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Math Help - gamma distribution 2

  1. #1
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    gamma distribution 2

    I had a similar problem and I was trying to solve this question based on reply (thank you so much!!) but im having a difficult time to do following question:

    For Z distributed on R^3 with Z1, Z2, Z3 IID Z ~ exp(1)

    If i have W = (Z1 ^2, Z2 ^2, Z1Z2) then how can I find expected value and variance?

    Do I still use gamma distribution? I am so confused....
    plz help me!!!
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  2. #2
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    W is the joint density of three i.i.d variates. What does this mean?
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  3. #3
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    []EDIT[]Corrected post. Please refer to Matheagle and Moo's expert posts below.
    Last edited by Anonymous1; March 17th 2010 at 08:25 AM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    are you asking for the mean and covariance matrix of a vector?
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  5. #5
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    Thank you so much !!!!

    Jacobian tranfomation? I am not familiar with it T-T

    I need to find E(W) and Var(W)........that's what question was asking does it help?
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  6. #6
    Super Member Anonymous1's Avatar
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    I think you can use my method. Unless there are any objections?

    Let me know if you get stuck.
    Anonymous
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  7. #7
    MHF Contributor matheagle's Avatar
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    E(W) = \left(E(Z_1 ^2), E(Z_2 ^2), E(Z_1)E(Z_2)\right)=(2,2,1)

    The variance/covariance matrix has in the ij^{th} position the covariance between Zi and Zj.
    Note that the covariance between Z1 and Z1 is its variance.
    It's a 3 by 3 symmetric matrix, where you have the 3 variances down the diagonal and the covariances in the other positions.
    Last edited by matheagle; March 17th 2010 at 12:15 AM.
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  8. #8
    Moo
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    Quote Originally Posted by Anonymous1 View Post
    Since they are i.i.d you can multiple the three densities f_{Z}^2(z) where f_{Z}(z) is the density of an exponential(1) to find your joint distribution. See what you get. Does it look gamma? Then simply compute the first and second moments to find the variance of W.

    \int_0^{\infty} z\times f_{Z_1}^2(z)*f_{Z_2}^2(z)*f_{Z_3}^2(z)dz

    \int_0^{\infty} z^2\times f_{Z_1}^2(z)*f_{Z_2}^2(z)*f_{Z_3}^2(z)dz

    wait a minute... You may have to use a Jacobian transform. Does anyone know?
    What the... ?

    The pdf of Z^2 when Z has the pdf f_Z is certainly not f_Z^2
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  9. #9
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    Thank you so much!!! so you mean that 3 by 3 symmetric matrix will look like

    var (Z1^2) cov (Z1^2, Z2^2) cov (Z1^2, Z1Z2^2)
    cov (Z2^2, Z1^2) var (Z2^2) cov (Z2^2, Z1Z2^2)
    cov (Z1Z2^2, Z1^2) cov (Z1Z2^2, Z2^2) var (Z1Z2^2)

    but aren't they all 1s for var(Z1^2) and var(Z2^2) and even var(Z1Z2^2)??? in this case??? im not sure....T-T
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  10. #10
    Moo
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    Quote Originally Posted by dymin3 View Post
    Thank you so much!!! so you mean that 3 by 3 symmetric matrix will look like

    var (Z1^2) cov (Z1^2, Z2^2) cov (Z1^2, Z1Z2)
    cov (Z2^2, Z1^2) var (Z2^2) cov (Z2^2, Z1Z2)
    cov (Z1Z2, Z1^2) cov (Z1Z2, Z2^2) var (Z1Z2)

    but aren't they all 1s for var(Z1^2) and var(Z2^2) and even var(Z1Z2^2)??? in this case??? im not sure....T-T
    False

    Var(Z^2)=E(Z^4)-[E(Z^2)]^2

    And it's easy, the moments of an exp(1) are : E[X^k]=k!

    So Var(Z^2)=4!-2^2=24-4=20

    For the covariance, we have for example cov(Z2^2,Z1^2)=0, because Z1 and Z2 are independent.


    And you messed up some things : you're dealing with Z1Z2, not Z1Z2^2. I corrected it.
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  11. #11
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    Wow, thank you so much. I have one last question

    Now i have Var(Z1Z2) = E[(Z1Z2)^2] - [E(Z1Z2)]^2 = ? - 1^2
    How can i find E[(Z1Z2)^2]? I can't use the moments of an exp(1) for this, can I?
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  12. #12
    Moo
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    ... It's algebra : (Z1Z2)^2=Z1^2 Z2^2
    And since Z1 and Z2 are independent, we'll have E[(Z1Z2)^2]=E[Z1^2]E[Z2^2]
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  13. #13
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    Quote Originally Posted by Moo View Post
    ... It's algebra : (Z1Z2)^2=Z1^2 Z2^2
    And since Z1 and Z2 are independent, we'll have E[(Z1Z2)^2]=E[Z1^2]E[Z2^2]
    oops.... i kept forgetting about the meaning of the independent
    Thanks!!

    Now I have EW = (2,2,1) and VarW = (20, 20, 3)
    Problem solved!!! YEY!!
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