Category Springer Texts in Business and Economics

Moment Generating Function Method

a. If Xi,.., Xn are independent Poisson distributed with parameters (Xi) respec­tively, then from problem 2.14c, we have

MXi (t) = eAi(e-1) for i = 1,2,… ,n

n n

Y = Xi has My(t) = П MXi (t) since the X/s are independent. Hence

i=i i=i

n

Ai (e‘-l)

MY(t) = ei=1

n

which we recognize as a Poisson with parameter ^ Xi.

i=i

b. IfXi, ..,Xn are IIN (^i, a2), then from problem 2.14b, we have MXi(t) = ew‘+ 1ai2‘2 for i = 1,2,.., n

nn

Y = ^ Xi has MY(t) = ]""[ MXi (t) since the X/s are independent. Hence

i=i i=i

image101

MY(t) = e 1=i

c. If Xi,.., Xn are IIN(|a,, a2),thenY = J2 Xi is N(np,,na2) from part b using

i=i

the equality of means and variances. Therefore, X = Y/nis N(p,, a2/n).

d. If Xi,.., Xn are independent x2 distributed with parameters (ri) respectively, then from problem 2...

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Multiple Regression Analysis

4.1 The regressions for parts (a), (b), (c), (d) and (e) are given below.

a. Regression of LNC on LNP and LNY Dependent Variable: LNC

Analysis of Variance

Sum of

Mean

Source

DF

Squares

Square

F Value

Prob>F

Model

2

0.50098

0.25049

9.378

0.0004

Error

43

1.14854

0.02671

C Total

45

1.64953

Root MSE

0.16343

R-square

0.3037

Dep Mean

4.84784

Adj R-sq

0.2713

C. V.

3.37125

Parameter Estimates

Parameter

Standard

T for HO:

Variable

DF

Estimate

Error Parameter=0

Prob>|T|

INTERCEP 1

4.299662

0.90892571

4.730

0.0001

LNP

1

-1.338335

0.32460147

-4.123

0.0002

LNY

1

0.172386

0.19675440

0.876

0.3858

Regression of LNC on LNP Dependent Variab...

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A Review of Some Basic Statistical Concepts

2.1 Variance and Covariance of Linear Combinations of Random Variables.

a. Let Y = a + bX, then E(Y) = E(a + bX) = a + bE(X). Hence,

var(Y) = E[Y – E(Y)]2 = E[a + bX – a – bE(X)]2 = E[b(X – E(X))]2 = b2E[X – E(X)]2 = b2 var(X).

Only the multiplicative constant b matters for the variance, not the additive constant a.

b. Let Z = a + bX + cY, then E(Z) = a + bE(X) + cE(Y) and

var(Z) = E[Z – E(Z)]2 = E[a + bX + cY – a – bE(X) – cE(Y)]2 = E[b(X – E(X)) + c(Y – E(Y))]2

= b2E[X-E(X)]2 + c2E[Y-E(Y)]2+2bc E[X-E(X)][Y-E(Y)]

= b2var(X) + c2var(Y) + 2bc cov(X, Y).

c. LetZ = a+bX+cY, andW = d+eX+fY, thenE(Z) = a+bE(X) + cE(Y) E(W) = d + eE(X) + fE(Y)

and

cov(Z, W) = E[Z – E(Z)][W – E(W)]

= E[b(X-E(X))+c(Y-E(Y))][e(X-E(X))+f(Y-E(Y))]

= be var(X) + cf var(Y) + (bf + ce) cov(X, Y).

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The back up regressions are given below. These are performed using SAS

a. Dependent Variable: EMP

Analysis of Variance

Sum of

Mean

Source

DF

Squares

Square

F Value

Prob>F

Model

1

2770902.9483

2770902.9483

4686.549

0.0001

Error

73

43160.95304

591.24593

C Total

74

2814063.9013

RootMSE 24.31555 R-square 0.9847

DepMean 587.94141 Adj R-sq 0.9845

C. V. 4.13571

Parameter Estimates

Variable

DF

Parameter

Estimate

Standard

Error

T for H0: Parameter=0

Prob> |T|

INTERCEP

1

-670.591096

18.59708022

-36.059

0.0001

RGNP.1

1

1.008467

0.01473110

68.458

0.0001

Durbin-Watson D 0.349

(For Number of Obs.) 75

1st Order Autocorrelation 0.788

c. Dependent Variable: RESID

Analysis of Variance

Sum of

Mean

Source

DF

Squares

Square

F Va...

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Simple Versus Multiple Regression Coefficients. This is based on Baltagi (1987)

The OLS residuals from Yi = у + 82v2i + 83v3i + wi, say tVi, satisfy the

following conditions:

n n n

wі = 0 ^]iViv2i = 0 ^^wiV 3i = 0

i=i i=i i=i

with Yi = у + 82V2i + 83V3i + vi.

n

Multiply this last equation by v2i and sum, we get P Yiv2i = °2 P v? i +

Подпись: i=iimage163
i=i

"ols = Y ^VEv.2 = E Yix^]xi2 = Y WExi2.

i=1 i=1 i=1 i=1 i=1 i=1

b. Regressing a constant 1 on Xi we get

n n n

b = ‘Y/ Ъ/Y Xi2 with residuals wi = 1 — I nX^ ^ XiM Xi i=1 i=1 i=1

nn

so that, regressing Yi on wi yields a = wiYi/ w2.

i=1 i=1

_ n _ n

n _ _ n n nY EXi2-n^XiYi

But P wiYi = nY — nX £ XiYi/ P Xi2 = —i=^-5———————— ^——- and

i=1 i=1 i=1 P Xi2

i=1

_ n n 2 n

n 2 2 2nX X Xi n X Xi2-n2X2 n X Xi2

P w2 = n + ———– ^ .

i=1 P Xi2 P Xi2 P Xi2 P Xi2

i = 1 i = 1 i = 1 i=1

xi2

Подпись: xi2 i=1 i=1

n

X XiYi – nXY

Подпись: Y

image166

a=4 = Y – " olsX

xi2

image167
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