Copyright 1996 Lawrence C. Marsh 0 PowerPoint Slides for Undergraduate Econometrics by
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i ) f(x i ) + Σ
2 (x i ) f(x i ) n i = 1 n i = 1 E
= E
1 (X)] + E
2 (X)] 2.23
Copyright 1996 Lawrence C. Marsh Adding and Subtracting Random Variables E(X-Y) = E(X) - E(Y) E(X+Y) = E(X) + E(Y) 2.24
Copyright 1996 Lawrence C. Marsh E(X+a) = E(X) + a Adding a
constant to a variable will add a constant to its expected value:
will multiply its expected value by that constant:
2.25
Copyright 1996 Lawrence C. Marsh var(X)
= average squared deviations around the mean of X. var(X) = expected value of the squared deviations around the expected value of X. var(X) =
E [(X - EX) ] 2 Variance 2.26
8 Copyright 1996 Lawrence C. Marsh var(X) =
E [(X - EX) ] =
E [X - 2XEX + (EX) ] 2 2 2 =
E(X ) - 2 EX EX + E (EX) 2 2 = E(X ) - 2 (EX) + (EX) 2 2
=
E(X ) - (EX) 2 2 var(X) = E [(X - EX) ] 2 var(X) = E(X ) - (EX) 2 2 2.27 Copyright 1996 Lawrence C. Marsh variance
of a discrete random variable, X: standard deviation is square root of variance var ( X ) = (x i
2 f(x
i ) i = 1 n ∑ 2.28 Copyright 1996 Lawrence C. Marsh x i f(x i ) (x i - EX) (x i - EX) f(x i ) 2 .1 2 - 4.3 = -2.3 5.29 (.1) = .529 3 .3 3 - 4.3 = -1.3 1.69 (.3) = .507 4 .1
.09 (.1) = .009 5 .2 5 - 4.3 = .7 .49 (.2) = .098 6 .3
2.89 (.3) = .867 Σ x i f(x
i ) = .2 + .9 + .4 + 1.0 + 1.8 = 4.3
Σ
i - EX) f(x i ) = .529 + .507 + .009 + .098 + .867 = 2.01
2 calculate the variance for a discrete random variable, X: i = 1 n n i = 1 2.29
Copyright 1996 Lawrence C. Marsh Z = a + cX var(Z) = var(a + cX) = E [(a+cX) - E(a+cX)] = c var(X) 2 2 var(a + cX) = c var(X) 2 2.30 Copyright 1996 Lawrence C. Marsh A
joint probability density function, f(x,y), provides the probabilities associated with the joint occurrence of all of the possible pairs of X and Y. Joint pdf 2.31
college grads in household .15
.05 .45
.35 joint pdf f(x,y) Y = 1
Y = 2 vacation
homes owned
X = 0 X = 1
Survey of College City, NY f (0,1) f (0,2)
f (1,1)
f (1,2)
2.32 9 Copyright 1996 Lawrence C. Marsh E[g(X,Y)] = Σ
g(x i ,y j ) f(x
i ,y j ) i j E(XY) = (0)(1)(.45)+(0)(2)(.15)+(1)(1)(.05)+(1)(2)(.35)= .75
E(XY) = Σ
Σ x
i y j f(x i ,y j ) i j Calculating the expected value of functions of two random variables. 2.33
Copyright 1996 Lawrence C. Marsh The
marginal probability density functions, f(x) and f(y), for discrete random variables, can be obtained by summing over the f(x,y) with respect to the values of Y to obtain f(x) with respect to the values of X to obtain f(y). f(x i
Σ f(x
i ,y j ) f(y
j ) =
Σ f(x
i ,y j ) i j Marginal pdf 2.34
Copyright 1996 Lawrence C. Marsh .15
.05 .45
.35 marginal
Y = 1 Y = 2
X = 0 X = 1
.60 .40
.50 .50
f (X = 1)
f (X = 0)
f (Y = 1)
f (Y = 2)
marginal pdf for Y: marginal pdf for X: 2.35
The
conditional probability density functions of X given Y=y , f(x | y), and of Y given X=x , f(y | x), are obtained by dividing f(x,y) by f(y) to get f(x | y) and by f(x) to get f(y | x).
f(x | y) = f(y | x) = f(x,y) f(x,y)
f(y) f(x)
Conditional pdf 2.36
Copyright 1996 Lawrence C. Marsh .15
.05 .45
.35 conditonal Y = 1 Y = 2
X = 0 X = 1
.60 .40
.50 .50
.25 .75
.875 .125
.90 .10
.70 .30
f (Y=2
| X= 0)=.25 f (Y=1
| X = 0)=.75 f (Y=2
| X = 1)=.875 f (X=0
| Y=2)=.30
f (X=1
| Y=2)=.70
f (X=0
| Y=1)=.90
f (X=1
| Y=1)=.10
f (Y=1
| X = 1)=.125 2.37
X and Y are independent random
variables if their joint pdf, f(x,y), is the product of their respective marginal pdfs, f(x) and f(y) . f(x
i ,y j ) = f(x i ) f(y j ) for independence this must hold for all pairs of i and j Independence 2.38
10 Copyright 1996 Lawrence C. Marsh .15
.05 .45
.35 not independent Y = 1 Y = 2
X = 0 X = 1
.60 .40
.50 .50
f (X = 1)
f (X = 0)
f (Y = 1)
f (Y = 2)
marginal pdf for Y: marginal pdf for X: .50x.60= .30
.50x.60= .30
.50x.40= .20
.50x.40= .20
The calculations in the boxes show the numbers required to have independence . 2.39 Copyright 1996 Lawrence C. Marsh The
covariance between two random variables, X and Y, measures the linear association between them. cov(X,Y) = E[(X - EX)(Y-EY)] Note that variance is a special case of covariance. cov(X,X) = var(X) = E[(X - EX) ]
Covariance 2.40
cov(X,Y) =
E [(X - EX)(Y-EY)] =
E [XY - X EY - Y EX + EX EY] =
E(XY) - 2 EX EY + EX EY =
E(XY) - EX EY cov(X,Y) =
E [(X - EX)(Y-EY)] cov(X,Y) = E(XY) - EX EY =
E(XY) - EX EY - EY EX + EX EY 2.41
Copyright 1996 Lawrence C. Marsh .15
.05 .45
.35 Y = 1
Y = 2 X = 0
X = 1 .60
.40 .50
.50 EX=0(.60)+1(.40)= .40 EY=1(.50)+2(.50)= 1.50 E(XY) = (0)(1)(.45)+(0)(2)(.15)+(1)(1)(.05)+(1)(2)(.35)= .75 EX EY = (.40)(1.50) = .60 cov(X,Y) = E(XY) - EX EY = .75 - (.40)(1.50) = .75 - .60 = .15 covariance 2.42 Copyright 1996 Lawrence C. Marsh The
correlation between two random variables X and Y is their covariance divided by the square roots of their respective variances. Correlation is a pure number falling between -1 and 1. cov(X,Y)
ρ (X,Y) =
var(X) var(Y) Correlation 2.43
.15
.05 .45
.35 Y = 1
Y = 2 X = 0
X = 1 .60
.40 .50
.50 EX=
.40 EY=
1.50 cov(X,Y) = .15 correlation EX=0(.60)+1(.40)= .40
2 2 2 var(X) = E(X ) - (EX)
= .40 - (.40) = .24 2
2 EY=1(.50)+2(.50) = .50 + 2.0 = 2.50 2
2 var(Y) = E(Y ) - (EY) = 2.50 - (1.50) = .25
2 2 2 ρ (X,Y) =
cov(X,Y) var(X) var(Y) ρ (X,Y) =
.61 2.44
11 Copyright 1996 Lawrence C. Marsh Independent random variables have zero covariance and, therefore, zero correlation. The converse is not true. Zero Covariance & Correlation 2.45
The expected value of the weighted sum of random variables is the sum of the expectations of the individual terms. Since expectation is a linear operator, it can be applied term by term. E[c 1
2 Y] = c
1 EX + c
2 EY E[c 1 X 1 +...+ c n X n ] = c
1 EX 1 +...+ c n EX n In general, for random variables X 1 , . . . , X n : 2.46 Copyright 1996 Lawrence C. Marsh The
variance of a weighted sum of random variables is the sum of the variances, each times the square of the weight, plus twice the covariances of all the random variables times the products of their weights. var(c 1
+ c 2 Y)=c 1 var(X)+c 2 var(Y)
+ 2c
1 c 2 cov(X,Y) 2 2 var(c 1 X −
c 2 Y) = c
1 var(X)+c
2 var(Y)
− 2c
1 c 2 cov(X,Y) 2 2 Weighted sum of random variables: Weighted
of random variables: 2.47
The Normal Distribution Y ~ N( β
σ 2 ) f(y) = 2
π
σ 2 1 exp β y f(y) 2 σ 2 (y - β ) 2 - 2.48 Copyright 1996 Lawrence C. Marsh The Standardized Normal Z ~ N( 0
