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I am trying to further reduce the following expression (if there's a nicer way to write it):

$\sum_{k=-\infty}^{+\infty}\sum_{p=-\infty}^{+\infty} \sum_{l=-\infty}^{+\infty} w(k) \varphi_\mathrm{y}(l-p)\varphi_\mathrm{x}(k-n-l+p)$

where $\varphi_\mathrm{y}$ and $\varphi_\mathrm{x}$ are the autocorrelations of the signals $y(i)$ and $x(i)$, respectively.

I know I can express $\sum_{k=-\infty}^{+\infty} w(k) \varphi_\mathrm{x}(k-n-l+p)$ as a convolution (or a correlation, using the evenness of $\varphi_\mathrm{x}$) but I left the expression expanded in case it can help.

If necessary, I could take the simplified case of finite (M-samples-)length signals.

Regards

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As you say, since $\varphi_x$ is an autocorrelation we have that $$\sum_{k=-\infty}^{\infty} w(k) \varphi_\mathrm{x}(k-n-l+p) = (w \star \varphi)\bigr |_{n+l-p} = g(n+l-p)$$ where $g = w \star \varphi$. So, since $\varphi_y$ is also an autocorrelation, we have $$\begin{align} \sum_{l=-\infty}^{\infty} g(n+l-p)\varphi_y(l-p) &= \sum_{l=-\infty}^{\infty} g(l+n-p)\varphi_y(p-l)\\ &=\sum_{m=-\infty}^{\infty} g(m)\varphi_y(n-m) & \text{upon setting}~l+n-p=m,\\ &= g \star \varphi_y\bigr|_n = h(n) \end{align}$$ where $h = g\star\varphi_y = w\star\varphi_x \star\varphi_y$. So now you are left with $\displaystyle \sum_{p=-\infty}^\infty h(n)$ which is unbounded. Are you sure that this last sum is with respect to $p$ and not with respect to $n$?

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Thanks for the reply Dilip. I can contextualize the starting point of my answer showing how I ended up there. It reads like this:

\begin{align} r(n) &= \mathbb{E}\left\{\left[w(i)\ast y(i) \ast x(i)\right]\left[y(i-n)\ast x(i-n)\right]\right\}\\ &= \mathbb{E}\left\{\left[\sum_{k=-\infty}^{\infty}\sum_{p=-\infty}^{\infty}w(k)y(p)x(i-k-p)\right]\left[\sum_{l=-\infty}^{\infty}y(l)x(i-n-l)\right]\right\}\\ &= \sum_{k=-\infty}^{\infty}\sum_{p=-\infty}^{\infty}\sum_{l=-\infty}^{\infty}w(k)\mathbb{E}\left\{\left[y(p)x(i-k-p)\right]\left[y(l)x(i-n-l)\right]\right\}\\ &= \sum_{k=-\infty}^{\infty}\sum_{p=-\infty}^{\infty}\sum_{l=-\infty}^{\infty}w(k)\mathbb{E}\left\{y(p)y(l)\right\}\mathbb{E}\left\{x(i-k-p)x(i-n-l)\right\}\\ &= \sum_{k=-\infty}^{\infty}\sum_{p=-\infty}^{\infty}\sum_{l=-\infty}^{\infty}w(k)\varphi_\mathrm{y}(l-p)\varphi_\mathrm{x}(-n-l+p+k)\\ \end{align}

where $\mathbb{E}\left\{\cdot\right\}$ represent the expectation operator and the random signals $x(i)$ and $y(i)$ are statistically independent and have zero mean (these properties allow the expression of the expected value as a product) and autocorrelations $\varphi_\mathrm{x}$ and $\varphi_\mathrm{y}$, respectively.

As you can see, $p$ is the dummy variable of one of the convolutions, while $n$ is the actual time-lag and it's the only parameter that should remain in the final expression (given that we get rid of $i$ through the expectation).

As I mentioned in the original question, I was trying to derive the expression for the general case of infinite-length signals, but given the unbounded summation I think I need to directly restrict it to the finite-length case (the signals are processed in $M$-length chunks), thus obtaining:

$$\sum_{p=-\infty}^{\infty}h(n) = \sum_{p=-M}^{M-1}h(n) = 2M\cdot h(n)$$

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  • $\begingroup$ The problem here seems to be one of bad notation which I have complained about elsewhere on this forum but everyone else seems to think that it is the gospel truth because a famous book uses it. If you look at the first quantity in square brackets on the second line, you will notice that it is a function of $i$ alone, and thus is a constant as far as the sum in the second square brackets is concerned. Thus, you will get an infinite sum of identical terms as in my answer. So, you need to re-write the convolutions that you need to do. $\endgroup$ Sep 11 '13 at 12:56
  • $\begingroup$ First of all, I should have specified that the signals I'm using are all real-valued, to avoid any misunderstanding. I read your comments in the other topic (as well as the one linked there) and I found it quite interesting given that I had never thought about this bad notation problems. Nevertheless, I don't quite get how in this particular case the mentioned problem plays a role. What I mean is that I don't get if you're saying that the result with the infinite sum is correct or only that it's correct given a wrong starting point (due to the wrong formulas used to express the convolutions) $\endgroup$
    – gbernardi
    Sep 11 '13 at 16:14
  • $\begingroup$ I am saying that the starting point may be wrong and that what you have written might not be what you want to compute. But, starting from your first displayed equation, the calculations are correct and give you that $r(n)$ is unbounded for all $n$ (except when, serendipitously, that $h(n)$ happens to equal $0$). If that is consistent with what you expect to see, then the original formulation is correct; if not, then maybe you need to start from an earlier point and work out what exactly it is you want to compute and how to express it in terms of convolutions. $\endgroup$ Sep 11 '13 at 18:18
  • $\begingroup$ Well, before this calculation I tried to do the same for a similar case, having $x(i)$ deterministic and, using a very similar line of calculations, that results in all bounded summations (and, I should mention, the right solution which is known in that case). So I'm quite confident that the result I obtained here is correct. Anyway, thanks again for the thoughtful replies! Cheers $\endgroup$
    – gbernardi
    Sep 12 '13 at 8:24

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