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Math Help - Interesting sum of series

  1. #1
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    Interesting sum of series

    Hi,

    I'm an engineer and recently, within my work on non uniform signal sampling, I encountered the following series :

      S(t)=\sum _{n=1}^{+\infty }{\frac{1}{\sigma \sqrt{2\pi n}}}\cdot \exp \left[-{\frac{(t-n\mu )^{2}}{2\sigma ^{2}n}}\right]

    with t, \mu and \sigma all positive constants.

    As can be seen, the series is a sum of normal distributions with means equal to n \mu and variances equal to n \sigma ^2

    I found numerically (using Matlab) that the limit of S(t) as t tends to +\infty is amazingly equal to \frac{1}{\mu}.

    I tried to proof this interesting result but with no success yet.

    If anybody could help me, I'd be grateful to him...
    Thanks for any help,
    Roland
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  2. #2
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    Quote Originally Posted by roland65 View Post
    Hi,

    I'm an engineer and recently, within my work on non uniform signal sampling, I encountered the following series :

      S(t)=\sum _{n=1}^{+\infty }{\frac{1}{\sigma \sqrt{2\pi n}}}\cdot \exp \left[-{\frac{(t-n\mu )^{2}}{2\sigma ^{2}n}}\right]

    with t, \mu and \sigma all positive constants.

    As can be seen, the series is a sum of normal distributions with means equal to n \mu and variances equal to n \sigma ^2

    I found numerically (using Matlab) that the limit of S(t) as t tends to +\infty is amazingly equal to \frac{1}{\mu}.

    I tried to proof this interesting result but with no success yet.

    If anybody could help me, I'd be grateful to him...
    Thanks for any help,
    Roland
    Hi, Roland,

    I've been looking for a probabilistic proof. The following is not a full proof, but it gives a result that is very close to what you want. Indeed, I shall prove that:
    [CENTER]for any h>0, \int_T^{T+h} S(t) dt\underset{T \to \infty}{\longrightarrow}\frac{h}{\mu}.
    [LEFT]
    Thus, for instance if S(t) converges, then it must be toward \frac{1}{\mu}. More generally, one would require to show that S(t) does not fluctuate too much as t\to\infty in order to complete the proof. Otherwise, there must be more analytical proofs.


    Now for the proof: Let (X_n)_{n\geq 1} be a family of independent Gaussian random variables with mean \mu and variance \sigma^2. Then, classicaly, X_1+\cdots+X_n is Gaussian with mean n\mu and variance n\sigma^2. Let h>0. For any T>0, we have (using Fubini)

    \int_T^{T+h}S(t)dt = \sum_{n=1}^\infty P(X_1+\cdots+X_n\in [T,T+h]) =  \sum_{n=1}^\infty E\left[{\bf 1}_{(X_1+\cdots+X_n\in [T,T+h])}\right] = E[N([T,T+h])],

    where N([T,T+h])=\sum_{n=1}^\infty {\bf 1}_{(X_1+\cdots+X_n\in [T,T+h])} is the number of indices n such that the sum X_1+\cdots+X_n falls inside [T,T+h]. If we remember that E[X_1]=\mu>0, the law of large numbers shows that N([T,T+h]) is finite. Moreover, it is a consequence of the renewal theorem that:

    [CENTER] E[N([T,T+h])]\underset{T \to \infty}{\longrightarrow}\frac{h}{\mu}.

    [LEFT]This can be understood intuitively: in average, the successive sums X_1, X_1+X_2,\ldots are distant of E[X_i]=\mu from the next one, so that the average number of such sums that fall inside a given interval of length h must be about \frac{h}{\mu}. This only holds for large T since some mixing must have happened for the starting point to be "forgotten"... I guess you can find ressource on the renewal theorem on the web.

    I hope you found this interesting.

    Laurent.


    Edit: In fact, I've just found there is a "renewal density theorem" proved by W.L.Smith in 1954, which is stronger that the "classical" renewal theorem I quoted, and which gives exactly your result. A reference is here (there is a proof, but it is complicated) or here (no proof, but simpler statement (n.b.: the definition m(t)=\frac{dM}{dt} is lacking)). So there exists a fully probabilistic proof, but it is not quite simple. (Maybe in the present case Smith's proof could be simplified, I didn't read it)
    Last edited by Laurent; March 21st 2009 at 05:50 AM.
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  3. #3
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    Hi Laurent,
    many thanks for your help. I appreciate very much your effort!
    I'll check today the material you indicate me. However, since I'm not a mathematician, it could be difficult for me to understand the renewal theorem but I'll check this carefully.
    Anyway, I'll contact you if I make some progress...
    Many thanks and best regards,
    Roland
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