Originally Posted by

**roland65** Hi,

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

$\displaystyle 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 $\displaystyle t$, $\displaystyle \mu$ and $\displaystyle \sigma$ all positive constants.

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

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

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 $\displaystyle h>0$, $\displaystyle \int_T^{T+h} S(t) dt\underset{T \to \infty}{\longrightarrow}\frac{h}{\mu}$.

[LEFT]

Thus, for instance *if* $\displaystyle S(t)$ converges, then it must be toward $\displaystyle \frac{1}{\mu}$. More generally, one would require to show that $\displaystyle S(t)$ does not fluctuate too much as $\displaystyle t\to\infty$ in order to complete the proof. Otherwise, there must be more analytical proofs.

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

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

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

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

[LEFT]This can be understood intuitively: in average, the successive sums $\displaystyle X_1, X_1+X_2,\ldots$ are distant of $\displaystyle E[X_i]=\mu$ from the next one, so that the average number of such sums that fall inside a given interval of length $\displaystyle h$ must be about $\displaystyle \frac{h}{\mu}$. This only holds for large $\displaystyle T$ 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 $\displaystyle m(t)=\frac{dM}{dt}$ 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)