From Brownian Increments to a Joint Law
A Brownian path is random at every time, but a finite collection of observations can be described using only the increments between successive observation times. Those increments are independent normal random variables, with variances equal to the lengths of their time intervals. Adding them cumulatively gives the observed Brownian values. This representation identifies the joint distribution without first needing to calculate a covariance matrix.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) be standard Brownian motion with state space \(\mathbb{R}\). In particular, \(B_0=0\) almost surely, paths are continuous, and for \(0\le s<t\), \(B_t-B_s\) has distribution \(N(0,t-s)\). Increments over disjoint time intervals are independent. We consider distinct ordered observation times \(0\le t_1<\cdots<t_k\), where \(k\) is a positive integer.
To construct the vector, put \(t_0=0\) and define the increments \(\Delta_r=B_{t_r}-B_{t_{r-1}}\) for \(1\le r\le k\). The intervals \((t_{r-1},t_r]\) are disjoint, so the \(\Delta_r\) are independent and \(\Delta_r\sim N(0,t_r-t_{r-1})\). The observations are their cumulative sums: \(B_{t_i}=\Delta_1+\cdots+\Delta_i\). If \(t_1=0\), then \(\Delta_1=B_0=0\) almost surely; that first coordinate is deterministic.
Proof. First, to prove Gaussianity, fix real numbers \(a_1,\ldots,a_k\). Using \(B_{t_i}=\sum_{r=1}^i\Delta_r\) and collecting the coefficient of each increment gives \[ \sum_{i=1}^k a_iB_{t_i} =\sum_{r=1}^k\left(\sum_{i=r}^k a_i\right)\Delta_r. \] For each \(r\), the coefficient in parentheses is deterministic. Since the increments are independent normal variables, the characteristic function of this linear combination at \(u\in\mathbb{R}\) is the product of their characteristic functions: \[ \prod_{r=1}^k \exp\!\left(-\frac{u^2}{2} \left(\sum_{i=r}^k a_i\right)^2(t_r-t_{r-1})\right). \] This product is the characteristic function of a centered normal variable, allowing variance zero. Thus every linear combination is normal, which proves that the vector is Gaussian, including when \(t_1=0\).
It remains to establish the density when \(t_1>0\). The increment vector \((\Delta_1,\ldots,\Delta_k)\) has independent normal coordinates with variances \(t_r-t_{r-1}>0\). Its density at \((y_1,\ldots,y_k)\) is the product of the \(k\) one-dimensional normal densities. The transformation from observations to increments is \(y_1=x_1\) and \(y_r=x_r-x_{r-1}\) for \(r\ge2\). Its matrix is lower triangular with diagonal entries all equal to \(1\), so its determinant is \(1\). The change-of-variables formula therefore gives the displayed density, with \(x_0=0\). The density integrates to \(1\), since it is the independent-increment density expressed under a transformation with Jacobian \(1\).
When \(t_1=0\), the first coordinate is fixed at zero. For \(r\ge2\), the increments have positive variances because the times are strictly ordered. Applying the same change of variables to the remaining coordinates gives their density. A distribution concentrated on the hyperplane whose first coordinate is zero cannot have a density with respect to \(k\)-dimensional volume. This covers both cases and completes the proof. \(\square\)
Working Directly with the Increments
The density formula has a useful interpretation: each successive difference \(x_r-x_{r-1}\) is assessed against the normal distribution for the corresponding time interval. Thus it is not a product of separate densities for the Brownian values themselves. The values share earlier increments; the increments, not the values, are independent. The examples below use this distinction to compute joint laws and check them in increment coordinates.
Worked Example: Three Observations with Unequal Gaps
Find the joint law of \((B_{1/2},B_{3/2},B_2)\). The observation times are \(1/2,3/2,2\), and their successive gaps are \(1/2,1,1/2\), respectively; they are unequal.
Define \(\Delta_1=B_{1/2}\), \(\Delta_2=B_{3/2}-B_{1/2}\), and \(\Delta_3=B_2-B_{3/2}\). By the Brownian increment properties, these variables are independent and have distributions \(N(0,1/2)\), \(N(0,1)\), and \(N(0,1/2)\). The observed vector is \((\Delta_1,\Delta_1+\Delta_2,\Delta_1+\Delta_2+\Delta_3)\). Writing its coordinates as \((x_1,x_2,x_3)\), the density is \[ \frac{1}{\pi\sqrt{2\pi}} \exp\!\left(-x_1^2-\frac{(x_2-x_1)^2}{2}-(x_3-x_2)^2\right). \] Indeed, the three normalizing factors are \(1/\sqrt{\pi}\), \(1/\sqrt{2\pi}\), and \(1/\sqrt{\pi}\), whose product is \(1/(\pi\sqrt{2\pi})\). The exponents are the increments squared divided by twice their variances: these denominators are \(1\), \(2\), and \(1\).
Check: Transform back using \(\Delta_1=x_1\), \(\Delta_2=x_2-x_1\), and \(\Delta_3=x_3-x_2\). The Jacobian determinant is \(1\), and the density becomes the product of the three independent normal densities with variances \(1/2,1,1/2\). Therefore it integrates to \(1\) and gives exactly the stated increment law.
Worked Example: A Positive-Increment Event at Three Times
Consider \((B_1,B_2,B_4)\). Find the probability that all three observations are positive and increasing in their listed order: \(0<B_1<B_2<B_4\).
