Successive Level Crossings via Strong Markov Decomposition

From Successive Crossings to Independent Intervals

Suppose Brownian motion must cross several increasing levels. The times at which it reaches those levels are not independent: reaching a later level requires reaching each earlier one. The intervals between successive crossings have a simpler structure. At each crossing, the strong Markov property restarts the motion from the level just reached, independently of its past. This turns a sequence of crossing times into a sum of independent first-passage intervals.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying a standard Brownian motion $(B_t)_{t\ge0}$, with its usual filtration $(\mathcal{F}_t)_{t\ge0}$. Fix levels $a_0<a_1<\cdots<a_m$, and assume $B_0=a_0$. For $1\le k\le m$, define the crossing time $T_k=\inf\{t\ge0:B_t=a_k\}$, and set $T_0=0$. Define the interval between crossings by $D_k=T_k-T_{k-1}$, and the gap by $g_k=a_k-a_{k-1}>0$. Brownian paths are continuous, so reaching $a_k$ from $a_0$ requires crossing every level below $a_k$, and $T_0\le T_1\le\cdots\le T_m$.

Definition: The successive passage intervals are $D_k=T_k-T_{k-1}$. They measure the elapsed times from reaching $a_{k-1}$ to first reaching $a_k$; they are distinct from the absolute crossing times $T_k$.

Each $T_k$ is a stopping time for the Brownian filtration. The one-sided first-passage result established in the earlier material gives, for Brownian motion started at $0$ and $g>0$, the Laplace transform $\mathbb{E}[e^{-sT_g}]=e^{-g\sqrt{2s}}$ for $s\ge0$, where $T_g=\inf\{t\ge0:B_t=g\}$. In particular, $T_g<\infty$ almost surely. The corresponding distribution function, from the first-passage law, is $\mathbb{P}(T_g\le t)=2[1-\Phi(g/\sqrt{t})]$ for $t>0$, where $\Phi(x)=\mathbb{P}(Z\le x)$ for $Z$ standard normal. We use these one-level formulas after establishing the independence structure.

Strong Markov Decomposition

Theorem: For the levels above, $D_1,\ldots,D_m$ are independent. For every $s_1,\ldots,s_m\ge0$, $$ \mathbb{E}\!\left[\exp\!\left(-\sum_{k=1}^m s_kD_k\right)\right] =\prod_{k=1}^m e^{-g_k\sqrt{2s_k}}. $$ Consequently, $T_j=\sum_{k=1}^jD_k$ has the same distribution as the time for Brownian motion started at $0$ to first reach $a_j-a_0$.

Proof. Assume the Brownian motion and levels are as specified. We show successively that each new interval is independent of all the preceding intervals and has the first-passage law for its gap. At $T_{k-1}$, the Brownian motion is at $a_{k-1}$, by continuity. The strong Markov property at the almost surely finite stopping time $T_{k-1}$ states that $(B_{T_{k-1}+u}-B_{T_{k-1}})_{u\ge0}$ is a standard Brownian motion independent of $\mathcal{F}_{T_{k-1}}$. Its first time to reach $g_k$ is exactly $D_k$, so, conditionally on $\mathcal{F}_{T_{k-1}}$, $D_k$ has the first-passage law for gap $g_k$, not depending on the past. Each $D_1,\ldots,D_{k-1}$ is measurable with respect to $\mathcal{F}_{T_{k-1}}$; therefore $D_k$ is independent of their collection. Applying this argument for $k=1,\ldots,m$ proves mutual independence. Applying the one-sided transform to each interval and multiplying the transforms of independent variables gives the displayed product. Finally, the definitions give $T_j=\sum_{k=1}^jD_k$, while telescoping the gaps gives $\sum_{k=1}^jg_k=a_j-a_0$. The transform of this sum is $e^{-(a_j-a_0)\sqrt{2s}}$, the one-sided passage transform across the total distance. The first-passage transform determines the distribution, proving the final assertion. $\square$

The theorem concerns the intervals $D_k$, not the absolute times $T_k$. In particular, the absolute times share the earlier intervals: $T_2=D_1+D_2$, while $T_1=D_1$. Their dependence is essential when calculating joint probabilities or transforms.