1 ) f(z) = 2
π 1 exp 2 z 2 - Z = (y - β )/
2.49 Copyright 1996 Lawrence C. Marsh P [ Y > a ] = P > = P Z > a - β
β Y -
β σ σ σ β y f(y) a Y ~ N( β , σ 2 ) 2.50 12 Copyright 1996 Lawrence C. Marsh P [ a < Y < b ] = P < <
= P < Z < a - β
β σ σ b - β σ a - β σ b - β σ β y f(y) a Y ~ N(
β , σ 2 ) b 2.51 Copyright 1996 Lawrence C. Marsh Y 1 ~ N( β 1 , σ 1 2 ), Y
2 ~ N(
β 2 , σ 2 2 ), . . . , Y n ~ N( β n , σ n 2 ) W = c
1 Y 1 + c 2 Y 2 + . . . + c n Y
Linear combinations of jointly normally distributed random variables are themselves normally distributed. W ~ N
[ E(W), var(W) ] 2.52
Copyright 1996 Lawrence C. Marsh mean: E[V] = E[ χ (m)
] = m
If Z 1 , Z
2 , . . . , Z m denote m independent N(0,1) random variables, and V = Z
1 + Z
2 + . . . + Z m , then V ~ χ (m)
2 2 2 2 V is
chi-square with m degrees of freedom. Chi-Square variance: var[V] = var[ χ (m)
] = 2m
If Z 1 , Z
2 , . . . , Z m denote m independent N(0,1) random variables, and V = Z
1 + Z
2 + . . . + Z m , then V ~ χ (m)
2 2 2 2 V is
chi-square with m degrees of freedom. 2 2 2.53
Copyright 1996 Lawrence C. Marsh mean: E[ t ] = E[
t (m)
] = 0 symmetric about zero
variance: var[ t ] = var[ t (m)
] = m / (m −
) If Z ~ N(0,1) and V ~ χ (m)
and if Z and V are independent then, ~
t (m) t is student-t with m degrees of freedom. 2 t =
Z V m Student - t 2.54
Copyright 1996 Lawrence C. Marsh If V
1 ~
χ (m
) and V
2 ~
χ (m
) and if V
1 and V
2 are independent, then ~ F (m
,m
) F is an F statistic with m
1 numerator degrees of freedom and m 2 denominator degrees of freedom. 2 F =
V 1 m
1 V 2 m 2 2 F Statistic 2.55
Copyright © 1997 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Copyright Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein. 3.1
13 Copyright 1996 Lawrence C. Marsh 1. Estimate a relationship among economic variables, such as y = f(x). 2. Forecast or predict the value of one variable, y, based on the value of another variable, x. Purpose of Regression Analysis 3.2
Copyright 1996 Lawrence C. Marsh Weekly Food Expenditures y = dollars spent each week on food items. x = consumer’s weekly income. The relationship between x and the expected value of y , given x, might be
: E(y|x) = β 1
+ β 2 x 3.3
Copyright 1996 Lawrence C. Marsh f(y|x=480) f(y|x=480) y µ y|x=480 Figure 3.1a Probability Distribution f(y|x=480) of Food Expenditures if given income x=$480. 3.4
Copyright 1996 Lawrence C. Marsh f(y|x)
f(y|x=480) f(y|x=800) y µ
µ y|x=800
Figure 3.1b Probability Distribution of Food Expenditures if given income x=$480 and x=$800. 3.5
{ β 1 ∆ x ∆ E(y|x)
E(y|x) Average
Expenditure x (income) E(y|x)= β
+ β 2 x β 2 = ∆ E(y|x) ∆ x Figure 3.2 The Economic Model: a linear relationship between avearage expenditure on food and income. 3.6
Copyright 1996 Lawrence C. Marsh . . x t x 1 =480 x 2 =800 y t f(y t ) Figure 3.3. The probability density function for y t at two levels of household income, x t expenditure Homoskedastic Case income
3.7 14 Copyright 1996 Lawrence C. Marsh . x
t x 1 x 2 y t f(y
t ) Figure 3.3+. The variance of y t increases
t , increases. expenditure Heteroskedastic Case x 3 . . income
3.8 Copyright 1996 Lawrence C. Marsh Assumptions of the Simple Linear Download 0.54 Mb. Do'stlaringiz bilan baham: |
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