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

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

Second solution . The discrete alternative

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

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.

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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 )

Determine P ( Z > 0 ) .

SOLUTION

First we determine N = { t : g ( t ) > 0 } . As in [link] , 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
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We consider, next, some important examples.

The normal distribution and standardized normal distribution

To show that if X N ( μ , σ 2 ) then

Z = g ( X ) = X - μ σ N ( 0 , 1 )

VERIFICATION

We wish to show the denity function for Z is

φ ( t ) = 1 2 π e - t 2 / 2

Now

g ( t ) = t - μ σ v iff t σ v + μ

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

F Z ( v ) = P ( Z v ) = P ( Z M ) = P ( X N ) = P ( X σ v + μ ) = F X ( σ v + μ )

Since the density is the derivative of the distribution function,

f Z ( v ) = F Z ' ( v ) = F X ' ( σ v + μ ) σ = σ f X ( σ v + μ )

Thus

f Z ( v ) = σ σ 2 π exp [ - 1 2 σ v + μ - μ σ 2 = 1 2 π e - v 2 / 2 = φ ( v )

We conclude that Z N ( 0 , 1 )

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Affine functions

Suppose X has distribution function F X . If it is absolutely continuous, the corresponding density is f X . Consider Z = a X + b ( a 0 ) . Here g ( t ) = a t + 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 ( a X + b v )

There are two cases

  • a > 0 :
    F Z ( v ) = P X v - b a = F X v - b a
  • a < 0
    F Z ( v ) = P X v - b a = P X > v - b a + P X = v - b a
    So that
    F Z ( v ) = 1 - F X v - b a + P X = v - b a

For the absolutely continuous case, P X = v - b a = 0 , and by differentiation

  • for a > 0 f Z ( v ) = 1 a f X v - b a
  • for a < 0 f 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 ) = 1 | a | f X v - b a
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Completion of normal and standardized normal relationship

Suppose Z N ( 0 , 1 ) . Show that X = σ Z + μ ( σ > 0 ) is N ( μ , σ 2 ) .

VERIFICATION

Use of the result of [link] on affine functions shows that

f X ( t ) = 1 σ φ t - μ σ = 1 σ 2 π exp - 1 2 t - μ σ 2
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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 )

In the absolutely continuous case

f Z ( v ) = F Z ' ( v ) = f X ( v a ) a v a - 1
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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 [link] ,

F Z ( t ) = F X ( t a ) = 1 - e - λ t a

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

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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 rangeof 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 = t 0 and b = t n . Let M i = [ t i - 1 , t i ) be the i th 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 n s i I E i

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 n s i I M i ( X ) , a function of X
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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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