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Math Help - Sum & difference of dependent normal variables?

  1. #1
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    Sum & difference of dependent normal variables?

    Given two normal random variables, which are not necessarily independent, with different means but the same variance, I need to show that:

    1) The sum of these two variables is normal
    2) The difference of these two variables is normal
    3) The sum and the difference are independent

    Can someone start me off/point me in the right direction/suggest a resource for this question? The "not necessarily independent" part has me stumped
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  2. #2
    MHF Contributor matheagle's Avatar
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    What if I let Y=-X, they have the same variance and the sum is ZERO, which is not normal.
    I'm confused.
    (3) Cov(X+Y,X-Y)=V(X)-V(Y)=0 which would imply indep IF we have normality.
    Last edited by matheagle; April 14th 2009 at 08:54 PM.
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  3. #3
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    Regarding (3) - I know independence impies zero correlation, but I thought the converse wasn't necessarily true?

    For (1) & (2)... Maybe I've missed something in the wording of the question. This is what I have, word for word:

    Let Z_{1}~ N(m_{1},\sigma^{2}), Z_{2}~ N(m_{2},\sigma^{2}) be Gaussian random variables (not necessarily independent).

    Prove that X_{1} = Z_{1}+Z_{2}, X_{2} = Z_{1}-Z_{2} are independent Gaussian random variables
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    MHF Contributor matheagle's Avatar
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    Look at the bivariate normal density.
    In THAT case if \rho =0, then the joint density factors, hence they are indep.
    see Multivariate normal distribution - Wikipedia, the free encyclopedia

    Again, what if X=Y, then the difference is not normal.
    IF Z_1 and Z_2 are indep, then this is correct.
    You can prove this via MGFs.
    Last edited by matheagle; April 14th 2009 at 10:14 PM.
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  5. #5
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    Yes, I agree. If Z1 and Z2 were independent then X1 and X2 would be JOINTLY normal and in that case zero correlation would imply independence. That's fine.

    BUT, I'm given specifically that Z1 and Z2 are NOT necessarily independent...

    Which throws into question whether X1 and X2 are even seperately normally distributed. The only clue seems to be the equal variances thing...

    Re: "zero is not normal" - could you not consider zero to be a normal random variable with zero mean and zero variance?
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    MHF Contributor matheagle's Avatar
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    That is a discrete rv known as P(X=0)=1.

    How can you write that as {1\over \sqrt{2\pi}\sigma}e^{-(x-\mu)^2\over 2\sigma^2}

    You can't divide by zero.
    Last edited by matheagle; April 16th 2009 at 10:12 PM.
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  7. #7
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    Well, the pdf wouldn't exist, but I've seen that convention adopted in a couple of textbooks... p247 of Bertsekas & Tsitsiklis for example

    So you're saying the question makes no sense?
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  8. #8
    MHF Contributor matheagle's Avatar
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    A normal rv is a continuous rv defined on (-infinity,infinity).
    P(X=0)=1 is certainly not defined on the entire real line.
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  9. #9
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    I'm not saying it is.

    Quote Originally Posted by HD09
    but I've seen that convention adopted in a couple of textbooks... p247 of Bertsekas & Tsitsiklis for example

    So you're saying the question makes no sense?
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  10. #10
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    Quote Originally Posted by HD09 View Post
    Given two normal random variables, which are not necessarily independent, with different means but the same variance, I need to show that:

    1) The sum of these two variables is normal
    2) The difference of these two variables is normal
    3) The sum and the difference are independent

    Can someone start me off/point me in the right direction/suggest a resource for this question? The "not necessarily independent" part has me stumped
    I think you must assume that the two normal variables have a joint bivariate normal distribution. Let's say the variables are X and Y. Define U = X + Y and V = X - Y. Then apply the change of variables theorem to find the joint distribution of U and V. Once that is done the rest of the problem should be easy.

    (Whether this works in the case where the correlation is -1 is a matter of definition, and I don't know whether the commonly accepted definition of the normal distribution is broad enough to encompass this case.)
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