# Showing that the sum of zero-mean noise is zero. Then computing the convolution of zero-mean noise with a given function

This is likely to be a quick fix for people with experience in stochastic processes.

Let $$\eta[k]$$ be a sequence of Uniform noise, $$\eta \sim U([-M,M])$$. I want to test if the following is correct theoretically $$\begin{equation*} \exists B>0\ s.t. \left|\sum_{k\in\mathbb{Z}} \eta[k] \right| < B. \end{equation*}$$

In the general case, I would also want to try to test if $$\begin{equation*} \exists B>0\ s.t. \left|\sum_{k\in\mathbb{Z}} \eta[k] \psi [n-k] \right|

What I tried:

Clearly, for $$\psi[k]=1$$ the two equations above are equal.

For the first one, I tried using ergodicity, which leads to $$\lim_{N\rightarrow\infty} \frac{1}{2N}\sum_{k=-(N-1)}^{N} \eta[k] =0,$$ but this does not imply $$\sum_{k\in\mathbb{Z}} \eta[k] =0.$$

For the second one, it's easy to show that $$\left|\sum_{k\in\mathbb{Z}} \eta[k] \psi [n-k] \right| but I don't want to assume that $$\sum_{k\in\mathbb{Z}} \left| \psi [n-k] \right|<\infty.$$

Even if the reader might not have an answer, it would be very helpful to know if this is a trivial result in random processes I am not aware of, or if it is a more complex problem. Thank you!

• Welcome to SE.SP! I think you will need to take the expectation on your first equation. Otherwise, that summation of just the random uniformly distributed values will never be identically zero (well, it'll have a vanishingly small probability of being zero). To make the question more sensible, I believe you'll need to put the expectation operator around the summation. – Peter K. Jul 2 at 16:22
• Thanks for the reply Peter! For me expectation would not be useful, because I want to show it is smaller than something in absolutely all cases. I updated saying I want to make it smaller than B in absolute value. – Dorian Jul 2 at 16:33
• My intuition is that even though a finite sequence of noise can take any values, when you take an infinite sequence it needs to behave somehow like the pdf, taking both positive and negative values. Maybe I am wrong. – Dorian Jul 2 at 16:34
• Also, if I am reading your notation right, there is no need to specify that $B>0$ when you are already specifying it is greater than an absolute value. I don't believe what you are trying to show is correct. Going to infinity can be a tricky process, which is why a strict limit approach is needed to do it properly. I don't see anything that would prove that B is a finite bound. – Cedron Dawg Jul 2 at 19:25
• As an afterthought, I don't think that even the condition that $\psi$ approaches zero as the domain goes to positive or negative infinity is strong enough to give you an overall finite bound. – Cedron Dawg Jul 2 at 23:27

Strictly: No such $$B$$ exists. You could simply have a sufficient streak of "bad luck" and draw positive $$\eta>\epsilon>0$$ continuously, for example. Obviously, $$B<\lim_{N\to\infty}\sum_{n=-N}^{N-1}\epsilon_n<\lim_{N\to\infty}\left\lvert\sum_{n=-N}^{N-1}\eta_n\right\rvert\,\forall B\in \mathbb R$$.

Granted, the event that every $$\eta_n>0$$ has probability 0 ($$=\lim_{N\to\infty}\prod_{n=-N}^{N-1} P(\eta_n>0)=\lim_{N\to\infty}\left(\frac12\right)^N$$). But you asked for the existence of such $$B$$, not for a proof of you being able to be sure that you'll be able to pick a finite $$B$$ which your sum-absolute never exceeds.

Sadly, there exist uncountably infinitely many cases for which the $$B$$ can't exist, and all have probability 0. Whether or not that makes a non-zero sum probability isn't trivial to say, far as I can tell.

What you could show is that there's a 100% probability that your sum is below some $$B$$; i.e. show that: \begin{align} 1&=P\left(\lim_{N\to\infty}\left\lvert\sum_{n=-N}^{N-1}\eta_n\right\rvert

which really isn't a probability of 1. Quite the opposite is true: You can be certain that if you let a random walk go on for infinity, that you're infinitely far from where you started.

By the way, the above proof doesn't use the fact that $$\eta$$ is uniform – it just "needs" that all $$\eta_n$$ are i.i.d. and that they have bounded variance.

You could construct the very same proof even without knowledge of the central limit theorem, just using Chebyshev's inequality.