Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
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Textbook Question
Chapter 6.3, Problem 1E

Let Y be a random variable with probability density function given by

f ( y ) = { 2 ( 1 y ) , 0 0 1 , 0 , elsewhere .

a Find the density function of U1 = 2Y − 1.

b Find the density function of U2 = 1 − 2Y.

c Find the density function of U3 = Y2.

d Find E(U1), E(U2), and E(U3) by using the derived density functions for these random variables.

e Find E(U1), E(U2), and E(U3) by the methods of Chapter 4.

a.

Expert Solution
Check Mark
To determine

Find the density function for U1=2Y1.

Answer to Problem 1E

The density function for U1=2Y1 is fU1(u)=1u21u1.

Explanation of Solution

Calculation:

From the given information, the probability density function for Y is f(y)=2(1y),0y1.

The distribution function for Y is,

FY(y)=0yf(t)dt=0y2(1t)dt=2[tt22]0y=2yy2

From the given information, the random variable U1 is defined as U1=2Y1.

Consider the distribution function for U1,

FU1(u1)=P(U1u)=P(2Y1u)=P(Yu+12)=2(u+12)(u+12)2

Limits for the random variable U1:

The range for the random variable Y is from  0 to 1 and U1=2Y1.

For Y=  0, the value of U1 is  ̶ 1.

For Y=1, the value of U1 is 1.

Hence, the range for the random variable U1 is from  ̶ 1 to 1.

The probability density function for U1 is,

fU1(u)=FU1(u)=ddu(2(u+12)(u+12)2)=(2×12)(2×u+12×12)=1u+12

             =1u2,    1u1

b.

Expert Solution
Check Mark
To determine

Find the density function for U2=12Y.

Answer to Problem 1E

The density function for U2=12Y is fU2(u)=1+u2,1u1.

Explanation of Solution

Calculation:

From the given information, U2=12Y.

Consider the distribution function for U2,

FU2(u)=P(U2u)=P(12Yu)=P(2Yu1)=P(Y>1u2)

             =1F(1u2)=1[2(1u2)(1u2)2]=1[1u(1u2)2]=u+(1u2)2

Limits for the random variable U2:

The range for the random variable Y is from  0 to 1 and U2=12Y.

For Y= 0, the value of U2 is 1.

For Y=1, the value of U2 is  ̶ 1.

Hence, the range for the random variable U2 is from  ̶ 1  to 1.

The probability density function for U2 is,

fU2(u)=FU2(u)=ddu(u+(1u2)2)=1+(21u2×12)=1+u2,1u1

c.

Expert Solution
Check Mark
To determine

Find the density function for U3=Y2.

Answer to Problem 1E

The density function for U3=Y2 is fU3(u)=1u1,0u1.

Explanation of Solution

Calculation:

From the given information, U3=Y2.

Consider the distribution function for U3,

FU3(u)=P(U3u)=P(Y2u)=P(Yu)=F(u)

             =2u(u)2=2uu

The probability density function for U3 is,

fU3(u)=FU3(u)=ddu(2uu)=1u1,0u1

d.

Expert Solution
Check Mark
To determine

Find the value of E(U1) by using the derived density function of U1.

Find the value of E(U2) by using the derived density function of U2.

Find the value of E(U3) by using the derived density function of U3.

Answer to Problem 1E

The value of E(U1) is 13.

The value of E(U2) is 13.

The value of E(U3)  is 16.

Explanation of Solution

Calculation:

The density function for U1 is fU1(u)=1u21u1

Consider,

E(U1)=11ufU1(u)du=11u×1u2du=12[u22u33]11=12×23

             =13

The density function for U2 is fU2(u)=1+u2,1u1.

Consider,

E(U2)=11ufU2(u)du=11u×1+u2du=12[u22+u33]11=12×23

             =13

The density function for U3 is fU3(u)=1u1,0u1.

Consider,

E(U3)=01ufU3(u)du=01u×(1u1)du=[u32(32)u22]01=2312

              =16

e.

Expert Solution
Check Mark
To determine

Find the value of E(U1) by the methods of Chapter 4.

Find the value of E(U2) by the methods of Chapter 4.

Find the value of E(U3) by the methods of Chapter 4.

Answer to Problem 1E

The value of E(U1) is 13.

The value of E(U2) is 13.

The value of E(U3)  is 16.

Explanation of Solution

Calculation:

Result:

Let X be the random variable, then E(aX+b)=aE(X)+b, where a and b are constants.

The density function for Y is f(y)=2(1y),0y1.

Consider,

E(y)=01yf(y)dy=01y×2(1y)dy=2[y22y33]01=13

E(U1)=E(2Y1)=2E(Y)1=2×131=13

E(U2)=E(12Y)=12E(Y)=12×13=13

Consider,

E(y2)=01y2f(y)dy=01y2×2(1y)dy=2[y33y44]01=16

E(U3)=E(Y2)=16

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