Results 1 to 4 of 4

Math Help - Sums and products of random variables

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    5

    Sums and products of random variables

    Hi all,

    Can anyone help me with the below question?

    for each of the following pairs of random variables X,Y, indicate
    a. whether X and Y are dependent or independent
    b. whether X and Y are positively correlated, negatively correlate or uncorrelated

    i. X and Y are uniformly distributed on the disk {(x,y) in R^2: 0<=x^2+y^2<=1}
    since this is a circle with R=1 the area is pi. Since x and y are uniform then fxy(xy) is 1/area = 1/pi. In order to see if x and y are independent i need to computer marginal densities and multiply to see if i get fxy(xy). This is where I am stuck. I know that to get fx i need to integrate fxy(xy) with respect to dy and for fy integrate with respect to dx. But what am i integrating? if fx = integral from -1 to 1 of 1/pi dy? and fy = to integral from -1 to 1 1/pi dx? For correlation i first need COV(XY) which equals = E(XY)-E(X)E(Y). To get E(XY) i integrate xy*fxy(xy) so the double integral from -1 to 1 of xy/pi? Is this correct? Now, since x and y are uniform then E(X) and E(Y) are just the interval over two so since each is between -1 and 1 we get 2/2=1=E(X)=E(Y). The VAR(X) and VAR(Y) is 2^/12, the radical of which gives standard deviation and from these value we can compute rho.

    ii. X and Y are uniformly distributed on the parallelogram {(x,y) in R^2: x-1<=y<=x+1, -1<=x<=1}. the area of the parallelogram is 4, so fxy(xy) is 1/4. to get fx i integrate 1/4 from x-1 to x+1 with respect to dy which yields 1/2. To get fy i integrate 1/4 from -1 to 1 with respect to dx which also yields 1/2. since fx*fy=1/4=fxy(xy) then x and y are independent and thus uncorrelated. correct?

    X and Y are uniformly distributed on the diamond {(x,y) in R^2: |x|+|y|<=1, |x|<=1}. i am not sure about this one. what will the graph of this one look like? is it a diamond on -1<=x<=1 and -1<=y<=1?
    Last edited by dizzle1518; May 16th 2011 at 05:47 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    uniformly distributed on a circle means that they are dependent.
    It's not a rectangle.

    Since P(.3<X<.5|Y=.5)\ne P(.3<X<.5|Y=.99)
    which both would have to be equal to P(.3<X<.5)

    The marginals are easy to compute

    f_X(x)=\int^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}}{dy\over \pi}

    ={2\sqrt{1-x^2}\over \pi}

    on -1<x<1 of course
    Last edited by matheagle; May 25th 2011 at 05:36 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2011
    From
    Sacramento, CA
    Posts
    165
    Quote Originally Posted by matheagle View Post
    uniformly distributed on a circle means that they are dependent. It's not a rectangle.
    Could you elaborate on this? I recall my instructor rambling something about geometry to the chalkboard, but she never actually explained it to the class. I sat in on a couple of classes with a professor that did mention it, but I didn't stay around long enough to gain an intuition with it. I suspect that if it is rectangular, then you can make a movement in one dimension (y-ward) without a change in the other dimension (x-ward). However, on a circle you cannot do that. For every y-ward movement you're also making an x-ward movement. Is this what you are getting at?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    For two rvs to indep it is necessary and suff that their joint density factors and they are defined on some rectangle.

    Look at the example of the unit circle.
    Draw a line where Y=0 (x-axis) and Y=.95.
    The probablity of X in the same interval,given these Y's, say .95<X<1 isn't the same.
    Just like sets, for X and Y to be independent
    Since P(.95<X<1|Y=0)= P(.95<X<1)....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sums and products...
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: November 17th 2011, 03:36 AM
  2. Probability: Sums and Products of Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 12th 2011, 04:44 AM
  3. Probability of Sums of Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 7th 2009, 09:31 PM
  4. sums and products of roots
    Posted in the Algebra Forum
    Replies: 10
    Last Post: September 2nd 2009, 09:32 AM
  5. Replies: 5
    Last Post: October 16th 2008, 01:52 PM

Search Tags


/mathhelpforum @mathhelpforum