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| <B, \end{equation*}

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 $$\left|\sum_{k\in\mathbb{Z}} \eta[k] \psi [n-k] \right|<M\sum_{k\in\mathbb{Z}} \left| \psi [n-k] \right|, $$ 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!

  • 2
    $\begingroup$ 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. $\endgroup$
    – Peter K.
    Jul 2, 2020 at 16:22
  • $\begingroup$ 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. $\endgroup$
    – Dorian
    Jul 2, 2020 at 16:33
  • $\begingroup$ 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. $\endgroup$
    – Dorian
    Jul 2, 2020 at 16:34
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 2, 2020 at 19:25
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 2, 2020 at 23:27

1 Answer 1


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<B\right)\\ &=P\left(\lim_{N\to\infty}\left\lvert\underbrace{\sum_{n=-N}^{N-1}\frac{\eta_n}{2N}}_{=:X_N\sim\mathcal N(0,\sigma_\eta^2)}\right\rvert<\frac{B}{2N} \right)\\ &=P\left(\lim_{N\to\infty}\left\lvert X_N \right\rvert<\frac{B}{2N} \right)\\ &=P\left(\lim_{N\to\infty}X_N <\frac{B}{2N} \right)-P\left(\lim_{N\to\infty}X_N <-\frac{B}{2N} \right)\\ &=F_{\mathcal N(0,\sigma_\eta^2)}\left(\lim_{N\to\infty}\frac{B}{2N}\right) - F_{\mathcal N(0,\sigma_\eta^2)}\left(\lim_{N\to\infty}-\frac{B}{2N}\right)\\ &=\frac12-\frac12\\ &=0\text, \end{align}

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.


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