Chapter 7. Functions of Random Variables
Sections 7.2 -- 7.4: Functions of Discrete Random
Variables, Method of Distribution Functions and
Method of Transformations in One Dimension
Jiaping Wang
Department of Mathematical Science
04/10/2013, Wednesday
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Outline
Functions of Discrete Random Variables
Methods of Distribution Functions
Method of Transformations in One Dimension
More Examples
Homework #10
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Part 1. Functions of Discrete Random
Variables
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Introduction
For example, there is a sample X1, X2, …, Xn from same
distribution, also there is a function denoted by Y=f(X1, X2, …,
Xn)=1/n∑Xi, which is a function of random variables {X1, X2, …,
Xn}.
Considering the discrete random variables, for example, X is a
discrete random variables with space S={0, 1, 2, 3}, C is a
function of X with C=150+50X, then we can have a mass
probability table
x
0
1
2
3
p(x)
0.5
0.3
0.15
0.05
c
150
200
250
300
p(c)
0.5
0.3
0.15
0.05
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Cont.
Considering X is a discrete random variables with space S={-1,
0, 1}, define Y=X2, then we can have a mass probability table
as follows
x
-1
0
1
p(x)
0.25
0.5
0.25
y
1
0
1
p(y)
0.25
0.5
0.25
y
0
1
p(y)
0.5
0.5
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Example 7.1
A quality control manager samples from a lot of items, testing
each item until r defectives have been found. Find the distribution
of Y, the number of items that are tested to obtain r defectives.
Answer: Assume that the probability p of obtaining a defective item is constant from trial
To trial, the number of good items X sampled prior to the r-th defective one is a negative
Binomial random variable. The mass function is
𝑃 𝑋 = 𝑥 = 𝑝 𝑥 = 𝑥+𝑟−1
𝑝𝑟𝑞𝑥, 𝑥 = 0, 1, 2, …
𝑟−1
The number of trials, Y, is equal to the sum of the number of good items and defective
Ones, that is, Y=X+r thus X=Y-r, with Y=r, r+1, r+2, … so the mass function for Y is
𝑃 𝑌 = 𝑦 = 𝑝 𝑦 = 𝑦−1
𝑝𝑟𝑞𝑦 − 𝑟, 𝑦 = 𝑟, 𝑟 + 1, 𝑟 + 2, …
𝑟−1
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Part 2. Method of Distribution Functions
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Introduction
If X has a probability density function 𝒇𝑿(𝒙), and Y is a
function of X, we are interested in finding 𝑭𝒀(𝒚) = 𝑷(𝒀 ≤
𝒚) or the density 𝒇𝒀(𝒚) by using the distribution of X.
For example, 𝒀 = 𝑿𝟐 with density 𝒇𝑿(𝒙). For y≥0,
𝑭𝒀 𝒚 = 𝑷 𝒀 ≤ 𝒚 = 𝑷 𝑿𝟐 ≤ 𝒚 = 𝑷 − 𝒚 ≤ 𝑿 ≤ 𝒚
= 𝑷 𝑿 ≤ 𝒚 − 𝐏 𝐗 ≤ − 𝒚 = 𝑭𝑿 𝒚 − 𝑭𝑿 − 𝒚 .
Then we can have the density function for Y:
𝒅
𝒅
𝒅
𝒇 𝒀 𝒚 = 𝑭𝒀 𝒚 = 𝑭𝑿 𝒚 − 𝑭𝑿 − 𝒚
𝒅𝒚
𝒅𝒚
𝒅𝒚
𝟏
𝟏
=
𝒇𝑿 𝒚 +
𝒇𝑿 − 𝒚
𝟐 𝒚
𝟐 𝒚
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Application in Normal Distribution
X is standard normal random variable, what is the probability density function
of Y=X2?
We know 𝑓𝑋 𝑥 =
𝑓𝑌 𝑦 =
1
2 𝑦
[
Recall that Γ
1
exp
2𝜋
1
exp
2𝜋
1
2
𝑥2
−
2
( 𝑦)2
−
2
+
, −∞ < 𝑥 < ∞, thus for y≥0,
1
exp
2𝜋
(− 𝑦)2
−
2
]=
−
1
y
2 𝜋
1
2
y
2
exp(− )
= 𝜋, we can see Y follows a gamma distribution with
parameters α=1/2 and β=2.
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Example 7.2
The proportion of time X that a lathe is in use during a typical 40-hour
workweek is a random variable whose probability density function is given by
3𝑥2, 0 ≤ 𝑥 ≤ 1
f x =
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.
The actual number of hours out of a 40-hour week that the lathe is not in use
then is Y=40(1-X). Find the probability density function for Y.
Answer:
𝑦
𝐹𝑌(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(40(1 − 𝑋) ≤ 𝑦) = 𝑃(𝑋 > 1 − 40) =
1
𝑦 3𝑥2𝑑𝑥
1−40
=1− 1−
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Example 7.5
Let X have the probability density function given by
𝑥+1
,
−1 ≤ 𝑥 ≤ 1
𝑓 𝑥 =
2
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the density function for Y=X2.
Answer: In the earlier section, we found that
𝟏
𝒇𝒀 𝒚 =
[𝒇
𝒚 + 𝒇𝑿 − 𝒚 ]
𝟐 𝒚 𝑿
By substituting into this equation, we have
𝒇𝒀 𝒚 = 𝟐
𝟏
𝒚
𝒚+𝟏
− 𝒚+𝟏
+
𝟐
𝟐
𝟏
=
𝟐 𝒚
, 𝟎≤𝒚≤𝟏
𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
As −𝟏 ≤ 𝒙 ≤ 𝟏, 𝒚 = 𝒙𝟐 𝟎 ≤ 𝒚 ≤ 𝟏
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Summary
Summary of the Distribution Function Method
Let Y be a function of the continuous random variables X1, X2, …, Xn. Then
1.
