Simplification of a convolution-like expression

Question

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

Answer 1

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 $n$?

Answer 2

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:

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