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Math Help - Is my proof correct?

  1. #1
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    Lightbulb Is my proof correct?

    I'm not sure if I proved the following correctly. (It is in a chapter on expectation, and I didn't use anything about expectation.) Any feedback would be greatly appreciated.

    Problem:
    Suppose that X, Y, and Z are nonnegative random variables such that P(X+Y+Z \leq 1.3)=1. Show that X, Y, and Z cannot possibly have a join distribution under which each of their marginal distributions is the uniform distribution on the interval [0,1].


    Work:
    \int_0^1\int_0^1 f(x,y,z)dxdy=z
    \int_0^1\int_0^1 f(x,y,z)dxdz=y
    \int_0^1\int_0^1 f(x,y,z)dydz=x

    \int_0^1 xdx=\frac{1}{2}
    \int_0^1 ydy=\frac{1}{2}
    \int_0^1 zdz=\frac{1}{2}

    Answer:
    Integrating f(x,y,z) over the cube gives p=\frac{1}{2} instead of 1, therefore we have a contradiction. So X, Y, and Z cannot possibly have a join distribution under which each of their marginal distributions is the uniform distribution on the interval [0,1].
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  2. #2
    MHF Contributor matheagle's Avatar
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    Re: Is my proof correct?

    I don't understand your integrals.

    IF you integrate out the other variables you obtain the marginal density.
    Hence

    \int_0^1\int_0^1 f(x,y,z)dxdy=f_Z(z)=1 IF you think that Z is U(0,1)
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  3. #3
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    Re: Is my proof correct?

    You're right. I had done \int_0^z\int_0^1\int_0^1 f(x,y,z)dxdydz=F_Z(z)=z but wrote \int_0^1\int_0^1 f(x,y,z)dxdy=z. So \int_0^1dz=1.

    Now I'm at a loss on how to prove this.
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  4. #4
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    Re: Is my proof correct?

    P(X+Y+Z \leq 1.3)=1 \Rightarrow E(X+Y+Z) \leq 1.3
    If X, Y, and Z are U(0,1) then E(X+Y+Z)=E(X)+E(Y)+E(Z)=1.5
    Contradiction. Therefore at least one of X, Y, or Z is not U(0,1).
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