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Math Help - Expectation with Double Summation

  1. #1
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    Expectation with Double Summation

    Hi Guys, I need help on this question.



    Where the conditions are shown below:



    The final answer given in the paper is


    Kindly enlighten me on this as I do not know how to derive to the final answer.

    Thanks in advance.

    Sincerely,
    James
    Attached Thumbnails Attached Thumbnails Expectation with Double Summation-picture1.jpg   Expectation with Double Summation-picture2.jpg   Expectation with Double Summation-picture3.jpg  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by nightvision1984 View Post
    Hi Guys, I need help on this question.



    Where the conditions are shown below:



    The final answer given in the paper is


    Kindly enlighten me on this as I do not know how to derive to the final answer.

    Thanks in advance.

    Sincerely,
    James
    First sort out the brackets in the first expression:

     <br />
E\left( \frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)^2<br />

    This is ambiguous and could plausibly mean either:

     <br />
\left[ E \left(\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)\right]^2<br />

    or:

     <br />
E\left( \left[\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right]^2 \right)<br />

    Given what you are asked to prove you probably want the latter. If so expand the square and use the linearity of expectation to simplify.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    First sort out the brackets in the first expression:

     <br />
E\left( \frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)^2<br />

    This is ambiguous and could plausibly mean either:

     <br />
\left[ E \left(\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)\right]^2<br />

    or:

     <br />
E\left( \left[\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right]^2 \right)<br />

    Given what you are asked to prove you probably want the latter. If so expand the square and use the linearity of expectation to simplify.

    CB
    Hi Captian Black,

    I've used the latter one just as you have predicted.

    But when I expand the square of the term, I got



    And I'm stuck with the double summation.

    Kindly help me with this. I'm quite weak in this.

    Thanks,
    James
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  4. #4
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    Quote Originally Posted by nightvision1984 View Post
    Hi Captian Black,

    I've used the latter one just as you have predicted.

    But when I expand the square of the term, I got



    And I'm stuck with the double summation.

    Kindly help me with this. I'm quite weak in this.

    Thanks,
    James
    E\left[ \frac{1}{T^2}\left( 2\sum_{s\ne t} \frac{u_yu_s}{x_t x_s} +\sum \frac{u_t^2}{x_t^2}\right)\right] = <br />
\frac{1}{T^2}\left[\sum_{s\ne t} E\left(\frac{u_tu_s}{x_tx_s}\right)\right]+\frac{1}{T^2}\left[\sum E\left(\frac{u_t^2}{x_t^2} \right)\right]<br />

    You are told that the all the terms in the first summation on the right are zero, and so:

    E\left[ \frac{1}{T^2}\left( 2\sum_{s\ne t} \frac{u_yu_s}{x_t x_s} +\sum \frac{u_t^2}{x_t^2}\right)\right] = <br />
\frac{1}{T^2}\left[\sum E\left(\frac{u_t^2}{x_t^2} \right)\right]<br />
<br />
=\frac{1}{T^2}\sum \frac{E(u_t^2)}{x_t^2}<br />

    and you should be able to finish from there.

    CB
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  5. #5
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    Captain Black,

    THANK YOU!!

    Sincerely,
    James
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