Two Transformations That Preserve Brownian Laws
Brownian motion has covariance \(\mathbb{E}[B_sB_t]=\min(s,t)\). That simple formula makes it possible to test transformations of its time parameter and values without calculating complicated path distributions. Two useful transformations are to read the increments over a fixed interval backward and to invert time while multiplying the process by time. In both cases, the transformed process has the covariance of Brownian motion; because the processes are Gaussian, this identifies all their finite-dimensional distributions.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Thus \(B_0=0\), its paths are continuous almost surely, and its increments are independent centered normal random variables with \(\mathbb{E}[(B_t-B_s)^2]=t-s\) for \(0\le s\le t\). In particular, \(\mathbb{E}[B_sB_t]=\min(s,t)\). The transformations below are statements about the law of \(B\); they do not assert that a transformed process is adapted to the original filtration \((\mathcal{F}_t)\).
Reversal of Brownian Increments
The subtraction in \(R_t\) matters: it makes \(R_0=0\) and records the increment accumulated over the final \(t\) units of the interval. The covariance calculation below gives the central identity.
Proof. It suffices first to take \(0\le s\le t\le T\). The vector \((R_s,R_t)\) is centered Gaussian because it is a linear transformation of Brownian values. Using \(\mathbb{E}[B_aB_b]=\min(a,b)\), and noting that \(T-t\le T-s\), we obtain
Symmetry in \(s,t\) gives \(\mathbb{E}[R_sR_t]=\min(s,t)\) for all pairs. Also \(R_0=0\), and the paths of \(R\) are continuous because those of \(B\) are continuous. To verify independent increments, let \(0\le s\le u\le v\le T\). Then
More generally, the same calculation gives zero covariance between \(R_v-R_u\) and each \(R_{s_j}\) for any finite collection \(s_j\le u\). These variables are jointly Gaussian, so zero covariance implies independence. Consequently \(R_v-R_u\) is independent of every finite collection of past values. By the definition of the sigma-field generated by those values, it is independent of \(\mathcal{G}_u\). Its mean is zero and its variance is
Thus the increments are independent of the past and have the required normal distribution \(N(0,v-u)\). Together with the initial value and path continuity, this is the Brownian property on \([0,T]\). \(\square\)
Worked Example: Reversing the Final Increments on \([0,5]\)
Take \(T=5\) and compare \(R_1\) and \(R_3\). Their covariance matrix, using the theorem’s formula, is
The later increment is \(R_3-R_1=(B_5-B_2)-(B_5-B_4)=B_4-B_2\), while \(R_1=B_5-B_4\). Their variances are \(2\) and \(1\), respectively, and their covariance is \(0\), since the Brownian increments on \([2,4]\) and \([4,5]\) are independent. The covariance-matrix calculation gives the same check: \(\operatorname{Var}(R_3-R_1)=3+1-2(1)=2\) and \(\operatorname{Cov}(R_1,R_3-R_1)=1-1=0\). Since the pair is Gaussian, these two increments are independent.
Check: the increment \(R_3-R_1\) has variance \(3-1=2\), exactly the length of its time interval. Both its direct Brownian representation and the covariance formula agree.
Time Inversion and Its Covariance
Time inversion sends large times of \(B\) to small times of \(I\), and small times of \(I\) to large times of \(B\). The multiplier \(t\) compensates for the change in variance. The covariance identity works for positive times; continuity at zero requires a separate pathwise argument.
Proof. For \(s,t>0\), the variables \(I_s,I_t\) are centered jointly Gaussian. If \(s\le t\), then \(1/t\le1/s\), and hence
Symmetry gives \(\mathbb{E}[I_sI_t]=\min(s,t)\) for all positive \(s,t\). This proves the desired covariance identity away from zero. To establish continuity at zero, we show that \(B_u/u\to0\) almost surely as \(u\to\infty\).
