Solutions Exercises Econometrics

PART ONE Solutions to Exercises

Chapter 2

Review of Probability

Solutions to Exercises

1. (a) Probability distribution function for Y Outcome (number of heads) probability Y=0 0.25 Y=1 0.50 Y=2 0.25

(b) Cumulative probability distribution function for Y Outcome (number of heads) Probability Y −1.96 and

< −1.96. Solving these inequalities yields n ≥ 9220.

18. Pr (Y = $0) = 0.95, Pr (Y = $20000) = 0.05. (a) The mean of Y is

μY = 0 × Pr (Y = $0) + 20, 000 × Pr (Y = $20000) = $1000.

The variance of Y is

2 σ Y = E ⎢ ( Y − μY ) ⎢ ⎡ ⎣ 2⎤ ⎥ ⎥ ⎦

= (0 − 1000)2 × Pr (Y = 0 ) + (20000 − 1000)2 × Pr (Y = 20000) = (−1000)2 × 0.95 + 19000 2 × 0.05 = 1.9 × 10 7, so the standard deviation of Y is σ Y = (1.9 × 107 ) 2 = $4359.

1

2 (b) (i) E (Y ) = μY = $1000, σ Y = σnY = 1.9×10 = 1.9 × 10 5. 100

2 7

(ii) Using the central limit theorem,

Pr (Y > 2000) = 1 − Pr (Y ≤ 2000) ⎛ Y − 1000 2, 000 − 1, 000 ⎞ = 1 − Pr ⎜ ≤ ⎟ 5 1.9 × 10 5 ⎠ ⎝ 1.9 × 10 ≈ 1 − Φ (2.2942) = 1 − 0.9891 = 0.0109.

Solutions to Exercises in Chapter 2

11

19. (a) Pr (Y = y j ) = ∑ Pr ( X = xi , Y = y j )

i =1 l l

= ∑ Pr (Y =y j|X =xi )Pr ( X =xi )

i =1

(b)

E (Y ) = ∑ y j Pr (Y = yj ) = ∑ yj ∑ Pr (Y = yj |X = xi ) Pr ( X = xi )

j =1 j =1 i =1 k k l

=∑

l

⎛ k ⎜ ⎜ ⎜ i =1 ⎜ j =1 ⎝ l

∑ yj Pr (Y = yj |X = xi ) ⎟ Pr ( X =xi )

⎟ ⎟ ⎠

⎞ ⎟

=∑ E (Y |X =xi )Pr ( X =xi ).

i =1

(c) When X and Y are independent, Pr (X = xi , Y = yj ) = Pr (X = xi )Pr (Y = yj ), so

σ XY = E[( X − μ X )(Y − μY )]

=∑ ∑ ( xi −μX )( y j −μY ) Pr ( X =xi , Y =y j )

i =1 j =1 l k l k

=∑ ∑( xi −μX )( y j −μY ) Pr ( X =xi ) Pr (Y =y j )

i =1 j =1

⎞ ⎛ l ⎞⎛ k = ⎜ ∑ ( xi − μ X ) Pr ( X = xi ) ⎟ ⎜ ∑ ( yj − μY ) Pr (Y = yj ⎟ ⎝ i=1 ⎠ ⎝ j =1 ⎠ = E ( X − μ X )E(Y − μY ) = 0 × 0 = 0,

cor (X , Y ) =

l m

σ XY 0 = = 0. σ XσY σ XσY

20. (a) Pr (Y = yi ) = ∑∑ Pr (Y = yi |X = xj , Z = zh ) Pr (X = xj , Z = zh )

j =1 h=1

(b)

E (Y ) = ∑ yi Pr (Y = yi ) Pr (Y = yi )

i =1 k

k...