutely continuous case

Suppose X∼X∼ exponential (λ)(λ). Then Z=X1/a∼Z=X1/a∼ Weibull (a,λ,0)(a,λ,0).

According to the result of Example 8,

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

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]I=[a,b]. Suppose I is partitioned into n subintervals by points t_i, 1≤i≤n-11≤i≤n-1, with a=t0a=t0 and b=tnb=tn. Let Mi=[ti-1,ti)Mi=[ti-1,ti) be the ith subinterval, 1≤i≤n-11≤i≤n-1 and Mn=[tn-1,tn]Mn=[tn-1,tn]. Let Ei=X-1(Mi)Ei=X-1(Mi) 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 si,ti-1≤si<tisi,ti-1≤si<ti. The simple random variable

approximates X to within the length of the largest subinterval M_i. Now IEi=IMi(X)IEi=IMi(X), since IEi(ω)=1IEi(ω)=1 iff X(ω)∈MiX(ω)∈Mi iff IMi(X(ω))=1IMi(X(ω))=1. We may thus write

X=10IA+18IB+10ICX=10IA+18IB+10IC with {A,B,C}{A,B,C} independent and P=[0.60.30.5]P=[0.60.30.5].

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

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

Suppose X has density function fX(t)=12(3t2+2t)fX(t)=12(3t2+2t) for 0≤t≤10≤t≤1. Then FX(t)=12(t3+t2)FX(t)=12(t3+t2). Let Z=X1/2Z=X1/2. 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)P(Z≤0.8). Now Z≤0.8Z≤0.8 iff X≤0.82=0.64X≤0.82=0.64. The desired probability may be calculated to be

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