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Mean of the square of a sum of exponential terms

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data

    Calculate [itex]\widehat{Y^{2}}[/itex]

    (i.e., the mean of the square of [itex]Y[/itex].

    2. Relevant equations

    [tex]Y=\sum_{k=0}^{N-1}y_{k}[/tex]

    where

    [tex]y_{k}=e^{-\gamma t}e^{\gamma \tau k}G_{k}[/tex]

    and

    [tex]t=N\tau[/tex]

    The quantities [itex]y_{k}[/itex] (or [itex]G_{k}[/itex]) are statistically independent.

    3. The attempt at a solution

    [tex]\widehat{Y^{2}}=\widehat{G^{2}}e^{-2\gamma t}\sum_{k=0}^{N-1}e^{2\gamma \tau k}=\widehat{G^{2}}e^{-2\gamma t}(\ \frac{1-e^{2\gamma t}}{1-e^{2\gamma \tau}} ) [/tex]

    However, the correct answer should be

    [tex]\widehat{Y^{2}}=\widehat{G^{2}}\frac{1}{2\gamma\tau} (\ 1-e^{-2\gamma \tau} )[/tex]

    so it seems like I'm doing something wrong, or is it possible to somehow simplify my answer in order to get the correct one?
     
  2. jcsd
  3. Oct 20, 2015 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Unless the quantities ##G_k## have mean zero (##E G_k = 0##) and are identically distributed for different ##k## (or, at least, all have the same second moments), the expression you give is incorrect; and in that case the one given in the book (or wherever) is also incorrect. As far as I can see, the only way you could obtain something like the second expression would be to have non-zero values of ##E G_j G_k## of the form ##E G_j G_k = E(G^2) c_{jk}##, where ##E(G^2) \equiv E(G_1^2) = E(G_2^2) = \cdots = E(G_{N-1}^2)##. This is because we have ##Y = \sum_{k=1}^{N-1} w_k G_k## (with ##w_k = e^{-\gamma t} e^{\gamma \tau k}##), so that
    [tex] Y^2 = \sum_{k=1}^{N-1} w_k^2 G_k^2 + 2 \sum_{1 \leq j < k \leq N-1} w_j w_k G_j G_k, [/tex]
    so you would need an appropriate value for the expectation of the second summation above, and that would depend on the form of the coefficients ##c_{jk}##.
     
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