Worked Example: Joint Transform for Unequal Gaps

Let Brownian motion start at $0$, and consider the levels $1$ and $3$. Thus $g_1=1$, $g_2=3-1=2$, $T_1=D_1$, and $T_2=D_1+D_2$. Find the joint Laplace transform of the two intervals, and the transform of the total time $T_2$.

The row-free one-sided formula applies because both gaps are strictly positive. For $s_1,s_2\ge0$, independence gives

\mathbb{E}[e^{-s_1D_1-s_2D_2}] =e^{-1\sqrt{2s_1}}e^{-2\sqrt{2s_2}} =e^{-\sqrt{2s_1}-2\sqrt{2s_2}}.

Setting both transform parameters equal to $s$ gives the transform of $T_2=D_1+D_2$: $\mathbb{E}[e^{-sT_2}]=e^{-\sqrt{2s}-2\sqrt{2s}}=e^{-3\sqrt{2s}}$. This agrees with a direct one-sided calculation from starting point $0$ to level $3$, which gives $e^{-(3-0)\sqrt{2s}}=e^{-3\sqrt{2s}}$.

Check: The gaps add to $1+2=3$, the total distance from $0$ to $3$. The transform therefore has the coefficient $3$ in its exponent, as the direct calculation requires.

Crossing Probabilities

The distribution of a single interval $D_k$ is the one-sided first-passage distribution for distance $g_k$. Thus, for $t>0$, $\mathbb{P}(D_k\le t)=2[1-\Phi(g_k/\sqrt{t})]$. Independence allows probabilities involving separate intervals to factor. By contrast, the event that the final level has been reached by time $t$ concerns their sum; the theorem identifies that sum with a single passage across the total distance.

Worked Example: Each Crossing Interval Meets Its Own Deadline

For Brownian motion starting at $0$, let $T_1$ and $T_2$ be the first crossing times of levels $1$ and $3$. Find the probability that the first crossing occurs within $1$ time unit and that the subsequent crossing interval is at most $2$ time units. The event is $D_1\le1,\ D_2\le2$; both inequalities are non-strict.

The gaps are $g_1=1$ and $g_2=2$. Since the intervals are independent, the event probability is the product of their one-passage probabilities:

\begin{aligned} \mathbb{P}(D_1\le1,\ D_2\le2) &=\mathbb{P}(D_1\le1)\mathbb{P}(D_2\le2)\\ &=2[1-\Phi(1/\sqrt{1})]\;2[1-\Phi(2/\sqrt{2})]\\ &=4[1-\Phi(1)][1-\Phi(\sqrt{2})]. \end{aligned}

Check: In the first factor, $g_1/\sqrt{1}=1$; in the second, $g_2/\sqrt{2}=2/\sqrt{2}=\sqrt{2}$. Both factors are probabilities of events with non-strict deadlines. Their product is valid because the two passage intervals are independent.

Worked Example: Reaching the Final Level by a Deadline

Let Brownian motion start at $0$, and let $T_1,T_2$ be its first crossing times of levels $1$ and $2$. Find the probability that both levels have been reached by time $1$. Since the levels are increasing, this event is exactly $T_2\le1$.

Here $D_1$ and $D_2$ each cross a gap of $1$, and $T_2=D_1+D_2$. It would be incorrect to multiply the probabilities that each interval is at most $1$: that would calculate $\mathbb{P}(D_1\le1,D_2\le1)$, not $\mathbb{P}(D_1+D_2\le1)$. Instead, the sum has the one-passage law across total distance $2$, so

\mathbb{P}(T_1\le1,\ T_2\le1) =\mathbb{P}(T_2\le1) =2[1-\Phi(2/\sqrt{1})] =2[1-\Phi(2)].

