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 $N\le 1$ requires the second sign, so $N$ is not a stopping time for the natural filtration. The four equally likely possibilities giveBoth 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.^[ref]