The successive increments are \(\Delta_1=B_1\), \(\Delta_2=B_2-B_1\), and \(\Delta_3=B_4-B_2\). Their intervals have lengths \(1,1,2\), so the increments are independent with distributions \(N(0,1)\), \(N(0,1)\), and \(N(0,2)\). The event in question is exactly \(\{\Delta_1>0,\Delta_2>0,\Delta_3>0\}\). Each centered normal increment has probability \(1/2\) of being strictly positive, and independence gives \[ \mathbb{P}(0<B_1<B_2<B_4)=\frac12\cdot\frac12\cdot\frac12=\frac18. \] The inequalities are strict; replacing them by non-strict inequalities does not change the probability because each increment has a continuous distribution and hence has zero probability of equaling zero.
The joint density also follows from the increment law. For coordinates \((x_1,x_2,x_3)\), it is \[ \frac{1}{4\pi\sqrt{\pi}} \exp\!\left(-\frac{x_1^2}{2} -\frac{(x_2-x_1)^2}{2} -\frac{(x_3-x_2)^2}{4}\right). \] The denominator is the product \(\sqrt{2\pi}\sqrt{2\pi}\sqrt{4\pi}=4\pi\sqrt{\pi}\); the three exponent denominators are twice the corresponding variances \(1,1,2\). The event becomes \(x_1>0,\ x_2-x_1>0,\ x_3-x_2>0\) under the transformation to increments. Integrating the density over that event therefore gives the product of the three positive-half probabilities, \(1/8\), verifying the calculation.
Worked Example: Including the Observation at Time Zero
Describe the joint law of \((B_0,B_{1/4},B_1)\), and calculate the probability that \(B_{1/4}>0\) while \(B_1-B_{1/4}<0\).
The first coordinate equals zero almost surely. The increments after time zero are \(\Delta_1=B_{1/4}\) and \(\Delta_2=B_1-B_{1/4}\). Their interval lengths are \(1/4\) and \(1-1/4=3/4\), so they are independent with distributions \(N(0,1/4)\) and \(N(0,3/4)\). For the two nonconstant coordinates \((x_1,x_2)=(B_{1/4},B_1)\), the density is \[ \frac{2}{\pi\sqrt{3}} \exp\!\left(-2x_1^2-\frac{2}{3}(x_2-x_1)^2\right). \] The normalizing factor is the reciprocal of \(\sqrt{2\pi(1/4)}\sqrt{2\pi(3/4)} =\sqrt{\pi/2}\sqrt{3\pi/2}=\pi\sqrt{3}/2\). The exponent terms use \(2(1/4)=1/2\) and \(2(3/4)=3/2\) as their denominators, respectively.
For the requested event, \(\Delta_1>0\) and \(\Delta_2<0\). Each event has probability \(1/2\) by symmetry of its centered, nondegenerate normal distribution, and the increments are independent. Hence the probability is \(1/2\cdot1/2=1/4\). This is a law on the plane \(B_0=0\) inside \(\mathbb{R}^3\), not a three-dimensional density. Integrating the displayed two-coordinate density over the event in increment coordinates also gives \(1/4\), since it becomes the product of the two independent increment densities.
Common Pitfalls and a Practical Check
Common mistake: treating Brownian values as independent. For example, an incorrect calculation might assign the joint density of \((B_1,B_2)\) as the product of an \(N(0,1)\) density at \(x_1\) and an \(N(0,2)\) density at \(x_2\). That would treat \(B_1\) and \(B_2\) as independent. But \(B_2=B_1+(B_2-B_1)\), where the second term is independent of \(B_1\); the first increment remains part of both values. The correct density is \[ \frac{1}{2\pi} \exp\!\left(-\frac{x_1^2}{2}-\frac{(x_2-x_1)^2}{2}\right), \] because the increments \(B_1\) and \(B_2-B_1\) are independent \(N(0,1)\) variables. Its normalization is \(1/(2\pi)\), the product of their two normalizing factors. The difference \(x_2-x_1\), not \(x_2\), is the second increment.
Another mistake is to use a full-dimensional density when one of the observation times is zero. Since \(B_0=0\) almost surely, that coordinate has no spread and the joint distribution is concentrated on a lower-dimensional set. Remove the deterministic coordinate, find the density of the remaining observations, and then state separately that the removed coordinate equals zero.
The increment representation is useful whenever a calculation concerns finitely many Brownian observations: it gives a direct way to express their joint law, compute probabilities of events defined by successive differences, and detect degeneracy at time zero. As a final check in a calculation, take the proposed observation coordinates back to their successive differences. The differences should have the correct interval variances and should be the independent variables specified by the Brownian increment property.
Check Your Understanding
Use the independent increments between successive times, and check each result in increment coordinates.
- For observation times \(1,3\), write both Brownian values as sums of independent increments and state the distribution of each increment.
- For times \(1/2,1,2\), list the successive interval lengths and write the density of the three observations.
- For times \(0,2\), explain why the joint vector has no two-dimensional density, and give the distribution of its nonconstant coordinate.
- For \(0<B_1<B_3\), express the event using successive increments and find its probability.
- In the density for \((B_1,B_2)\), identify which expression represents the second increment and explain why it is not \(x_2\).