hints

# Hints - x − μ 2 2 σ 2 Then du = 1 σ 2 x − μ dx and your Frst integral now will be equal to σ √ 2 π ° e − u du This can be easily

This preview shows page 1. Sign up to view the full content.

Let f ( x ):= 1 2 πσ e ( x μ ) 2 2 σ 2 , then note that ° + −∞ f ( x ) dx =1 , since f ( x ) is a density function. We now split the integral up into two EX = ° + −∞ x · f ( x ) dx = ° + −∞ ( x μ ) · f ( x ) dx + ° + −∞ μ · f ( x ) dx. The second integral is just equal to μ using the previous fact. So you are left to prove that the Frst integral is zero, for this use the substitution u :=
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( x − μ ) 2 2 σ 2 . Then du = 1 σ 2 ( x − μ ) dx and your Frst integral now will be equal to σ √ 2 π ° e − u du. This can be easily solved and you then only need to get the correct integral bounds for u and then it should be fairly obvious that this integral is zero. 1...
View Full Document

## This note was uploaded on 01/21/2011 for the course MATH 3C taught by Professor Schonmann during the Spring '07 term at UCLA.

Ask a homework question - tutors are online