1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Probability density functions

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data

    f(x)=&(x-a)exp((-(x-a)^2)/b) where a and b are constants

    2. Relevant equations

    find & in terms of b:

    show that the expected value of X is given by
    X=a + sqrt(pi*b/4)
    identity given
    x(x-a)=(x-a)^2+a(x-a)
    and integral from 0 to infinity of x^2*exp-x^2 dx=sqrt (pi) /4

    3. The attempt at a solution

    i found &=2/b and thought my solution was coherent but seeing as i cant answer the next question im confused as to where i went wrong .
    i manage to find X= a + sqrt(pi/4) but cant get that b into the square root no matter what i try .
    i separated into 2 integrals using the first identity then set Y=(x-a)/sqrt b and used the second identity to get sqrt (pi /4)( the other integral giving the expected a)
     
  2. jcsd
  3. Apr 17, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Did you remember to write dx in terms of dy when you did the substitution?

    By the way, it would help in the future if you provide the complete problem statement. You didn't tell us what the domain of f(x) was, for instance.
     
  4. Apr 17, 2010 #3
    im sorry the domain of fx is the function provided for x>=a and 0 otherwise
     
  5. Apr 17, 2010 #4
    also the probability density function is a Rayleigh distribution
     
  6. Apr 17, 2010 #5
    using wikipedia i found the correct answer (using the formulas that use the variance and such) but id still like to know how to recalculate it using the identities given so my question still stands :D
     
  7. Apr 17, 2010 #6
    thx vela problem solved :D
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook