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Homework Help: Probability theory.

  1. Dec 12, 2005 #1
    Probability theory.......

    So we were given some practise problems for the exam....
    in it, we got a question that we have NEVER seen in class nor can I find it in the text book.

    The question is:

    Find P(min(X, Y ) > x) and hence give the probability density function of U =
    min(X, Y ).

    Okay.. so i know how to answer it.. i think.. he gave us a solution...

    But just what on earth does P(min(X, Y ) > x) mean? could their be P(max(X, Y ) > x) and if so, what does that mean?

    Somebody care to take a moment and explain please?

    Thanks
     
  2. jcsd
  3. Dec 12, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    It's the probability that the minimum of X and Y is larger than x. What is it you not understand? min(X,Y) is clearly larger than x if both X and Y are.
     
  4. Dec 12, 2005 #3
    okay... i see...

    what about P(max(X,Y)>x) can that exist

    How about P(min(X,Y)<x)... thats just P(X<x)P(Y<x) right??
     
  5. Dec 12, 2005 #4

    Dale

    Staff: Mentor

    If you have the joint pdf of X and Y then you can get P(min(X, Y ) > x) by integrating the joint pdf over the proper region.

    [tex]\int _{x}^{\infty}\int _{x}^{\infty}
    \Muserfunction{pdf}(X,Y)\,dY\,dX[/tex]

    The other ones that you mention are the same idea, just integrating over different regions of the (X,Y) plane. For example, P(max(X,Y)>x) would be

    [tex]1 - \int _{{-\infty}}^{x}
    \int _{{-\infty}}^{x}
    \Muserfunction{pdf}(X,Y)\,dY\,dX[/tex]

    I am sure you can get the rest of these similarly.

    -Dale
     
  6. Dec 12, 2005 #5
    yes thanks this helps!
     
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