ps1_ak - Problem Set 1 Introduction to Econometrics - Fall...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem Set 1 Introduction to Econometrics - Fall 2006 Due Date: Thursday, September 21st, BEFORE CLASS. Each student is responsible for handing one handwritten solution. If working in groups, indicate the members of the group (to facilitate grading). PLEASE SHOW YOUR WORK . 1. The random variable X is distributed normally with N & & X 2 X ± (a) What is the Pr( X & & X ) ? What is the Pr( X & & X + 1 : 96 ± X ) ? Solution: 0 : 5 , 0 : 975 (b) Let Y = X & & X ± X . Show that Y is distributed normally with N (0 ; 1) ; that is E ( Y ) = 0 and V ar ( Y ) = 1 . Solution: Just use formulas at page 26: E [ Y ] = E ² X ± & X ± X ³ = 1 ± X E [ X ± & X ] since 1 ± X is a constant. Moreover: E [ Y ] = 1 ± X E [ X ± & X ] = 1 ± X ( E [ X ] ± & X ) since & X is another constant. We know E [ X ] = & X ; then: E [ Y ] = 1 ± X ( E [ X ] ± & X ) = 1 ± X 0 Consequently: E [ Y ] = 0 Next, regarding V ar ( Y ) ; by de&nition: V ar ( Y ) = E h ( Y ± E [ Y ]) 2 i = E " ´ X ± & X ± X ± 0 µ 2 # = E " ´ X ± & X ± X µ 2 # = E ² 1 ± 2 X ( X ± & X ) 2 ³ = 1 ± 2 X E h ( X ± & X ) 2 i since 1 ± 2 X is a constant. We know that ± 2 X = E h ( X ± & X ) 2 i by de&nition, so: V ar ( Y ) = 1 ± 2 X E h ( X ± & X ) 2 i = ± 2 X ± 2 X = 1 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(c) Show that E & Y 2 ± = 1 Solution: either use the de&nition of Y = X & & X ± X and compute E ² ³ X & & X ± X ´ 2 µ , or use the fact that V ar ( Y ) = E & Y 2 ± & [ E ( Y )] 2 By de&nition: E & Y 2 ± = E " ² X & & X ± X µ 2 # = E 1 ± 2 X ( X & & X ) 2 · = 1 ± 2 X E h ( X & & X ) 2 i since 1 ± 2 X is a constant. We know that ± 2 X = E h ( X & & X ) 2 i by de&nition, so: E & Y 2 ± = 1 ± 2 X E h ( X & & X ) 2 i = ± 2 X ± 2 X = 1 On the other side, we could have noticed that V ar ( Y ) = E & Y 2 ± & [ E ( Y )] 2 In this case, E ( Y ) = 0 ; so [ E ( Y )] 2 = 0 : Then: E & Y 2 ± = V ar ( Y ) = 1 2. Let the random variable X be the temperature in Madison, WI at 12:00pm on September 21th. You know that X is distributed normally with N (72 ; 100) . (a) What is the probability the temperature is lower than 46 ± F ? What is the probability the temperature is higher than 90 ± F ? What is the probability the temperature is in between 65 ± F and 80 ± F ? Higher than 80 ± F ?
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

ps1_ak - Problem Set 1 Introduction to Econometrics - Fall...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online