Worked Problems in Brownian Laws and Markov Calculations

A Toolkit for Multi-Step Brownian Problems

Many Brownian-motion calculations require more than one fact at a time. A useful approach is to identify what is being conditioned on, determine whether the relevant time is deterministic or random, and then choose the appropriate tool: a joint Gaussian law, a transition density, scaling, a Markov property, or reflection. The examples below make those choices explicit and check the resulting formulas.

Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\), adapted to \((\mathcal{F}_t)\). Thus \(B_0=0\), paths are continuous, and for \(0\le s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). No stochastic integrals are needed in these calculations.

Definition: Write \(\Phi(x)=\mathbb{P}(Z\le x)\) for the standard normal CDF, where \(Z\sim N(0,1)\). For \(t>0\), the Brownian transition density from \(x\) to \(y\) in time \(t\) is \(p_t(x,y)=\dfrac{1}{\sqrt{2\pi t}}\exp\!\left(-\dfrac{(y-x)^2}{2t}\right)\).

For an event conditioned on another event, assume throughout that the conditioning event has positive probability. Conditional distributions given \(B_t=x\) are understood through their Gaussian conditional-density versions; individual values \(B_t=x\) have probability zero. All distributional identities below are understood almost surely where appropriate.

Conditioning on a Brownian Endpoint

Proposition: If \(0<s<t\), then the conditional distribution of \(B_s\) given \(B_t=x\) is Gaussian with mean \(\dfrac{s}{t}x\) and variance \(\dfrac{s(t-s)}{t}\). At the endpoints, \(B_0=0\) and \(B_t=x\) under this conditioning.

Proof. The vector \((B_s,B_t)\) is jointly Gaussian, with \(\mathbb{E}[B_s]=\mathbb{E}[B_t]=0\), \(\operatorname{Var}(B_s)=s\), \(\operatorname{Var}(B_t)=t\), and \(\operatorname{Cov}(B_s,B_t)=s\). Define \(R=B_s-\dfrac{s}{t}B_t\). Then \(\operatorname{Cov}(R,B_t)=s-\dfrac{s}{t}t=0\). Jointly Gaussian variables with zero covariance are independent, so \(R\) is independent of \(B_t\). Its mean is zero and its variance is \(\operatorname{Var}(R)=s-2\dfrac{s}{t}s+\dfrac{s^2}{t^2}t=\dfrac{s(t-s)}{t}\). Since \(B_s=\dfrac{s}{t}B_t+R\), conditioning on \(B_t=x\) gives the stated Gaussian distribution. The endpoint statements follow directly from \(B_0=0\) and the conditioning value at time \(t\). \(\square\)

Worked Example: A Midpoint Conditional on the Endpoint

Find the conditional law of \(B_1\) given \(B_2=1\), and calculate the conditional probability that \(0<B_1<1\).

Here \(s=1\), \(t=2\), and \(x=1\). The conditional mean is \(\dfrac{s}{t}x=\dfrac12\), and the variance is \(\dfrac{s(t-s)}{t}=\dfrac12\). Thus \(B_1\mid(B_2=1)\sim N(\tfrac12,\tfrac12)\). The conditional standard deviation is \(\sqrt{1/2}\). Standardizing the lower and upper endpoints gives \(\dfrac{0-\tfrac12}{\sqrt{1/2}}=-\dfrac{1}{\sqrt2}\) and \(\dfrac{1-\tfrac12}{\sqrt{1/2}}=\dfrac{1}{\sqrt2}\). Therefore

\mathbb{P}(0<B_1<1\mid B_2=1) =\Phi\!\left(\frac{1}{\sqrt2}\right)-\Phi\!\left(-\frac{1}{\sqrt2}\right) =2\Phi\!\left(\frac{1}{\sqrt2}\right)-1.

Check: the conditional mean \(1/2\) is the midpoint of the two endpoints \(B_0=0\) and \(B_2=1\), and the conditional distribution is symmetric about \(1/2\). The probability of the open interval \((0,1)\) equals the probability of the closed interval \([0,1]\), since a nondegenerate Gaussian variable assigns probability zero to either endpoint.

Transition Densities and Deterministic-Time Conditioning

The deterministic-time Markov property says that for \(u>0\) and a bounded measurable function \(g\), \(\mathbb{E}[g(B_{t+u})\mid\mathcal{F}_t]=\int_{\mathbb{R}}g(y)p_u(B_t,y)\,dy\). This follows because \(B_{t+u}=B_t+(B_{t+u}-B_t)\), and the increment is independent of \(\mathcal{F}_t\) with distribution \(N(0,u)\). Transition densities also compose: for \(s,t>0\), \(\int_{\mathbb{R}}p_s(x,z)p_t(z,y)\,dz=p_{s+t}(x,y)\). This is the Gaussian convolution identity: the sum of independent \(N(0,s)\) and \(N(0,t)\) increments is \(N(0,s+t)\).

Worked Example: A Conditional Transition Probability

Calculate \(\mathbb{P}(0<B_2<2\mid\mathcal{F}_1)\) on the event \(B_1=1\), using the transition density.

From time \(1\) to time \(2\), the increment \(B_2-B_1\) is independent of \(\mathcal{F}_1\) and has distribution \(N(0,1)\). At \(B_1=1\), the event \(0<B_2<2\) is equivalent to \(-1<B_2-B_1<1\). Consequently the conditional probability, evaluated at \(B_1=1\), is

\int_0^2p_1(1,y)\,dy =\int_0^2\frac{1}{\sqrt{2\pi}}\exp\!\left(-\frac{(y-1)^2}{2}\right)dy =\Phi(1)-\Phi(-1)=2\Phi(1)-1.

Check: the interval \((0,2)\) is centered at the starting value \(1\), and the one-unit increment is symmetric about zero. The answer is therefore the probability that a standard normal variable lies between \(-1\) and \(1\), exactly as the density calculation gives. The conditional-probability statement is a formula as a function of \(B_1\); evaluating it at \(B_1=1\) specifies the requested conditional value.

Scaling and Reflection for Maximum Probabilities

Brownian scaling states that for \(c>0\), the process \(\widetilde B_u=c^{-1/2}B_{cu}\), \(u\ge0\), is itself standard Brownian motion. In particular, a maximum over a time interval of length \(T\) can be rescaled to one over a unit interval. The reflection principle gives, for \(a>0\) and \(T>0\), \(\mathbb{P}(\sup_{0\le s\le T}B_s\ge a)=2\mathbb{P}(B_T\ge a)=2[1-\Phi(a/\sqrt T)]\). The reflection principle is a standard result from the earlier material on Brownian maxima. Its probability formula follows by reflecting a path after its first hit of \(a\): paths that hit \(a\) and end below \(a\) correspond to paths ending above \(a\). Since \(\mathbb{P}(B_T=a)=0\), the endpoint boundary does not affect the formula.

Worked Example: Rescaling a Maximum-Crossing Probability

Compute the probability that Brownian motion reaches level \(3\) by time \(4\), and express the calculation on a unit time interval.

Set \(\widetilde B_u=B_{4u}/2\) for \(0\le u\le1\). By Brownian scaling, \(\widetilde B\) is standard Brownian motion. The event \(\sup_{0\le s\le4}B_s\ge3\) is the same as \(\sup_{0\le u\le1}\widetilde B_u\ge3/2\). Reflection on the unit interval therefore gives

\mathbb{P}\!\left(\sup_{0\le s\le4}B_s\ge3\right) =2\left[1-\Phi\!\left(\frac32\right)\right].

Check: applying the formula directly at time \(T=4\) gives \(2[1-\Phi(3/\sqrt4)]=2[1-\Phi(3/2)]\), the same expression. The event uses a non-strict crossing, \(\sup B_s\ge3\); replacing it by a strict crossing does not change the probability because the maximum has a continuous distribution at \(3>0\).

Reflection can also answer questions that restrict the endpoint. For \(a>0\) and \(y<a\), the density of paths that hit \(a\) by time \(T\) and end at \(y\) is \(p_T(0,2a-y)\). The reflected endpoint is \(2a-y>a\). Integrating this density over an endpoint interval gives a crossing probability with that endpoint restriction.

Worked Example: Crossing a Barrier and Ending Below Zero

Find \(\mathbb{P}(\sup_{0\le s\le1}B_s\ge1,\ B_1\le0)\).