2.
3.
4.
Find the region Y=y in the (X1, X2, …, Xn) space.
Find the region Y≤y.
Find 𝐹𝑌(𝑦) = 𝑃(𝑌 ≤ 𝑦) by integrating 𝑓(𝑋1, 𝑋2, … , 𝑋𝑛) over the region Y≤y.
Find the density function fY(y) by differentiating FY(y). That is,
𝑑
𝑓𝑌 𝑦 = 𝐹𝑌 𝑦 .
𝑑𝑦
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Part 3. Method of Transformation in One
Dimension
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Introduction
The transformation method for finding the probability distribution of a function
of a random variable is simply a generalization of the distribution function
method.
Consider a random variable X with the distribution function FX(x). Suppose that
Y is a function of X, say, Y=g(X) which is an increasing function with the
inverse X=g-1(Y)=h(Y). Then
We have
𝐹𝑌 𝑦 = 𝑃 𝑌 ≤ 𝑦 = 𝑃 𝑔 𝑋 ≤ 𝑦 = 𝑃 𝑋 ≤ ℎ 𝑦 = 𝐹𝑋[ℎ 𝑦 ]
Then the density function is
𝑑
𝑑
𝑓𝑌 𝑦 = 𝐹𝑌 𝑦 = 𝐹𝑋 ℎ 𝑦 = 𝑓𝑋 ℎ 𝑦 ∙ |ℎ′ 𝑦 |.
𝑑𝑦
𝑑𝑦
Similarly, we can have the same result for g(X) is a decreasing function.
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Theorem 7.1
Transformation of Random Variable. Let X be an absolute continuous random
variable with probability density function
> 0, 𝑥 ∈ 𝐴 = (𝑎, 𝑏)
𝑓𝑋 𝑥 =
0, 𝑥 ∈ 𝐴
Let 𝑌 = 𝑔(𝑋) with inverse function 𝑋 = ℎ(𝑌) such that h is a one-to-one,
continuous function from 𝐵 = (𝛼, 𝛽) onto A. If ℎ’(𝑦) exists and ℎ’(𝑦) ≠ 0 for all
y ∈ 𝐵. Then 𝑌 = 𝑔(𝑋) determines a new random variable with density
𝑓𝑋 ℎ 𝑦 |ℎ′ 𝑦 |, 𝑦 ∈ 𝐵 = (𝛼, 𝛽)
𝑓𝑌 𝑦 =
0, 𝑦 ∈ 𝐵
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Example 7.6
Let X have the probability density function given by
2𝑥,
0≤ 𝑥≤1
𝑓 𝑥 =
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the density function for Y=-2X+5.
5−𝑌
Answer: Here 𝑌 = 𝑔(𝑋) = −2𝑋 + 5  the inverse function 𝑋 = ℎ 𝑌 = 2
where h is a continuous and one-to-one function from B=(3,5) onto A=(0,1).
So ℎ’(𝑦) = −1/2 for any y ∈ 𝐵 .
Then we can have
𝑓𝑌 𝑦 = 𝑓𝑋 ℎ 𝑦
ℎ′ 𝑦
=
2 5−𝑦
2
1
−2 =
5−𝑦
,
2
3 < 𝑦 < 5.
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Summary
Summary of the Univariate Transformation Method
Let Y be the function of the continuous random variables X, Y=g(X). Then
1. Write the probability density function of X.
2. Find the inverse function h such that X=h(Y). Verify that h is a continuous
one-to-tone function from B=(α, β) onto A=(a, b) where for 𝑥 ∈ 𝐴, 𝑓 𝑥 >
0.
𝑑
3. Verify ℎ 𝑦 = ℎ′(𝑦) exists, and is not zero for any 𝑦 ∈ 𝐵.
𝑑𝑦
4. Find 𝑓𝑌(𝑦) by calculating 𝑓𝑋 ℎ 𝑦 |ℎ′ 𝑦 |
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Part 4. Additional Examples
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Additional Example 1
Let X be a random variable having a continuous c.d.f., F(x). Let Y=F(X).
Show that Y has a uniform distribution on (0,1). Conversely, if U has a uniform
distribution on (0,1), show that X = F-1(U) has the c.d.f, F(x).
Answer: F(X) is non-decreasing and has domain 0<F(X)<1, that is, 0<Y<1.
Suppose F(x) has inverse function, ie., y=F(x)x=F-1(y).
Then FY(y)=P(Y ≤ y)=P[F(X) ≤ y]=P[X ≤F-1(y)]=F(F-1(y))=y  fY(y)=1, for 0<y<1.
FX(x)=P(X ≤x)=P(F-1(U) ≤x)=P(U ≤F(x))=F(x).
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Additional Example 2
Show that if U is uniform on (0,1), then X=-log(U) has an exponential
distribution Exp(1).
Answer: The density function for U is fU(u)=1.
X=-log(U) U=exp(-X), so h(x)=e-x,
which is continuous and one-to-one function with B=(0, ∞) as A=(0, 1).
The derivative of h(x) is h’(x)=-e-x which is not zero in the domain.
So we can have
fX(x) =fU[h(x)]|h’(x)|=(1)(|-e-x|)=e-x.
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Homework #10
Page
Page
Page
Page
275:
354:
362:
366:
5.138, 5.140
7.3, 7.4
7.6, 7.8
7.18, 7.20
Due Monday, 04/22/2013
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