Results 1 to 10 of 10
Like Tree4Thanks
  • 1 Post By chiro
  • 1 Post By chiro
  • 1 Post By chiro
  • 1 Post By chiro

Math Help - Probability question

  1. #1
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Probability question

    For two random variables, the "distance" between two distributions is defined to be the maximum, \text{ max all x}|F_{1}(x)-F_{2}(x)| over the range which  F_1 and F_2 are defined, where f(x) is the cumulative distribution function. Find the distance between the following two distributions:

    \text{(i) uniform on the interval }[0,1]
    \text{(ii) pdf is }f(x)=\frac{1}{(x+1)^2}\text{ for }0<x<\infty

    So:
    F_1=x
    F_2=1-\frac{1}{1+x}
    |F_1(x)-F_2(x)|=\Big{|}x-1+\frac{1}{1+x}\Big{|}

    From this point I'm not sure of what to do. Any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: Probability question

    Hey downthesun01.

    You will have to look at two situations: the first is when x is in [0,1] and the second is when x > 1. You will have two situations and you need to find the maximum value of both situations separately.

    In the second situation F1 will be a constant (i.e. 1).
    Thanks from downthesun01
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Re: Probability question

    Thanks. I see what you're saying about the two situations. How do I go about finding the maximum value of each situation? Am I taking the derivative of each situation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: Probability question

    Expand out the absolute value function so that it's analytic (i.e. |x| = x if x > 0 and -x if x < 0) and then use standard calculus to find the maximum point.
    Thanks from downthesun01
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Re: Probability question

    Hmm.. For the situation involving [0,1], I've tried, but I'm getting x= 0, -1 while the book gives x=0, 1

    The solution in the manual lacks any real intermediate steps, I'm left kind of stumped right now
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: Probability question

    You can't get -1 since the range is in [0,infinity). Can you show us how you got -1 as a solution?
    Thanks from downthesun01
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Re: Probability question

    I can't find where I wrote everything down at the moment, but since the range is [0,\infty) shouldn't absolute value function be

    x+\frac{1}{x+1}-1?

    After that, I thought that I take the derivative, which is

    1-\frac{1}{(x+1)^2}

    Set it equal to 0, and find the values x=c such that f'(c)=0 \text{ or }f'(c)\text{ doesn't exist and }f(c) \text{ exists}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: Probability question

    You can only use that function in the interval [0,1]. In the interval (1,infinity) you will use |1 - 1/(1+x) - 1| = |1/(x+1)| = dist(F1,F2) for x > 1.
    Thanks from downthesun01
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Oct 2009
    Posts
    295
    Thanks
    9

    Re: Probability question

    Sorry, I meant [0,1]

    The manual's solution says that it's maximized at either x=0,1, or a critical point; critical points occur where 1-\frac{1}{(x+1)^2}=0 or x=0.

    So, I guess:

    f(0)=1-\frac{1}{(0+1)^2}=0

    I don't see how they got 1 though
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,648
    Thanks
    601

    Re: Probability question

    That takes care of x = 0.

    For x = 1 case we look at maximizing |1 - 1/(x+1)^2 - 1| or |1/(x+1)^2|

    If x > 1 then |1/(x+1)^2| = 1/(x+1)^2.

    We can show that the function is decreasing for all x > -1 since d/dx 1/(x+1)^2 = -2/(x+1)^3 < 0 for x >= 1.

    This implies that the maximum value for x >= 1 is at x = 1 and thus we have the second solution of x = 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 6th 2013, 10:29 AM
  2. Replies: 0
    Last Post: May 5th 2013, 07:32 PM
  3. Replies: 5
    Last Post: February 10th 2013, 02:11 PM
  4. Probability question involving (A given B type question )
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 9th 2009, 09:08 AM
  5. A probability question and an Expectations question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 29th 2006, 06:13 PM

Search Tags


/mathhelpforum @mathhelpforum