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Math Help - Bivariate random variable

  1. #1
    Junior Member NoFace's Avatar
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    Bivariate random variable

    First the question:

    Let X and Y be two independent variables. If X and Y are both standard normal, then what is the distribution of \frac{1}{2} (X^2-Y^2)?

    So I've tried to use the fact that X and Y are standard normal to tell me that \mu_x=\mu_y=0 and {\sigma}_{x}=\sigma_y=1.
    So,
    {\sigma}^2=E[X^2]-E[X]^2
    1=E[X^2]-0^2 \implies E[X^2]=1

    I found that E[\frac{1}{2} (X^2-Y^2)]= \frac{1}{2} (E[X^2] -E[Y^2]) = \frac{1}{2}(1^2-1^2) = 0

    I'm not really sure what to do with this.
    Maybe I am wrongly assuming something.

    Thanks in advance!
    Last edited by NoFace; April 7th 2009 at 10:14 AM. Reason: Add info
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,
    Quote Originally Posted by NoFace View Post
    First the question:

    Let X and Y be two independent variables. If X and Y are both standard normal, then what is the distribution of \frac{1}{2} (X^2-Y^2)?

    So I've tried to use the fact that X and Y are standard normal to tell me that \mu_x=\mu_y=0 and {\sigma}_{x}=\sigma_y=1.
    So,
    {\sigma}^2=E[X^2]-E[X]^2
    1=E[X^2]-0^2 \implies E[X^2]=1

    I found that E[\frac{1}{2} (X^2-Y^2)]= \frac{1}{2} (E[X^2] -E[Y^2]) = \frac{1}{2}(1^2-1^2) = 0

    I'm not really sure what to do with this.
    Maybe I am wrongly assuming something.

    Thanks in advance!
    Unfortunately, expectations and variances are not characteristic of a distribution !
    So it may help you check if you got the correct distribution, but it won't help you find which one it is.

    Have you ever dealt with problems that required finding the pdf of a function of X and Y (e.g. \tfrac 12(X^2-Y^2)), while given the pdf of X and Y ?
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  3. #3
    Junior Member NoFace's Avatar
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    Quote Originally Posted by Moo View Post
    Have you ever dealt with problems that required finding the pdf of a function of X and Y (e.g. \tfrac 12(X^2-Y^2)), while given the pdf of X and Y ?
    Unfortunately, no.

    I began using the fact that M_{aX+bY}(t)=M_X(at)M_Y(bt) to find the moment generating function then match that up, but the fact that I am working with X^2 and Y^2 is tripping me up.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Both X^2 and Y^2 are \chi^2 random variables. Their difference has support on all of R, so it's not a gamma or anything obvious. Hence you need to play with the densities. NOW the sum of X^2 and Y^2 is a different matter.
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  5. #5
    Junior Member NoFace's Avatar
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    Figured it out. After consulting with the professor, we established that the answer in the back of the book is wrong. Way wrong.

    Thanks.
    Last edited by NoFace; April 8th 2009 at 04:52 PM.
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  6. #6
    Junior Member NoFace's Avatar
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    Solution if anyone is interested

    Here is the solution:
    X^2 and Y^2 are \chi^2 random variables, so we already know the moment generating functions M_{X^2}(t) and M_{Y^2}(t).


    M_{\frac{1}{2}(X^2-Y^2)}(t) <br /> <br />
\\ \\ = E\big[e^{\frac{t}{2}(X^2-Y^2)}\big]

    \\ \\ = E\big[e^{\frac{t}{2}X^2}]E\big[e^{\frac{-t}{2}Y^2}\big]
    <br />
\\ \\ = M_{X^2}\bigg(\frac{t}{2}\bigg)M_{Y^2}\bigg(\frac{-t}{2}\bigg)
    \\ \\ = \big((1-t)^{-\frac{1}{2}}\big)\big((1+t)^{-\frac{1}{2}}\big)
    <br />
\\ \\ = \big((1-t^2)^{-\frac{1}{2}}\big)

    Which is the MGF of a function which has no name and is certainly not what my book says.
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  7. #7
    MHF Contributor matheagle's Avatar
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    You can derive the density, but as I said before the MGF technique won't help you recognize the probability distribution.
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