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  • joint density problem

    Can anybody try to solve and explain these problems? Thanks!

    The joint density of the random variables X and Y is ¼ in the square 0<x<2, 0<y<2. Let W=min(X, Y). Calculate E (W).
    Answer: 2/3


    A random variable X has the following CDF.
    F(x)= x/2 0<x<1
    1 x>(or equal to) 1
    Let M(t) be the moment generating function of X. Calculate M(ln2).
    Answer: 1/ln2+1
    Last edited by Rup; November 12 2005, 08:35 AM.

  • #2
    The first problem is a double integral problem. Because the joint density is uniform, you only have to do one integral and multiply the result times 2.
    So please forgive my notation, but here's how you're supposed to set up the integral:

    2 double integral (x ranging from 0 to 2 and y ranging from 0 to x) of x/4 dydx.

    The reason it's only x in the numerator and not xy is because of W=min(X,Y), so in the triangle where we integrated X is the min so W=X.

    I hope this helps!

    Comment


    • #3
      It is not necessary to do this the most complicated way

      Originally posted by Rup
      Can anybody try to solve and explain these problems? Thanks!
      The joint density of the random variables X and Y is &#188; in the square 0<x<2, 0<y<2. Let W=min(X, Y). Calculate E (W).
      Answer: 2/3
      Sorry, yall, but you are really trying to do this the most complicated way, unnecessarily so.

      E (W) = integral from 0 to 2 of the survival function of W =
      = integral from 0 to 2 of (1 - 0.5w)^2 =
      = (-2/3)(1 - 0.5w)^3 evaluated from 0 to 2 = 2/3

      Simple and direct. I will post this and the other way with my weekly exercises. Neat problem.

      Yours,
      Krzys'
      Last edited by krzysio; November 15 2005, 07:28 PM.
      Want to know how to pass actuarial exams? Go to: smartURL.it/pass

      Comment


      • #4
        Re:Joint density problem

        Thanks for your help!

        Comment


        • #5
          Re:joint density problem

          Mr. Ostaszewski,
          I am just wondering did you use ¼ at all in the solution.

          Thanks

          Rup

          Comment


          • #6
            No 1/4 in the problem

            Originally posted by Rup
            Mr. Ostaszewski,
            I am just wondering did you use &#188; at all in the solution.
            Rup
            My solution was probably confusing with the typo I had in it, so I corrected it now. So sorry.

            No 1/4 in it. I know you wanted 1/4 because it is joint density, but there is no need to us it.

            Yours,
            Krzys'
            Want to know how to pass actuarial exams? Go to: smartURL.it/pass

            Comment

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