Why a Time Shift Needs Recentring
Brownian motion changes its location as time passes: \(B_h\) is generally random, and the later value \(B_{h+t}\) includes that accumulated position. The increment \(B_{h+t}-B_h\), by contrast, measures only the change after time \(h\). Brownian motion has the same increment distribution over every interval of a fixed length. More strongly, after subtracting \(B_h\), the whole shifted process has the same finite-dimensional distributions as Brownian motion starting at zero.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((B_t)_{t\ge0}\) be a standard Brownian motion, adapted to a filtration \((\mathcal{F}_t)_{t\ge0}\). We use the defining properties that \(B_0=0\) almost surely, paths are continuous almost surely, and increments over disjoint time intervals are independent, with \(B_t-B_s\) having the normal distribution with mean \(0\) and variance \(t-s\) whenever \(0\le s\le t\). In particular, \(B_t\) has mean \(0\) and variance \(t\). The covariance formula \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\) follows from these properties and will also be useful below.
The case \(t=0\) is included: the increment is then \(0\) almost surely. For \(t>0\), the increment has a nondegenerate normal distribution. The starting time \(h\) may be zero or positive; the definition imposes no additional restriction on it.
The Shifted Process
Fix \(h\ge0\), and define \(Y_t=B_{h+t}-B_h\) for \(t\ge0\). It starts at zero almost surely, and its paths are continuous almost surely because they are differences of shifted continuous paths. For the filtration, one may use \(\mathcal{G}_t=\mathcal{F}_{h+t}\); then \(Y_t\) is \(\mathcal{G}_t\)-measurable. The conclusion below concerns the joint distributions of its values at finitely many times, not equality of the random variables \(Y_t\) and \(B_t\).
Proof. Fix \(h\ge0\) and \(0\le t_1<\cdots<t_m\). Set \(t_0=0\). The successive differences of the shifted vector are \(Y_{t_1}-Y_{t_0}=B_{h+t_1}-B_h\), and, for \(r=2,\ldots,m\), \(Y_{t_r}-Y_{t_{r-1}}=B_{h+t_r}-B_{h+t_{r-1}}\). These are Brownian increments on the pairwise disjoint intervals \([h,h+t_1]\) and \([h+t_{r-1},h+t_r]\). They are independent, and their respective distributions are \(N(0,t_1)\) and \(N(0,t_r-t_{r-1})\). The successive differences of \((B_{t_1},\ldots,B_{t_m})\) are independent with exactly those same distributions, by the Brownian increment properties. Both vectors are obtained from their successive differences by the same cumulative-sum map: the \(r\)th coordinate is the sum of differences \(1\) through \(r\). Applying this map to vectors with the same distribution gives vectors with the same distribution. This proves the finite-dimensional assertion, including \(t_1=0\), when the first difference is the constant zero. Taking \(m=1\) and \(t_1=u\) gives the stated single-increment distribution. \(\square\)
The proof shows why the whole vector statement is stronger than checking one time at a time. A matching distribution for each individual increment does not by itself establish a matching joint distribution. Here, the successive increments are independent and have the same respective distributions for both vectors, which determines the joint distribution.
A Covariance Verification
There is another useful check. Brownian finite-dimensional vectors are Gaussian, and a Gaussian vector is determined by its mean vector and covariance matrix, as established in the earlier material on Gaussian vectors. Each \(Y_t\) has mean zero. For \(s,t\ge0\), bilinearity of covariance and \(\operatorname{Cov}(B_u,B_v)=\min(u,v)\) give
For the last equality, if \(s\le t\), the four minimum terms are \(h+s\), \(h\), \(h\), and \(h\), so their signed sum is \(s\). If \(t\le s\), they are \(h+t\), \(h\), \(h\), and \(h\), so the signed sum is \(t\). Thus the covariance is \(\min(s,t)\) in both cases, including \(s=0\) or \(t=0\). The shifted vector is Gaussian because it is a linear transformation of a Brownian Gaussian vector. Its means and covariances therefore match those of the corresponding Brownian vector, providing a second verification of the theorem.
Worked Example: Two Increments of Length Three
Compare the increments over \([2,5]\) and \([7,10]\). Define \(U=B_5-B_2\) and \(V=B_{10}-B_7\). Each interval has length \(3\). Brownian increments therefore give \(U\sim N(0,3)\) and \(V\sim N(0,3)\): both have mean \(0\), since each increment is centered, and variance \(3\), since the variance of an increment is its interval length. This verifies stationary increments for these two intervals.
The intervals are disjoint, so \(U\) and \(V\) are independent by the independent-increments property. Their joint distribution is consequently the product of two \(N(0,3)\) distributions. The equal marginal distributions alone would not establish this independence; disjointness and the Brownian independent-increments property are also being used.
Check: The length calculations are \(5-2=3\) and \(10-7=3\). Thus each variance is \(3\), and both means are \(0\). The disjoint intervals \([2,5]\) and \([7,10]\) have no overlapping positive-length portion, consistent with the independence conclusion.