For endpoints \(y\le0\), we have \(y<1\), so the reflected endpoint density applies. Integrating over all such endpoints gives

\begin{aligned} \mathbb{P}\!\left(\sup_{0\le s\le1}B_s\ge1,\ B_1\le0\right) &=\int_{-\infty}^{0}p_1(0,2-y)\,dy\\ &=\int_{-\infty}^{0}\frac{1}{\sqrt{2\pi}} \exp\!\left(-\frac{(2-y)^2}{2}\right)dy\\ &=\mathbb{P}(B_1\ge2)=1-\Phi(2). \end{aligned}

In the last equality, the change of variable \(z=2-y\) turns the integration range into \(z\ge2\). Check: every path ending at \(B_1\le0\) that hits \(1\) is included in the unrestricted crossing event, whose probability is \(2[1-\Phi(1)]\). Since \(1-\Phi(2)\le2[1-\Phi(1)]\), the restricted probability has the required ordering. The endpoint condition is non-strict; equality at zero has probability zero.

Restarting at a Hitting Time

The strong Markov property at bounded stopping times is a standard result from the earlier material on Brownian motion at stopping times. In the form used here, if \(\tau\) is a bounded stopping time, then the process \((B_{\tau+u}-B_\tau)_{u\ge0}\) is Brownian motion independent of \(\mathcal{F}_\tau\). For an unbounded hitting time, one applies this result after stopping at deterministic bounds and uses the corresponding hitting-time result; the restart statement for Brownian motion at a hitting time is the standard strong Markov property. This is distinct from the deterministic-time property: the restart time is random, and the strong Markov property is what permits the same independent-increment calculation after that time.

Worked Example: A Post-Hit Probability

Let \(\tau_1=\inf\{s\ge0:B_s=1\}\). Find \(\mathbb{P}(\tau_1\le1,\ B_{\tau_1+1}>2)\).

On \(\{\tau_1<\infty\}\), continuity gives \(B_{\tau_1}=1\). The event \(B_{\tau_1+1}>2\) is thus the event that the Brownian increment over the next unit of time exceeds \(1\). By the strong Markov property, this increment is standard normal and independent of \(\mathcal{F}_{\tau_1}\). The event \(\{\tau_1\le1\}\) is measurable at \(\tau_1\), so

\begin{aligned} \mathbb{P}(\tau_1\le1,\ B_{\tau_1+1}>2) &=\mathbb{P}(\tau_1\le1)\,\mathbb{P}(N(0,1)>1)\\ &=2[1-\Phi(1)]\,[1-\Phi(1)]\\ &=2[1-\Phi(1)]^2. \end{aligned}

Here the reflection formula gives \(\mathbb{P}(\tau_1\le1)=\mathbb{P}(\sup_{0\le s\le1}B_s\ge1)=2[1-\Phi(1)]\). The strict inequality in the post-hit event can be used because a normal variable has no atom at \(1\); using a non-strict inequality gives the same probability. Check: the product is valid because the post-hit increment event is independent of the information at \(\tau_1\), including the event that the hit occurred by time \(1\). Each factor lies between zero and one, so the result is also a valid probability.

Common Pitfall: Conditioning at the Wrong Time

A frequent error is to use the deterministic-time independent-increment property as if it automatically applied at a random time. For example, writing \(\mathbb{P}(B_{\tau+1}-B_\tau>1\mid\mathcal{F}_\tau)=1-\Phi(1)\) is justified when \(\tau\) is a Brownian stopping time by the strong Markov property, but not merely because \(\tau\) is random. If \(\tau\) were an arbitrary random time depending on future Brownian values, the future increment need not be independent of the information used to choose \(\tau\). The correction is to check that the time is a stopping time and invoke the strong Markov property under its hypotheses.

Also keep distinct the two Gaussian calculations used here. Conditioning on an endpoint such as \(B_t=x\) produces a bridge-type conditional law whose mean depends on the endpoint. Conditioning on \(\mathcal{F}_s\) at a deterministic time instead uses the independent increment \(B_{s+u}-B_s\), with variance \(u\). Choosing the conditioning information first prevents mixing up these formulas.

Check Your Understanding

For each question, state the conditioning assumptions and use strict or non-strict event boundaries as written.

Find the conditional mean and variance of \(B_2\) given \(B_4=3\).
Write \(\mathbb{P}(1<B_3<3\mid\mathcal{F}_2)\) as a function of \(B_2\).
Use scaling and reflection to express \(\mathbb{P}(\sup_{0\le s\le9}B_s\ge6)\) in terms of \(\Phi\).
For \(a>0\), explain why the reflected endpoint corresponding to \(y<a\) is \(2a-y\).
State which property justifies treating the Brownian increments after a hitting time as independent of the past at that time.