Adaptive stochastic paths and brownian bridge coupling — Examples

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.

Applying [F1] once with $b(y)=y$ and $\sigma (y)=1$ gives $Y_c=y_0+h{y}_0+\mathrm{\Delta }W=(1+h)y_0+\mathrm{\Delta }W$.

Applying [F1] on the first fine half-step gives $Y_{1\mathrm{/}2}=(1+h\mathrm{/}2)y_0+\mathrm{\Delta }{W}_1$. The second half-step gives $Y_f=(1+h\mathrm{/}2)Y_{1\mathrm{/}2}+\mathrm{\Delta }{W}_2$, so substitution yields the displayed fine formula.

Substitute [F2] into $Y_f$ and subtract $Y_c$ from step 1.1: $Y_f-Y_c=((1+h\mathrm{/}2{)}^2-(1+h))y_0+((1+h\mathrm{/}2)(\mathrm{\Delta }W\mathrm{/}2+Z)+(\mathrm{\Delta }W\mathrm{/}2-Z)-\mathrm{\Delta }W)=\frac{h^2}{4}{y}_0+\frac{h}{4}\mathrm{\Delta }W+\frac{h}{2}Z\mathrm{.}$ For $y_0=0$, independence and the variances in [F2] give $\mathrm{Var}(Y_f-Y_c)=(h\mathrm{/}4{)}^{2}h+(h\mathrm{/}2{)}^2(h\mathrm{/}4)=h^3\mathrm{/}8$. The identity $\mathrm{\Delta }{W}_1+\mathrm{\Delta }{W}_2=\mathrm{\Delta }W$ makes this a path-consistent coupling, not two independent simulations.

The case $h=0$ is excluded; for every $h>0$ the fine difference has strictly positive variance at $y_0=0$. A nonzero initial value changes the mean by $h^2{y}_0\mathrm{/}4$ but not that variance. This example checks an exact finite-dimensional coupling only; it is not a general MLMC or convergence rate theorem.

Apply the crossing theorem with $x=y=1$, $c=0$, and $h=1$. Its hypotheses $x,y\ge c$ and $h>0$ hold, and it gives $\mathbb{P}\{{\inf }_{0\le s\le 1}{B}_s\le 0\}=\exp (-2(1-0)(1-0)\mathrm{/}1)=e^{-2}$.

Since $e^{-2}>0$, the endpoint-only check “both endpoints exceed zero” does not certify that the path stayed above zero. If either endpoint were exactly zero, the crossing event would instead have probability one by the boundary case of the same theorem.