Redrawing a rejected Brownian increment biases the path

Put $I=[-c,c]$ and $p=\mathbb{P}\{G_1\in I\}>0$. For each Borel $A$, the event that the first accepted draw lies in $A$ is the disjoint union over $k\ge 0$ of the events that the first $k$ draws lie outside $I$ and draw $k+1$ lies in $A\cap I$. Independence therefore gives $\mathbb{P}\{R\in A\}={\sum }_{k=0}^{\mathrm{\infty }}(1-p{)}^k\mathbb{P}\{G_1\in A\cap I\}=\frac{\mathbb{P}\{G_1\in A\cap I\}}{p}\mathrm{.}$ This is exactly the stated truncated-normal law.

The law is symmetric, so $\mathbb{E}R=0$. Both $I$ and its complement have positive probability. On $I$, $G_1^2\le c^2$ with strict inequality on a positive-probability subset; on $I^c$, $G_1^2>c^2$. Hence $\mathbb{E}[G_1^2\mid G_1\in I]<c^2<\mathbb{E}[G_1^2\mid G_1\mathrm{\notin }I]\mathrm{.}$ The unconditional variance $h=\mathbb{E}{G}_1^2$ is the strict weighted average of these two conditional second moments. Therefore $\mathrm{Var}(R)=\mathbb{E}[G_1^2\mid G_1\in I]<h$.

The resampled increment is thus not $N(0,h)$, whereas bridge refinement in [F2] preserves the already sampled parent increment by exact sum consistency. This example concerns endpoint-dependent rejection; it does not contradict the predictable-grid error theorem, whose step lengths are fixed from current information.