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This may be an easy question, but I am just not sure so I am asking the community.

Suppose I have a wide-sense (second order) stationary (WSS) continuous random process with autocorrelation function $\mathcal{R}(\tau)$ such that it's square-integrable: $\int_{\mathbb{R}}|\mathcal{R}(\tau)|^2d\tau<\infty$.

I am interested in the discrete observations of this process. Suppose I am sampling it with some period $T_s$. Is the discrete autocorrelation function $R[i]=\mathcal{R}(iT_s)$ square-summable in that case, that is, is the following statement true?

$$\sum_{i=-\infty}^{i=\infty}|R[i]|^2<\infty$$

Seems to me that it should be, but I can't prove it (maybe because it is not).

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    Suggestion: Can you massage your sum into looking like a lower Riemann sum for the integral?2012-06-28
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    @DilipSarwate I tried that which is why I wrote "seems to me that it should [square-summable]". However, I've been taught in my calc class that Riemann sums don't generalize to integrals with unbounded limits: one may not be able to write this: $\int_{\mathbb{R}}|\mathcal{R}(\tau)|^2d\tau=\lim_{\delta\rightarrow 0}\delta\sum_{t=-\infty}^{\infty}|\mathcal{R}(\delta i)|^2$. And I know that [the sampled absolutely integrable function isn't necessarily absolutely summable](http://math.stackexchange.com/questions/137932/is-sampled-absolutely-integrable-function-absolutely-summable/137935#137935).2012-06-29
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    @DilipSarwate So, that's the reason for my post...2012-06-29
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    While samples of an absolutely integrable function need not be absolutely summable in general as the examples provided to you in response to your other question show, here you have a particular kind of function (autocorrelation) that has many properties that are not enjoyed by arbitrary absolutely integrable functions, and maybe these can be used to settle the result.2012-06-29

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