Results 1 to 11 of 11

Math Help - CDF's integration

  1. #1
    Newbie
    Joined
    Apr 2011
    Posts
    6

    CDF's integration

    how would i find the density function which generates the cdf F(x) = x^(N):

    Suppose that x and y are two independent random variables, each of which is uni-
    formly distributed on the unit interval [0; 1]. Find the probability that x + y =z for
    any z between 0 and 2?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    Just take the derivative and get f(x) = \frac{x^{n-1}}{n}, valid on [0, 1].

    X and Y are both continuous random variables, so P(X + Y = z) = 0 for all z. It would be more meaningful to ask for the distribution of X + Y, which involves a little bit of work.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    If I recall correctly f(z)=z on (0,1) and 2-z on (1,2)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Please I'm kind of desperate I've tried to do the calculations, but can't find where my mistake is !

    P(X+Y\leq z)=E[P(X+Y\leq z|Y)]

    And P(X+Y\leq z| Y=y)=z-y, if z-y\in(0,1), that is y\in(z-1,z) ; 0 otherwise.

    - if z<1, there's no problem, I find zē/2 as the cdf
    - if z>1, \displaystyle P(X+Y\leq z)=E[(z-Y)\bold{1}_{y>z-1}]=\int_{z-1}^1 z-y ~dy=2z-z^2-\frac 12+\frac{(z-1)^2}{2}
    by differentiating, it gives the pdf on this part : 1-z, which is negative...

    So what's wrong ?
    Last edited by Moo; April 12th 2011 at 11:32 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    You forgot to deal with the case where z - y \in (1, 2) I think.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Well the probability would be 0, wouldn't it ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    Quote Originally Posted by Moo View Post
    Well the probability would be 0, wouldn't it ?
    More like 1 minus that
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by theodds View Post
    More like 1 minus that
    Oh yeah right, now I can see through it ! Thanks !
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    It's easier to use geometry and see the triangles,
    including the ones for the complement when 1<z<2
    This is a uniform distribution on the unit square.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by matheagle View Post
    It's easier to use geometry and see the triangles,
    including the ones for the complement when 1<z<2
    This is a uniform distribution on the unit square.
    I like conditional stuff and dislike graphic stuff, hence the tortuous method
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Apr 2011
    Posts
    6
    so its 0 or 1?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 05:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum