elemprob-fall2010-page29

elemprob-fall2010-page29 - Changing to polar coordinates,...

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

View Full Document Right Arrow Icon
Changing to polar coordinates, I 2 = Z π/ 2 0 Z 0 re - r 2 / 2 dr = π/ 2 . So I = p π/ 2, hence R -∞ e - x 2 / 2 dx = 2 π as it should. Note Z xe - x 2 / 2 dx = 0 by symmetry, so E Z = 0. For the variance of Z , we use integration by parts: E Z 2 = 1 2 π Z x 2 e - x 2 / 2 dx = 1 2 π Z x · xe - x 2 / 2 dx. The integral is equal to - xe - x 2 / 2 i -∞ + Z e - x 2 / 2 dx = 2 π. Therefore Var Z = E Z 2 = 1. We say X is a N ( μ,σ 2 ) if X = σZ + μ , where Z is a N (0 , 1). We see that F X ( x ) = P ( X x ) = P ( μ + σZ x ) = P ( Z ( x - μ ) ) = F Z (( x - μ ) ) if σ > 0. (A similar calculation holds if σ < 0.) Then by the chain rule X has density f X ( x ) = F 0 X ( x ) = F 0 Z (( x - μ ) ) = 1 σ
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online