= Peeking ahead in a random sum
Let $X_1,X_2$ be <independent> fair signs, each taking $-1$ and $1$ with equal <probability>. Set $N=1$ if $X_2=1$, and $N=2$ if $X_2=-1$. Deciding whether $N\leq1$ requires the second sign, so $N$ is not a <stopping time> for the <natural filtration>. The four equally likely possibilities give
$$
\begin{array}{c|c|c|c}
X_1&X_2&N&S_N\\\hline
1&1&1&1\\
-1&1&1&-1\\
1&-1&2&0\\
-1&-1&2&-2
\end{array}
\qquad
\mathbb E[N]=\frac32,\quad \mathbb E[S_N]=-\frac12,\quad \mu\mathbb E[N]=0.
$$
Both the <increments> and the count are bounded, but the identity fails because the second contribution is retained only when it is negative. This is selection using an unobserved outcome, rather than a legitimate sequential decision. Boundedness alone cannot replace the <stopping time> hypothesis.<Lecture notes for Introduction to Stochastic Processes by Gordan Žitković/More about Random Walks>[§4.3, Problem 4.8 and its solution]{ref}
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