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!

Mean and variance of a probability density

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    find μ and σ^2 for the following probability density
    f(x) = x for 0<x<1
    f(x) = 2-x for 1 <= x < 2
    f(x) = 0 elsewhere

    2. Relevant equations
    μ = ∫xf(x)dx

    σ^2 = ∫x^2 f(x)dx - μ^2

    3. The attempt at a solution
    first we find the mean. we split up the integral into sums of different integrals
    since f(x) = 0 from -∞ to 0 and from 2 to ∞ we have
    μ = ∫(0to1)x^2dx + ∫(1to2)x(2-x)dx
    = ∫x^2dx + ∫2xdx - ∫x^2dx
    where the first integral is from 0 to 1 and the others are from 1 to 2
    = (1/3)x^3|(0to1) + x^2|(1to2) - (1/3)x^3|(1to2)
    = 1/3 + (4-1) - (8/3-1/3) = 1/3 + 3 -7/3 = 3-6/3 = 3-2 = 1

    so the mean is 1. now to find the variance
    σ^2 = ∫x^3dx + ∫x^2(2-x)dx - 1^2
    = ∫x^3dx + ∫2x^2dx - ∫x^3dx - 1
    = (1/4)x^4|(0to1) + (2/3)x^3|(1to2) - (1/4)x^4|(1to2) - 1
    = 1/4 + (16/3-2/3) - (16/4-1/4) - 1
    = 1/4 + 14/3 - 15/4 - 1
    = 1.667

    am i doing this correctly?
     
  2. jcsd
  3. Oct 11, 2015 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No. Check your arithmetic for the last line.
     
  4. Oct 11, 2015 #3
    oops. It would be .1667
     
  5. Oct 11, 2015 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, actually it would be 1/6 exactly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Mean and variance of a probability density
  1. Mean and variance (Replies: 1)

Loading...