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  • joint probability problem

    X1, X2, X3, X4 are continuous unif(0, 1)

    let T=max(X1, X2, X3, X4)
    S=min(X1, X2, X3, X4)

    find the joint Pr(T=t, S=s)

    i can find the marginal probabilities, but since they are not independent, i cant just multiply them

  • #2
    Originally posted by akirika View Post
    X1, X2, X3, X4 are continuous unif(0, 1)

    let T=max(X1, X2, X3, X4)
    S=min(X1, X2, X3, X4)

    find the joint Pr(T=t, S=s)

    i can find the marginal probabilities, but since they are not independent, i cant just multiply them
    are X1, X2, X3 AND X4 independent?

    Comment


    • #3
      yes they are independent

      Comment


      • #4
        Originally posted by akirika View Post
        yes they are independent
        If X1, X2, X3 AND X4 are independent, then, i think, their max and min are independent as well. Can someone confirm that?

        Comment


        • #5
          Originally posted by KHC831 View Post
          If X1, X2, X3 AND X4 are independent, then, i think, their max and min are independent as well. Can someone confirm that?
          Why is that true? Think of the 2 variable case and see if you can prove/disprove that.

          Comment


          • #6
            their max/min are not independent because once a max is given, the min cannot be greater than the max

            so...anyone know how to solve this problem?

            Comment


            • #7
              Have you tried googling "order statistics"? The wikipedia article is rather informative - especially regarding the answer to this exact question...

              Comment


              • #8
                is the answer just (t-s)^4?

                Comment


                • #9
                  does this make sense?

                  F(t,s) = P(all 4 are less than t)* P(not all 4 are greater than s/all 4 are less than t)
                  t^4 * (1-P(all 4 greater than s/all 4 are less than t)
                  t^4 * (1-(t-s)^4/t^4)
                  t^4 * (t^4 -(t-s)^4)/t^4

                  = t^4-(t-s)^4

                  then take the derivative?
                  f(t,s) = 12 (t-s)^2
                  CAStudy
                  Actuary.com - Level II Poster
                  Last edited by CAStudy; May 30 2009, 09:30 PM. Reason: chnaged my mind

                  Comment


                  • #10
                    Originally posted by akirika View Post
                    X1, X2, X3, X4 are continuous unif(0, 1)

                    let T=max(X1, X2, X3, X4)
                    S=min(X1, X2, X3, X4)

                    find the joint Pr(T=t, S=s)

                    i can find the marginal probabilities, but since they are not independent, i cant just multiply them
                    Pr(T=t, S=s) is of course zero, but that's clearly not the question you meant to ask. You said one thing and meant another: f_T,S (s,t), joint PDF. You really need to understand that difference.
                    Yours,
                    Krzys'
                    Want to know how to pass actuarial exams? Go to: smartURL.it/pass

                    Comment


                    • #11
                      Originally posted by CAStudy View Post
                      does this make sense?

                      F(t,s) = P(all 4 are less than t)* P(not all 4 are greater than s/all 4 are less than t)
                      t^4 * (1-P(all 4 greater than s/all 4 are less than t)
                      t^4 * (1-(t-s)^4/t^4)
                      t^4 * (t^4 -(t-s)^4)/t^4

                      = t^4-(t-s)^4

                      then take the derivative?
                      f(t,s) = 12 (t-s)^2
                      P(not all 4 are greater than s/all 4 are less than t)

                      i dont get that part

                      Comment


                      • #12
                        He meant to put | not / which is used for "given" i.e. P(A|B) = P(A and B)/P(B)

                        Comment


                        • #13
                          yes, that's what I meant, thank you.
                          and does it make sense?
                          my reasoning was
                          first take the probability that you have a max less than t (so all of them are less than t) t^4.
                          Then the second qualification (so I multiplied) is that at least one of them has to be <= s which is the opposite of them all being between s and t (so 1-(t-s)) but I need to divide it by t^4 since I already accounted for the fact that every x is less than t.
                          I didn't make it (1-(t-s)^4) / t^4 because that would include the probability of s being greater that t also

                          And this is much easier to talk about if you draw a number line.

                          I didn't mean that f(x) was the probability. I just guessed that was what you were referring to.

                          I didn't read the wiki article , is it useful?
                          CAStudy
                          Actuary.com - Level II Poster
                          Last edited by CAStudy; May 31 2009, 12:54 PM.

                          Comment


                          • #14
                            Originally posted by CAStudy View Post
                            yes, that's what I meant, thank you.
                            and does it make sense?
                            my reasoning was
                            first take the probability that you have a max less than t (so all of them are less than t) t^4.
                            Then the second qualification (so I multiplied) is that at least one of them has to be <= s which is the opposite of them all being between s and t (so 1-(t-s)) but I need to divide it by t^4 since I already accounted for the fact that every x is less than t.
                            I didn't make it (1-(t-s)^4) / t^4 because that would include the probability of s being greater that t also

                            And this is much easier to talk about if you draw a number line.

                            I didn't mean that f(x) was the probability. I just guessed that was what you were referring to.

                            I didn't read the wiki article , is it useful?
                            I followed your logic but not the math. If you do min first, would you get the same answer?

                            Comment


                            • #15
                              Yes, but when I redid it,I realized I could say it more simply.

                              F(t,s) =P(max<t, min<s)= P(max<t)- P(all Xi between t and s)
                              t^4 - (t-s)^4

                              Comment

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