Lecture 06-2007

# Lecture 06-2007 - LectureVI 1 x2 2 f x = e 2 First,...

This preview shows pages 1–8. Sign up to view the full content.

Lecture VI

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

View Full Document
The order of proof of the normal distribution  function is to start with the standard normal: ( 29 2 2 2 1 x e x f - = π
First, we need to demonstrate that the distribution  function does integrate to one over the entire sample  space, which is -  to  .  This is typically accomplished  by proving the constant. Let us start by assuming that - - = dy e I y 2 2

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

View Full Document
Squaring this expression yields ∫ ∫ - - + - - - - - = = dx dy e dx e dy e I x y x y 2 2 2 2 2 2 2 2
Polar Integration: The notion of polar integration is  basically one of a change in variables.  Specifically, some integrals may be ill-posed in the  traditional Cartesian plane, but easily solved in a  polar space. By polar space, any point ( x , y ) can be written in a  trigonometric form:

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

View Full Document
( 29 ( 29 ( 29 θ θ θ sin cos tan 1 2 2 r x r y y x y x r = = = + = -
As an example, take Some of the results for this function are: ( 29 2 1 15 2 f x x x = + -

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