Check: The total distance is $2-0=2$, so the formula uses $2/\sqrt{1}=2$. Also, $T_2\le1$ implies $T_1\le1$ because $T_1\le T_2$; thus the two-level event really is just the final crossing event.

Worked Example: Which Equal-Gap Interval Is Shorter?

Brownian motion starts at $0$, and the successive levels are $1$ and $2$. Find $\mathbb{P}(D_1<D_2)$, where the inequality is strict.

Both gaps equal $1$. The strong Markov decomposition shows that $D_1,D_2$ are independent and have the same first-passage distribution. This distribution has a density for $t>0$, namely $f_1(t)=\frac{1}{\sqrt{2\pi}t^{3/2}}e^{-1/(2t)}$; in particular, it assigns probability zero to any single value. Hence $\mathbb{P}(D_1=D_2)=0$. Interchanging the two identically distributed independent variables leaves their joint law unchanged, so $\mathbb{P}(D_1<D_2)=\mathbb{P}(D_2<D_1)$. These two disjoint events have total probability $1$, giving

\mathbb{P}(D_1<D_2)=\mathbb{P}(D_2<D_1)=\tfrac12.

Check: The strict-order events are disjoint, and their complement is the tie event of probability $0$. Their probabilities therefore sum to $1$, consistent with each being $\tfrac12$.

Infinite Mean and a Common Pitfall

Finite crossing times do not imply finite expected crossing times. For a gap $g>0$, let $T_g$ be the one-sided first-passage time. For $s>0$, the identity $(1-e^{-sT_g})/s$ increases to $T_g$ as $s$ decreases to $0$. Monotone convergence and the passage transform therefore give

\mathbb{E}[T_g] =\lim_{s\downarrow0}\frac{1-e^{-g\sqrt{2s}}}{s} =\infty.

For the last equality, put $x=g\sqrt{2s}$. As $s\downarrow0$, $x\downarrow0$, and $(1-e^{-x})/x\to1$; hence the displayed ratio is asymptotic to $g\sqrt{2}/\sqrt{s}$, which diverges. Since each $D_k$ has this one-sided law with $g=g_k>0$, every interval has infinite mean. The decomposition provides distributional and transform calculations, but not a finite average time between levels.

Common mistake: treating absolute crossing times as independent. For levels $1$ and $2$ from $0$, $T_1=D_1$ and $T_2=D_1+D_2$. A mistaken independence calculation would give $\mathbb{E}[e^{-sT_1-sT_2}]=\mathbb{E}[e^{-sT_1}]\mathbb{E}[e^{-sT_2}]=e^{-2\sqrt{2s}}$. The correct calculation uses the shared interval $D_1$: $sT_1+sT_2=2sD_1+sD_2$. Independence applies to $D_1,D_2$, so the joint transform is $e^{-\sqrt{2(2s)}-\sqrt{2s}}=e^{-2\sqrt{s}-\sqrt{2s}}$. The two expressions are not equal for $s>0$. The safe procedure is to express absolute times as sums of passage intervals before using independence.

Takeaway: Successive upward passage intervals are independent, with each interval’s law determined by its gap. Their sum across several levels has the same distribution as one passage across the total distance. Use interval independence for separate interval events, and the total-distance law for deadlines on the final crossing.

Check Your Understanding

In each question, distinguish the interval times from the absolute crossing times.

Brownian motion starts at $0$ and crosses levels $2$ and $5$. What are the two gap sizes, and what is the joint Laplace transform of the passage intervals?
For levels $1$ and $4$ from $0$, express the transform of the time to reach $4$ both as a product over intervals and as a single-passage transform.
For levels $1$ and $2$ from $0$, explain why $\mathbb{P}(D_1\le1,D_2\le1)$ is not the probability that level $2$ is reached by time $1$.
For a positive gap $g$, what does the one-sided passage transform imply about $\mathbb{E}[T_g]$, and why does almost-sure finiteness not contradict that conclusion?
When can the probabilities of two passage-interval events be multiplied? State the independence fact that justifies the calculation.