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: Expectation and Variance for Continous Uniform RV

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data

    8. Suppose that X and Y are independent continuous random variables, and each is uniformly distributed on the interval [0,1] (thus the pdfs for X and Y are zero outside of this interval and equal to one on [0,1]).

    (a) Find the mean and variance for X+Y.

    (b) Calculate and graph the pdf for X+Y (using the convolution formula, and be careful to remember that the pdf of X vanishes outside of [0,1], etc).

    2. Relevant equations

    fX+Y = fX * fY

    X + Y = Z


    3. The attempt at a solution

    I know that the density functions for X and Y are both 1 since the continuous distribution is 1/(b-a). To get the density function for Z, I calculated [tex]\int[/tex] fX(x) * fY(z-x) dx from 0 to z. Since this simplifies to [tex]\int[/tex] 1 dx from 0 to z. I get fZ(z) = z.

    Using the density function of Z, I calculate the mean, which is E(Z = z) = [tex]\int[/tex] z2 dz from 0 to 1. I get 1/3 for my mean when I evaluate the integral.

    To get the second moment, i.e. E(Z2) : [tex]\int[/tex] z3 dz from 0 to 1. When I evaluate this I get 1/4. To get the variance I then get E(Z2) - E(Z)2 which is 1/4 - 1/9 = 5/36.

    I don't know if this is right though.
     
  2. jcsd
  3. Apr 5, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Remember that X and Y take on values between 0 and 1. So X+Y can take on values between 0 and 2. Draw some lines for z = x+y = various constants on your square. If you calculate [itex] P(Z\le z)[/itex] directly from your uniform distribution on the square, you will probably see what the instructors caveat about the densities being 0 off the square has to do with it.
     
  4. Apr 5, 2010 #3
    I calculated the mean of X+Y to be 1. The variance resulted in 1/6. I am not getting the same answer when I use the density function for Z that I tried calculating. Even when I use the limits from 0 to 2, I get a different mean and variance is negative which tells me I am wrong somewhere.

    I'm quite sure I did part a correctly, but I am still struggling with part b.
     
  5. Apr 5, 2010 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, you don't have part a correct. Just check it:

    [tex]\hbox{Does }\int_0^2 f_Z(z) dz = 1\hbox{ ?}[/tex]

    If you show your calculations it might be possible to point out your mistake. If you are using the convolution formulas, are you starting something like this, using the convolution formula:

    [tex]f_Z(z) = \int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx = \int_{0}^{1}1\cdot f_Y(z-x)dx [/tex]

    If so, you need to take care about where the values of z lie and the value of fY
     
  6. Apr 5, 2010 #5
    Here is my work for part a. I got the answer without finding the pdf for X+Y. When I tried using pdf for X+Y which I called Z, the answers weren't matching. [PLAIN]http://img338.imageshack.us/img338/1226/partaprob.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Apr 5, 2010 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That is because you haven't calculated the pdf of X+Y correctly. I have given you two different suggestions how to do that.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook