Showing that the sum of zero-mean noise is zero. Then computing the convolution of zero-mean noise with a given function
Strictly: No such $B$ exists. You could simply have a sufficient streak of &quot;bad luck&quot; and draw positive $\eta&gt;\epsilon&gt;0$ continuously, for example. Obviously, $B&lt;\lim_{N\to\infty}\sum_{n=-N}^{N-1}\epsilon_n&lt;\lim_{N\to\infty}\left\lvert\sum_{n=-N}^{N-1}\eta_n\right\rvert\,\forall B\in \mathbb R$.
Granted, the event that every $\eta_n&gt;0$ has probability 0 ($=\lim_{N\to\infty}\prod_{n=-N}^{N-1} P(\eta_n&gt;0)=\lim_{N\to\infty}\left(\frac12\right)^N$). 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&amp;=P\left(\lim_{N\to\infty}\left\lvert\sum_{n=-N}^{N-1}\eta_n\right\rvert&lt;B\right)\\
&amp;=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&lt;\frac{B}{2N} \right)\\
&amp;=P\left(\lim_{N\to\infty}\left\lvert X_N \right\rvert&lt;\frac{B}{2N} \right)\\
&amp;=P\left(\lim_{N\to\infty}X_N &lt;\frac{B}{2N} \right)-P\left(\lim_{N\to\infty}X_N &lt;-\frac{B}{2N} \right)\\
&amp;=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)\\
&amp;=\frac12-\frac12\\
&amp;=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 &quot;needs&quot; 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.