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Math Help - moments of the sum of inverse ratio distribution

  1. #1
    Per
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    moments of the sum of inverse ratio distribution

    I've had problems working this out:

    Let z_1,...,z_N be N independent uniform distributed variables that is z_1,...,z_N are all iid uniform[0,1].

    Define:  s_i = z_i/(z_1+...+z_N) for all i=1,...,N

    That is,  s_i is the ratio of one of the variables divided by the sum of all variables, so that  s_1+...+s_N = 1

    I'm looking for the
    - expected value:  E[s_i] . I'm 99% sure this is equal to 1/N.

    - variance:  VAR[s_i]

    - covariance:  COV[s_i,s_j]


    I would be very grateful for all help!
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Per View Post
    I've had problems working this out:

    Let z_1,...,z_N be N independent uniform distributed variables that is z_1,...,z_N are all iid uniform[0,1].

    Define:  s_i = z_i/(z_1+...+z_N) for all i=1,...,N

    That is,  s_i is the ratio of one of the variables divided by the sum of all variables, so that  s_1+...+s_N = 1

    I'm looking for the
    - expected value:  E[s_i] . I'm 99% sure this is equal to 1/N.

    - variance:  VAR[s_i]

    - covariance:  COV[s_i,s_j]


    I would be very grateful for all help!
     E[s_i]  = \frac{\frac{1}{2}}{\frac{1}{2}N} Now, your 100% sure. []Edit[]Oops.

    As for variance, try obtaining the second moment.

     E[s_i^2] = E[\frac{z_i ^2}{(z_1 +... + z_N)^2}]

    Then you can find the variance and covariance from the definition.
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  3. #3
    MHF Contributor

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    Quote Originally Posted by Per View Post
    I've had problems working this out:

    Let z_1,...,z_N be N independent uniform distributed variables that is z_1,...,z_N are all iid uniform[0,1].

    Define:  s_i = z_i/(z_1+...+z_N) for all i=1,...,N

    That is,  s_i is the ratio of one of the variables divided by the sum of all variables, so that  s_1+...+s_N = 1

    I'm looking for the
    - expected value:  E[s_i] . I'm 99% sure this is equal to 1/N.
    Let me add the final percent: you have 1=E[s_1+\cdots+s_N]=E[s_1]+\cdots+E[s_N]=NE[s_i] by symmetry hence E[s_i]=1/N
    . (To Anonymous: you can't compute the expectation like you did!)


    - variance:  VAR[s_i]
    So you need E[(s_1)^2]. I'm not sure you can't avoid computing the N-fold integral \int_0^1\cdots\int_0^1\frac{(x_1)^2}{(x_1+\cdots+x  _N)^2}dx_1\cdots dx_N. Did you already try?

    - covariance:  COV[s_i,s_j]
    This one can follow from the previous one, in a similar way to E[s_1].
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  4. #4
    Per
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    I thought of doing the N-fold integral but wanted to convince myself that there isn't any easier way of doing it.
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