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Math Help - [SOLVED] normal random variable

  1. #1
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    [SOLVED] normal random variable

    if

    f_{XY}(x,y) = f_{X}(x)f_{Y}(y) = \frac{1}{2\pi \sigma^2} e^{-\frac{x^2+y^2}{2\sigma^2}}

    I have to show that Z is also a normal random variable

    Z = \frac{(X-Y)^2-2Y^2}{\sqrt{X^2+Y^2}}

    maybe using these substitions

    A = \sqrt {X^2+Y^2}

    \tan B = \frac{Y}{X}
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  2. #2
    MHF Contributor matheagle's Avatar
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    I don't recognize any tricks here, like using the MGF.

    One thing for sure is that X\sim N(0,\sigma^2) and Y\sim N(0,\sigma^2) are independent

    So  \left({X\over \sigma}\right)^2 + \left({Y\over \sigma}\right)^2\sim\chi^2_2


    IF you're going to transform from X,Y to something you should USE Z and something basic like A equal to X or Y OR maybe your A or maybe A squared? BUT not A and B. Try Z and A.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    I don't recognize any tricks here, like using the MGF.
    what does MGF stand for?

    can you please explain this:
    Quote Originally Posted by matheagle View Post
    So  \left({X\over \sigma}\right)^2 + \left({Y\over \sigma}\right)^2\sim\chi^2_2
    Quote Originally Posted by matheagle View Post
    IF you're going to transform from X,Y to something you should USE Z and something basic like A equal to X or Y OR maybe your A or maybe A squared? BUT not A and B. Try Z and A.
    this is really hard... I don't have the idea how to do it...

    thank you very much
    Last edited by akane; May 26th 2010 at 09:43 AM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    MGF stands for moment generating function.
    YOU may have to transform from X,Y to Z,A and then integrate out the A
    to obtain the density of Z.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    MGF stands for moment generating function.
    YOU may have to transform from X,Y to Z,A and then integrate out the A
    to obtain the density of Z.
    What would be A?
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by akane View Post
    What would be A?
    You can try any possibility for A.
    I was suggesting the one you had already selected, that square root of X and Y squared

    Though, I must say, this is a weird problem
    I don't see any tricks, are you sure you copied it correctly?
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  7. #7
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    Quote Originally Posted by matheagle View Post
    You can try any possibility for A.
    I was suggesting the one you had already selected, that square root of X and Y squared

    Though, I must say, this is a weird problem
    I don't see any tricks, are you sure you copied it correctly?
    unfortunately I copied it correctly
    that substitution won't work because I have to rewrite Z as a function of these two new variables, in the case you mentioned I would have Z = Z (X,Y,A)
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