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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?


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? – Dilip Sarwate Jun 28 '12 at 20:01
@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. – M.B.M. Jun 29 '12 at 16:00
@DilipSarwate So, that's the reason for my post... – M.B.M. Jun 29 '12 at 16:01
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. – Dilip Sarwate Jun 29 '12 at 16:12

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