1. Jiaping Wang Department of Mathematical Science 04/10/2013, Wednesday
Chapter 7. Functions of Random Variables Sections : 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

2. Outline Functions of Discrete Random Variables
Methods of Distribution Functions
Method of Transformations in One Dimension
More Examples
Homework #10

3. Part 1. Functions of Discrete Random Variables

4. 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 1 2 3
p(x)
0.5
0.3
0.15
0.05 c
150
200
250
300
p(c)

5. 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 1
p(x)
0.25
0.5 y
p(y) y 1
p(y)
0.5

6. 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 𝑟−1 𝑝𝑟𝑞𝑥, 𝑥=0, 1, 2, …
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 𝑟−1 𝑝𝑟𝑞𝑦−𝑟, 𝑦=𝑟, 𝑟+1, 𝑟+2, …

7. Part 2. Method of Distribution Functions

8. 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:
𝒇𝒀 𝒚 = 𝒅 𝒅𝒚 𝑭𝒀 𝒚 = 𝒅 𝒅𝒚 𝑭𝑿 𝒚 − 𝒅 𝒅𝒚 𝑭𝑿 − 𝒚
= 𝟏 𝟐 𝒚 𝒇𝑿 𝒚 + 𝟏 𝟐 𝒚 𝒇𝑿 − 𝒚

9. Application in Normal Distribution
X is standard normal random variable, what is the probability density function of Y=X2?
We know 𝑓𝑋 𝑥 = 1 2𝜋 exp − 𝑥2 2 , −∞<𝑥<∞, thus for y≥0,
𝑓𝑌 𝑦 = 1 2 𝑦 [ 1 2𝜋 exp − ( 𝑦 ) 𝜋 exp − (− 𝑦 )2 2 ]= 𝜋 y− 1 2 exp(− y 2 )
Recall that Γ = 𝜋 , we can see Y follows a gamma distribution with parameters α=1/2 and β=2.

10. 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
f x = 3𝑥2, 0≤𝑥≤1 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:
𝐹𝑌(𝑦)=𝑃(𝑌≤𝑦)=𝑃(40(1−𝑋)≤𝑦)=𝑃(𝑋>1− 𝑦 40 )= 1− 𝑦 𝑥2𝑑𝑥=1− 1− 𝑥 40 3, 0≤𝑦≤40. 1− 𝑦 𝑥2𝑑𝑥=1− 1− 𝑥 40 3, 0≤𝑦≤40.
For density function, we can obtain it by differentiating the distribution function
𝑓𝑌 𝑦 = − x 40 2, 0≤𝑦≤40.

11. Example 7.5 Let X have the probability density function given by
𝑓 𝑥 = 𝑥+1 2 , −1≤ 𝑥≤1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the density function for Y=X2.
Answer: In the earlier section, we found that
𝒇𝒀 𝒚 = 𝟏 𝟐 𝒚 [𝒇𝑿 𝒚 +𝒇𝑿 − 𝒚 ]
By substituting into this equation, we have
𝒇𝒀 𝒚 = 𝟏 𝟐 𝒚 𝒚 +𝟏 𝟐 + − 𝒚 +𝟏 𝟐 = 𝟏 𝟐 𝒚 , 𝟎≤𝒚≤𝟏 𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
As −𝟏≤𝒙≤𝟏, 𝒚=𝒙𝟐 𝟎≤𝒚≤𝟏

12. Summary Summary of the Distribution Function Method
Let Y be a function of the continuous random variables X1, X2, …, Xn. Then
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, 𝑓𝑌 𝑦 = 𝑑 𝑑𝑦 𝐹𝑌 𝑦 .

13. Part 3. Method of Transformation in One Dimension

14. 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
𝑓𝑌 𝑦 = 𝑑 𝑑𝑦 𝐹𝑌 𝑦 = 𝑑 𝑑𝑦 𝐹𝑋 ℎ 𝑦 =𝑓𝑋 ℎ 𝑦 ∙|ℎ ′ 𝑦 |.
Similarly, we can have the same result for g(X) is a decreasing function.

15. Theorem 7.1
Transformation of Random Variable. Let X be an absolute continuous random variable with probability density function
𝑓𝑋 𝑥 = >0, 𝑥∈𝐴=(𝑎,𝑏) 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
𝑓𝑌 𝑦 = 𝑓𝑋 ℎ 𝑦 | ℎ ′ 𝑦 |, 𝑦∈𝐵=(𝛼,𝛽) 0, 𝑦∈ 𝐵

16. Example 7.6 Let X have the probability density function given by
𝑓 𝑥 = 2𝑥, 0≤ 𝑥≤1 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Find the density function for Y=-2X+5.
Answer: Here 𝑌=𝑔(𝑋)=−2𝑋+5  the inverse function 𝑋=ℎ 𝑌 = 5−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.

17. Summary Summary of the Univariate Transformation Method
Let Y be the function of the continuous random variables X, Y=g(X). Then
Write the probability density function of X.
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.
Verify 𝑑 𝑑𝑦 ℎ 𝑦 =ℎ′(𝑦) exists, and is not zero for any 𝑦∈𝐵.
Find 𝑓𝑌(𝑦) by calculating 𝑓𝑋 ℎ 𝑦 | ℎ ′ 𝑦 |

18. Part 4. Additional Examples

19. 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).

20. 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.

21. Homework #10
Page 275: 5.138, 5.140
Page 354: 7.3, 7.4
Page 362: 7.6, 7.8
Page 366: 7.18, 7.20
Due Monday, 04/22/2013