Worked Example: A Shifted Vector at Three Times
Take \(h=3\) and the shifted times \(t_1=1,t_2=2,t_3=4\). The shifted vector is \((B_4-B_3,B_5-B_3,B_7-B_3)\). Its mean vector is \((0,0,0)\). By the covariance formula for shifted values, its covariance matrix has entries \(\min(t_i,t_j)\), hence is
For a direct check, the variances are \(4-3=1\), \(5-3=2\), and \(7-3=4\). The covariance of the first and second coordinates is \(\min(4,5)-\min(4,3)-\min(3,5)+3=4-3-3+3=1\). The first and third covariance is \(\min(4,7)-\min(4,3)-\min(3,7)+3=4-3-3+3=1\). The second and third covariance is \(\min(5,7)-\min(5,3)-\min(3,7)+3=5-3-3+3=2\). These give all off-diagonal entries; covariance symmetry gives the entries below the diagonal.
The vector is centered Gaussian, so this mean vector and covariance matrix determine its distribution. They are precisely the mean and covariance matrix of \((B_1,B_2,B_4)\). Thus the two vectors have the same distribution, as the theorem states.
Check: The increments between successive coordinates of the shifted vector are \(B_4-B_3\), \(B_5-B_4\), and \(B_7-B_5\). They are independent and have distributions \(N(0,1)\), \(N(0,1)\), and \(N(0,2)\), respectively. Their cumulative sums have variances \(1\), \(1+1=2\), and \(1+1+2=4\), agreeing with the covariance matrix diagonal.
Worked Example: Equal-Length Overlapping Increments
Let \(U=B_4-B_1\) and \(V=B_6-B_3\), corresponding to intervals \([1,4]\) and \([3,6]\). Both intervals have length \(3\), so \(U\) and \(V\) each have distribution \(N(0,3)\). They overlap on \([3,4]\), which has length \(1\). Their covariance can be computed from the Brownian covariance formula:
Thus the joint covariance matrix is \(\begin{bmatrix}3&1\\1&3\end{bmatrix}\). In particular, the increments are not independent: they are jointly Gaussian, and independent Gaussian coordinates would have covariance zero, whereas this covariance is \(1\). Equal-length intervals guarantee equal marginal distributions, not independence.
Check: Both diagonal entries are \(3\), the variance of an increment over length \(3\). The off-diagonal value \(1\) agrees with the overlap length \(4-3=1\). This confirms both the stationary-increment calculation and the dependence caused by overlap.
What the Shift Does—and Does Not—Preserve
Recentring is essential. The process \((B_{h+t})_{t\ge0}\) does not have the same finite-dimensional distributions as \((B_t)_{t\ge0}\) when \(h>0\). Already at one time \(t\), \(B_{h+t}\) has mean \(0\) and variance \(h+t\), whereas \(B_t\) has mean \(0\) and variance \(t\). The increment \(B_{h+t}-B_h\), instead, has variance \(t\). Stationary increments concern changes over intervals, not the absolute position of the path.
A common mistake is to treat an uncentered shifted value as if it were an increment. For example, with \(h=2\) and \(t=1\), the incorrect calculation is to assign \(B_{h+t}=B_3\) the distribution \(N(0,1)\), by using the interval length \(t=1\). But the variance of \(B_3\) is \(3\), not \(1\), so \(B_3\sim N(0,3)\). The correctly centered quantity is \(B_3-B_2\), and its variance is
This is \(N(0,1)\), as required for an increment over an interval of length \(1\). The correction is to subtract the value at the interval's starting time before applying the stationary-increment statement.
The key distinctions are:
- For any \(h,u\ge0\), \(B_{h+u}-B_h\sim N(0,u)\); the starting time does not change this marginal distribution.
- For any finite collection of times, the vector of recentered shifted values has the same joint distribution as the Brownian vector at those times.
- Neither statement says that \(B_{h+t}\) has the distribution of \(B_t\), or that increments on overlapping intervals are independent.
The finite-dimensional result is useful whenever a calculation involves a Brownian path observed after a fixed time. One can translate the time origin to \(h\), subtract \(B_h\), and use the familiar joint laws starting at zero. The subtraction preserves the random fluctuations after \(h\) while removing the random position accumulated before \(h\).
Check Your Understanding
Use the definitions and calculations above to answer the following questions. When computing a variance or covariance, show the time endpoints substituted into the formula.
- What is the distribution of \(B_{h+5}-B_h\) for a fixed \(h\ge0\)? State its mean and variance.
- For \(h=2\) and times \(t=1,3\), write the covariance matrix of \((B_3-B_2,B_5-B_2)\).
- Do \(B_4-B_1\) and \(B_6-B_3\) have the same distribution? Are they independent? Give the reason for each answer.
- For \(h=4\) and \(t=2\), compare the variances of \(B_{h+t}\) and \(B_{h+t}-B_h\). Explain why only the latter is determined by the interval length \(t\).
- In the finite-dimensional theorem, why is it necessary to compare joint distributions rather than only the marginal distribution at each time?