Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Density of probability/function of random variables question

  1. Apr 20, 2013 #1
    Hi everyone,
    I have the following exercise.
    Fx(x)=0, x<-1 or x>1
    Fx(x)=1/2, x=[-1;1]
    g(x)=x^2+1 --- this is the function of random variable
    I must calculate Fy which is the sum of solutions of g(xk)=y , Fy(y)=sumFx(xk)/|g`(xk)|
    g(x) is bijective on [-1;1]
    y=x^2+1=> x=+sqrt(y-1) or x=-sqrt(y-1), since x=[-1;1] both are posible solutions.
    And my question is on what interval is Fy defined... to find the intervals i use the formula [g(-1);g(1)] but i dont know if its right and in this case g(-1)=g(1)=2;
    What i am doing wrong???

    On a similar exercise i had
    Fx(x)=1/2, x=[0;2]
    Fx=0, out of the interval
    g(x)=x^2+3
    g`(x)=2x
    x=sqrt(y-3) and x=-sqrt(y-3), since x=[0;2], x=-sqrt(y-3) is not a posible solution.
    Fy(y)=sumFx(xk)/|g`(xk)|=1/4*1/sqrt(y-3))
    So
    Fy(y)=1/4*1/sqrt(y-3) for y=[g(0)=3;g(2)=7];
    Fy(y)=0 for y =[-infinite;3]U[7;+infinite]
    On seminars we did only with g(x)=a*x+b which was easy and these are for homework.
     
    Last edited: Apr 20, 2013
  2. jcsd
  3. Apr 20, 2013 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    The values of Y that that have non-zero probability will range from the minimum to the maxium value of the function X^2 + 1 for X in the interval [-1, 1]. You should be able to determine these values "by inspection". If not, you could work it as max-min problem using calculus. You method doesn't work because g(-1) is not the minumum value that g(x) attains.
     
  4. Apr 21, 2013 #3
    So there are more methods???
    For the second exercise the solution is good??? The condition for density gives me 1. http://www.wolframalpha.com/input/?i=integral+from+3+to+7+from+1/4*1/sqrt(y-3)dy
    Thanks for tip, ill look on seminars, i think we did somthing similar and ill post my solution.

    Later edit.
    This is the solution:
    http://www.wolframalpha.com/input/?i=integral+from+1+to+2+from+1/2*1/sqrt(y-1)dy
    It seems good :). Thanks very much. Its more logical like this because in theory y range on oy axis of graph x^2+1 for x=[-1;1].
    Its everything corect, right????
    The minimum valuea is 1 and the maximum value is 2 for x=[-1;1]
     
    Last edited: Apr 21, 2013
  5. Apr 21, 2013 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    Yes. I'm not going to check all your work. If you want that much help. you should post your questions in the sections on homework help. (You phrased the question well and showed your work, so your post would be OK for that section.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook