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!

Cumulative distribution problem Statistics

  1. Sep 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Let X be a continuous random variable with cumulative distribution function given by F(x) = P(X<x).
    Define a new random variable U=F(X).

    2. Relevant equations


    3. The attempt at a solution
    OK so to solve this problem I first say U=F(X).

    F(u)=P(F(X)<u)

    which means

    F(u)=P(U<u)

    is this how i should solve this problem?
     
    Last edited: Sep 2, 2015
  2. jcsd
  3. Sep 2, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You haven't stated what question is being asked of you.

    Also, you will need to use more careful notation to avoid getting confused between the cdf of X and the cdf of U.
    I suggest using a subscript, so that ##F_U## is the cdf of U, ie ##F_U(u)=Pr(U<u)##.
     
  4. Sep 2, 2015 #3
    OK so to solve this problem I first say U=F(X).

    F(u)=P(U<u)

    which means

    F(u)=P(F(X)<u)

    is this how i should solve this problem?
     
  5. Sep 2, 2015 #4
    ignore that top post mis clicked.

    However the questions asks to find the supports of U and the median.

    I'm a little confused how to approach but can i say that u<x?

    If thats not the case can you point me in the right direction.

    Thanks
     
  6. Sep 2, 2015 #5

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, start with the support. What are the maximum and minimum possible values for F(X)? The support of U will be the interval between those.

    By the way, there's a big hint in the choice of the variable name U. Can you think of a simple distribution whose name starts with the letter U?
     
  7. Sep 3, 2015 #6
    the possible values for F(X) is negative infinity to infinity.

    Uniform distribution?
     
  8. Sep 3, 2015 #7
    Also would i be able to say that U ranges from negative infinity to infinity also?
     
    Last edited: Sep 3, 2015
  9. Sep 3, 2015 #8

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    F(X) is a probability. What is the range of possible values for a probability?

    What is the range of possible values for a random variable whose distribution is the standard uniform distribution? How does that compare to the range of F(X)?
     
  10. Sep 3, 2015 #9
    oh i see, so F(X) the range would be 0 to 1.

    the range of values for a random variable whose distribution is the standard uniform distribution is negative infinity to infinity.

    not should what you mean when you say compare the two?
     
  11. Sep 3, 2015 #10

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The range of possible values for a standard uniform random variable is the real interval [0,1]. If you've never come across standard uniform random variables before then it won't help you to realise that U is one, so perhaps don't worry about that side of it. Go back to your equation:
    $$F_U(u)=Pr(F_X(X)<u)$$

    Now since ##F_X## is a monotonic increasing function, you can invert it. What happens if you apply ##{F_X}^{-1}## to both sides of the inequality inside the parentheses?

    After that you'll have one more step, which involves using the fact that ##Pr(X<x)=F_X(x)##.
     
  12. Sep 3, 2015 #11
    Fu(u)=Pr(F(U)<x)?
     
  13. Sep 3, 2015 #12

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No.

    What is ##{F_X}^{-1}\big(F_X(X)\big)##?
     
  14. Sep 3, 2015 #13
  15. Sep 3, 2015 #14
    inverse of u equals F(u)?
     
  16. Sep 3, 2015 #15
    ok so would it be correct to say that U=F(X)

    then the support of this must be U[0,1]

    and thus the median will be 1/2

    and the 25th percentile would be 1/4 and the third percentile would be 3/4?
     
  17. Sep 3, 2015 #16

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You tell us.
     
  18. Sep 3, 2015 #17
    well because U is essentially the height of the function and F(X) can only be between 0 and 1. then it must be so that the height U must be equal to [0,1]? is that sound reasoning?
     
  19. Sep 3, 2015 #18
  20. Sep 4, 2015 #19
    or should i say that because F(U) ranges from 0 to 1 then u must range from 0 to 1?
     
  21. Sep 4, 2015 #20

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Always start with a picture; it will help you straighten out your thinking. Look at the correspondence between ##X = x_0## and ##U = u_0 = F(x_0)##:
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Cumulative distribution problem Statistics
Loading...