{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# DMtutorial4s - Tutorial 4 Suggested solutions 1 comparing...

This preview shows pages 1–2. Sign up to view the full content.

Tutorial 4: Suggested solutions 1. comparing the estimators on the top of page 4 (chapter 2 part 1 ) and (2.3) on page 3 of part 2. Explain why there is not weight function in the estimator of page 4. Because the estimation on the top of page 2 (chapter 2 part 1 ) is to estimate a global parameter, we don’t need to estimate them locally. 2. Based on the notation in Lecture note (Chapter 2, part 2), eqns between (2.2) and (2.3), prove (2.3) First, we prove that for any Y and X , then the solution to minimize ( Y - X β ) ( Y - X β ) with respect to β is ˆ β = ( X X ) - 1 X Y (1) Write ( Y - X β ) ( Y - X β ) = ( Y - X ˆ β + X ˆ β - X β ) ( Y - X ˆ β + X ˆ β - X β ) = { ( I - X ( X X ) - 1 X ) Y - X ( ˆ β - β ) } { ( I - X ( X X ) - 1 X ) Y - X ( ˆ β - β ) } = Y ( I - X ( X X ) - 1 X ) Y + ( ˆ β - β ) X X ( ˆ β - β ) Therefore, the minimum point achieved when the second term is 0, i.e. β = ˆ β Let β = ( a 0 , a 1 , ..., a q , b 0 , ..., b q ) . Because n i =1 { Y i - [ a 0 + a 1 x i 1 + ... + a q x nq + b 0 ( Z n - z ) + b n ( Z n - z ) x n 1 + · · · + b q ( Z n - z ) x nq ] } 2 K h ( Z i - z ) = { Y - X β } W { Y - X β } = { W 1 / 2 Y - W 1 / 2 } { W 1 / 2 Y - W 1 / 2 X β } By (1), we have (2.3) 3. Consider the air-pollution and health data in Hong Kong. Suppose we consider the effect of pollutants (NO 2 , SO 2 , O 3 , Particulate matters (PM)) on the number of hospital admission suffering respiratory diseases Y . We fit each pollutant a linear regression model to find the dependence. For example Y = a + b * NO 2 + ε. (2)

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern