• wonderlic tests
  • EXAM REVIEW
  • NCCCO Examination
  • Summary
  • Class notes
  • QUESTIONS & ANSWERS
  • NCLEX EXAM
  • Exam (elaborations)
  • Study guide
  • Latest nclex materials
  • HESI EXAMS
  • EXAMS AND CERTIFICATIONS
  • HESI ENTRANCE EXAM
  • ATI EXAM
  • NR AND NUR Exams
  • Gizmos
  • PORTAGE LEARNING
  • Ihuman Case Study
  • LETRS
  • NURS EXAM
  • NSG Exam
  • Testbanks
  • Vsim
  • Latest WGU
  • AQA PAPERS AND MARK SCHEME
  • DMV
  • WGU EXAM
  • exam bundles
  • Study Material
  • Study Notes
  • Test Prep

Chapter 2 Answer Key

Testbanks Dec 29, 2025 ★★★★★ (5.0/5)
Loading...

Loading document viewer...

Page 0 of 0

Document Text

Chapter 2 Answer Key

  • A random variable is random because its numerical value is unknown until the
  • variable is realized or observed.

  • A discretely distributed random variable can, with some probability, take any
  • one of a discrete set of values.

    3.m X i=1 Pi=P1+P2+P3+:::::::::::+Pm1+Pm= 1 4.1 6 + 1 6 + 1 6 = 3 6 = 1 2

  • Here there must be onlym=4 possible states, since the given probabilities add
  • up to one. The expected value for this discretely distributed random variable is thus P 4 j=1 cjpjwherec1= 5; c2= 8; c3= 9; c4= 11 andp1= 0:1; p2= 0:4; p3= 0:2; p4= 0:3 So = 4 X j=1 cjpj= (5)(0:1) + (8)(0:4) + (9)(0:2) + (11)(0:3) = 8:8 and V ar(Y) =E[(Y) 2 ]

= (0:1)[58:8]

2

+ (0:4)[88:8]

2

+ (0:2)[98:8]

2

+ (0:3)[118:8]

2

= 3:156

  • The expected value of a random variable is a xed number that can be calcu-
  • lated with a given probability distribution. It will be a xed number across any set of samples taken. Whereas,

Yis the random variable whose realization is the xed value yin a particular sample.

7.

E[Y] =

1 8

[1 + 2 + 3 + 4] +

1 4

[7 + 8] = 5:0

var[Y] = 8 X i=1 pi[ci] 2

= [(15)

2

+ (25)

2

+ (35)

2

+ (45)

2

](1=8) + [(75)

2

+ (85)

2

](1=4)

= 7 1 There is no Answer Key for Chapter 1 (Fundamentals of Applied Econometrics 1e Richard A. Ashley) (Solution Manual, For Complete File, Download link at the end of this File) 1 / 3

  • E[Y] is usually unknown because a probability distribution of the random vari-
  • able is usually unknown.

  • Thek
  • th central moment around 0 is equal toE[(Y0) k

] =E[Y

k ]. Thek th central moment around the mean is equal toE[(Y)] k . Thek th central moment around 0 measures how any distribution of a random variable varies around the specic value of zero. The dierence between thek th central moment around the mean is a measure how the random variable Y varies around the mean.

10.E[Y 3

] = (:2)(1

3

)+(:2)(2

3

)+(:2)(3

3

)+(:1)(4

3

)+(:1)(5

3

)+(:1)(6

3

)+(:1)(7

3

) = 82

var[Y 3

] = (:2)(1

3 82) 2

+ (:2)(2

3 82) 2

+ (:2)(3

3 82) 2

+ (:1)(4

3 82) 2 +

(:1)(5

3 82) 2

+ (:1)(6

3 82) 2

+(:1)(7

3 82) 2

= 11836:6

  • A realized value y can take on any real value from [1;1]
  • 12.

E[Y3] =E[Y]3 = 11:343 = 8:34

13.

E[3Y] = 3E[Y] = (3)(25) = 75

14.

var[Y8] = 11:34

15.var[3Y] = (9)var[Y] = (9)(25) = 225 16.

E[3Y+ 7] = 3E[Y] + 7 = (3)(11) + 7 = 40

var[3Y+ 7] =var[3Y] = 9var[Y] = (9)(3) = 27:

  • Letcov(Y; Z)<0. This means that X and Y have an inverse linear relationship.
  • Note that the covariance does not sensibly quantify the strength of the linear relationship, since the covariance depends on the units in which X and Z are measured, but it indicates that thereisa relationship and that (approximated by a linear relation) it would have a negative slope.

  • The correlation between two random variables quanties the sign and strength
  • of a linear approximation to the relationship, if any, between them. The co- variance does the same thing with regard to the sign, but not the strength of the relationship.

2 2 / 3

  • A correlation can take on any real number between [1;1].
  • 20.c1::::cnmust not be random, they must be xed constants. I.e they cannot depend in anyway on the random variablesY1:::::Yn

  • The covariance of Y and Z isEf[YE(Y)][ZE(Z)]g
  • =EfY ZY EfZg ZEfYg+EfYgEfZgg =EfY Zg EfYgEfZg EfZgEfYg+EfYgEfZg =EfY Zg EfYgEfZg

= 13(4)(3) = 1

  • For Y and Z to be uncorrelated, their covariance must be zero. Therefore,
  • E[Y Z] =E[Y]E[Z], soE[Y Z] = 12.

  • They are not related at all.
  • Since Y and Z are assumed to be independently distributed,p1;4in this case
  • equals the probability of observingYiin state 1 regardless of the value of Z, multiplied by the value ofZiin state 4 regardless of the value of Y.

  • Since Y and Z are independently distributed,
  • E[g(Y)w(Z)] =E[g(Y)]E[w(Z)] Therefore,E[w(z)] = 12 4 = 3

  • Let Y and Z be independent random variables. So thecorrf(Y; Zg= 0. It
  • must follow that thecovfY; Zg= 0.

  • No, ifcovfY; Zg= 0 then it implies that there does not exist a linear relation-
  • ship, but there very well could be a non-linear relationship between the two variables.

  • When two random variables, Y and Z, are identically distributed then this
  • means that both of the variables are realizations of `picks' from the same dis- tribution. Thus, for example, all of the population moments (mean, variance, and higher moments) of Y are equal to the corresponding population moments for Z.

  • If the mean and variance of anormallydistributed variable are given, then
  • everything about its distribution is completely known. If it is not known that the variable is normally distributed, then its distribution is not completely known { e.g., in terms of how symmetrically the variable is distributed around its mean, how thick the tails of its distribution are, etc.

  • / 3

User Reviews

★★★★★ (5.0/5 based on 1 reviews)
Login to Review
S
Student
May 21, 2025
★★★★★

With its step-by-step guides, this document made learning easy. Definitely a impressive choice!

Download Document

Buy This Document

$1.00 One-time purchase
Buy Now
  • Full access to this document
  • Download anytime
  • No expiration

Document Information

Category: Testbanks
Added: Dec 29, 2025
Description:

Chapter 2 Answer Key 1. A random variable is random because its numerical value is unknown until the variable is realized or observed. 2. A discretely distributed random variable can, with some pro...

Unlock Now
$ 1.00