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  • krzysio
    replied
    No 1/4 in the problem

    Originally posted by Rup
    Mr. Ostaszewski,
    I am just wondering did you use ¼ 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'

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  • Rup
    replied
    Re:joint density problem

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

    Thanks

    Rup

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  • Rup
    replied
    Re:Joint density problem

    Thanks for your help!

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  • krzysio
    replied
    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.

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  • sszpakowski
    replied
    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!

    Leave a comment:


  • Rup
    started a topic joint density problem

    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.
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