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Math Help - A question related to Derived Distributions

  1. #1
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    Question A question related to Derived Distributions

    Can anyoner help me to solve this question... I'm really stuck on this one

    Andy is vacationing in Las Vegas. The amount X (in Dolars) he takes to casino each evening is a random variable
    with a PDF of the form
    fX(x)=ax if 0≤x≤23 and
    0 otherwise.
    At the end of each night, the amount Y that he has when leaving the casino is uniformly distributed between
    zero and twice the amount that he came with.
    Determine the joint PDF fX,Y(x,y).
    fX,Y(8,14)=
    What is the probability that on a given night Andy makes a positive profit at the casino?
    Find the PDF of Andy's profit Z=Y-X on a particular night.
    fZ(0)=
    Determine the expected value of Z=Y-X.
    E[Z]=


    Thank you very much.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'm not sure I follow all of this.
    Parts seem hazy, especially dollars being continuous.
    Here's what I get...

    f_X(x)={x\over 264.5} on (0,23)

    Next you have Y given X as uniform on 2X....

    So f_{X,Y}(x,y)=f_X(x)f_{Y|X}(x,y)= \left({x\over 264.5}\right)\left({1\over 2x}\right)

    ={1\over 529} where we have 0<x<23 and 0<y<2x

    which I would rewrite as either 0<y<2x<46 or 0<y/2<x<23

    IS this what you're looking for?
    I can get the density of Z=Y-X if you wish, but I would only do that if this makes any sense to you.
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  3. #3
    MHF Contributor matheagle's Avatar
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    If you want the probability Y>X, then you need to integrate over that region.
    But since the joint density is constant you can use geometry to find that volume.
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  4. #4
    MHF Contributor matheagle's Avatar
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    In order to obtain the density of Z=Y-X, you can first obtain the cummulative of Z, then differentiate.

    P(Z\le z)=P(Y-X\le z)

    Now draw the triangle where X,Y exists, that's inside the region, 0<x<23, 0<y<46 and y=2x.
    To integrate, but as I said before you can use geometry since the joint density is constant, 1/529,
    you find the probabiity of Y<X+z, for any z.
    First you draw y=x+z and then figure out where you need to integrate.
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  5. #5
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    I got fX(x) and fX,Y(x,y) the same way you did, matheagle.
    and then I tried to do this FZ(z) = P(Y-X≤z) = ∫∫ {X,Y|Y-X≤z} fX,Y(x,y) dx dy.
    But couldn't figure it out.

    For you method, isn't gonna be like ∫(2x-x)dx?
    Pardon me if I'm wrong, my calculus is rusty at best.
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  6. #6
    MHF Contributor matheagle's Avatar
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    There is no need for calculus, the density is constant, use geometry.
    You must always draw your region in the xy-plane.
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  7. #7
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    Got it!! T__T
    Dude, you're an angel.
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  8. #8
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    What if I want to use calculus? Can you give me a hint?
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