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Math Help - inequality

  1. #1
    Rog
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    inequality

    If the expected $/-rate for a 1 year’s time is 1.50, is the expected rate /$-rate 2/3, < 2/3 or > 2/3?

    (Hint: the expected yearly $/-rate is defined as the average $/-daily
    rate is: 1/n . sum 1 to n of Xi for i = 1,...n=365 .The expected yearly /$-rate is 1/n sum 1 to n of 1/Xi
    WHERE Xi is the $/-Rate on the ith day.

    My first impression is that obviously the expected rate should be the inverse which = 2/3. Im having trouble proving it though:

    1/365 ( X1 + X2 + ... + X365) = 3/2

    1/365 (1/X1 + 1/X2 + ... + 1/X365) = y

    Can equate them by setting 1/365 = 3/2 . 1/(X1 + X2 + ... + X365) = y/(1/X1 + 1/X2 + ... + 1/X365)

    But I just can not get anywhere! Any help welcome!
    Last edited by Rog; September 15th 2007 at 11:42 AM.
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  2. #2
    Rog
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    I think ive figured it out:

    Using Jensens Inequality:

    F(E[X]) >= (less than or equal to) E[F(X)]

    So, Let F be the function: F(Z) = 1/Z

    Then,

    F( (1/n) Sum Xi ) >= Sum ( F(X) ) / n
    F( (1/n) Sum Xi ) >= Sum ( 1/Xi ) / n
    F ( 3/2 ) >= Sum ( F(X) ) / n
    2/3 >= Expected value of /$-Rate

    Since the function F(Z) is 1-1 then we have equality in the above inequality.

    Thoughts and opinions welcome!
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  3. #3
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    Using Jensen is complete overkill.

    Try using AM-HM
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  4. #4
    Rog
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    Never heard of this AM-HM, but after some googling - it seems to be exactly Jensens Inequality with F(X)=1/X. Or just some special case of Jensens anyway.
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  5. #5
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    Quote Originally Posted by Rog View Post
    Never heard of this AM-HM, but after some googling - it seems to be exactly Jensens Inequality with F(X)=1/X. Or just some special case of Jensens anyway.
    Let a_1,a_2,...,a_n be positive numbers.
    Then,
    \frac{a_1+...+a_n}{n} \geq \frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}
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