Results 1 to 5 of 5

Math Help - change of variable: divide by zero?

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    2

    change of variable: divide by zero?

    Say X is a uniform random var on [-1,1], so  f_X(x)=1/2 on [-1,1]. I want to find the density of Y=g(X)=X^3 over [g(-1),g(1)]=[-1,1]. g is a monotonic function so everything should be kosher. I look up the change of variable formula on wikipedia:

    f_Y(y)=|1/g'(g^{-1}(y))|f_X(g^{-1}(y))

    So I do some algebra:
     g(x)=x^3
     g'(x)=3x^2
     g^{-1}(y)=y^{1/3}
    so g'(g^{-1}(y))=g'(y^{1/3})=3y^{2/3}
    thus
    f_Y(y)=|1/3y^{2/3}|*(1/2)

    But there's a singularity/asymtopes at y=0!... am I doing something wrong? The books I have conveniently choose random vars so there is no divide by zero (for example, if X was on [1,2] or some other interval where g(x) doesn't cross the x-axis, everything works fine).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    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
    Hello,

    X is uniform over [-1,1]. So the probability that it equals 0 is 0. We say that almost surely, X is different from 0. So it's not a problem if there is a singularity at a given point.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    2
    Ah, ok. In fact, the integral of f_Y on [-1,1] seems to be 1. So it is alright for density functions to have asymptotes, as long as the integral over the interval doesn't explode to infinity?
    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
    Well actually, there's just one discontinuity, so it doesn't affect the value of the integral.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2010
    Posts
    2
    The condition for that equality should be "strict monotocity". For a rigirous solution of that problem you can partition the function x^3 into two (for x>0 and x<0) where strict monotocity holds.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. change of variable
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 17th 2009, 05:02 PM
  2. change of variable
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2008, 06:41 AM
  3. change of variable
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 25th 2008, 02:26 AM
  4. change of variable in a PDE
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 8th 2008, 11:43 AM
  5. Change of variable
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 22nd 2006, 08:50 AM

Search Tags


/mathhelpforum @mathhelpforum