For any fixed \(\varepsilon>0\), the Gaussian tail bound gives \(\mathbb{P}(|B_n|>\varepsilon n)\le2e^{-\varepsilon^2n/2}\), and the right-hand side is summable over positive integers \(n\). On each interval \([n,n+1]\), the process \(M_u=B_{n+u}-B_n\), \(0\le u\le1\), is a square-integrable martingale with \(\mathbb{E}[M_1^2]=1\). Doob’s \(L^2\) maximal inequality therefore gives
This bound is also summable. By the Borel–Cantelli lemma, for each fixed \(\varepsilon>0\), almost surely both \(|B_n|\le\varepsilon n\) and \(\sup_{0\le u\le1}|B_{n+u}-B_n|\le\varepsilon n\) hold for all sufficiently large \(n\). For \(x\in[n,n+1]\), \(n\ge1\), it follows that \[ \frac{|B_x|}{x}\le\frac{|B_n|+\sup_{0\le u\le1}|B_{n+u}-B_n|}{n}\le2\varepsilon. \] Applying this argument to \(\varepsilon=1/k\) for each positive integer \(k\) proves \(B_x/x\to0\) almost surely. With \(x=1/t\), we have \(I_t=tB_{1/t}=B_x/x\to0=I_0\) as \(t\downarrow0\). Away from zero the paths are continuous by continuity of \(B\).
It remains to identify the finite-dimensional laws and increment property. For any finite collection of positive times, the corresponding vector of \(I\)-values is centered Gaussian and has covariance entries \(\min(s,t)\); including time zero only adds a deterministic zero coordinate. These are precisely the centered Gaussian finite-dimensional laws of Brownian motion. For \(0\le u\le v\), the covariance identity gives \(\operatorname{Var}(I_v-I_u)=v-u\) and, for every \(s\le u\), \(\operatorname{Cov}(I_v-I_u,I_s)=s-s=0\). Joint Gaussianity and the sigma-field argument used in the reversal theorem imply that each increment is independent of the natural past. Together with path continuity and \(I_0=0\), this proves the claim. \(\square\)
Worked Example: Covariances at Times \(1/2\), \(2\), and \(3\)
For the time-inverted process, the covariance of two positive-time values is \(\min(s,t)\). Substituting each pair of times gives
For instance, the off-diagonal entry at times \(2\) and \(3\) is \(2\cdot3\min(1/2,1/3)=6(1/3)=2\). The diagonal entry at time \(3\) is \(3^2\min(1/3,1/3)=9(1/3)=3\). The increments over \([1/2,2]\) and \([2,3]\) have variances \(2-1/2=3/2\) and \(3-2=1\). Their covariance is \(2-2-1/2+1/2=0\), as follows by expanding \(\operatorname{Cov}(I_2-I_{1/2},I_3-I_2)\). They are jointly Gaussian, so they are independent.
Check: the increment variances \(3/2\) and \(1\) equal the lengths of the two time intervals. The displayed covariance matrix gives the same values when the increments are expanded.
Worked Example: Three Independent Increments of the Inverted Process
Consider \(I_1\), \(I_2-I_1\), and \(I_4-I_2\). Their respective variances, computed from \(\mathbb{E}[I_sI_t]=\min(s,t)\), are
The three pairwise covariances are also obtained by expanding products: \[ \begin{aligned} \operatorname{Cov}(I_1,I_2-I_1)&=1-1=0,\\ \operatorname{Cov}(I_1,I_4-I_2)&=1-1=0,\\ \operatorname{Cov}(I_2-I_1,I_4-I_2)&=2-2-1+1=0. \end{aligned} \] The last line uses \(\mathbb{E}[I_2I_4]=2\), \(\mathbb{E}[I_2^2]=2\), \(\mathbb{E}[I_1I_4]=1\), and \(\mathbb{E}[I_1I_2]=1\). The vector of these increments is jointly Gaussian, so zero pairwise covariances imply mutual independence. Their distributions are \(N(0,1)\), \(N(0,1)\), and \(N(0,2)\), matching the lengths of the intervals \([0,1]\), \([1,2]\), and \([2,4]\).
Check: adding the three interval lengths gives \(1+1+2=4\), the total elapsed time from \(0\) to \(4\); the sum of the three increment variances is also \(1+1+2=4\).
Why the Exact Transformations Matter
These identities are useful because they let one replace a transformed process by a standard Brownian motion in distribution when studying questions that depend on its finite-time values. Time reversal is especially easy to misstate: the process \(B_{T-t}\) is not the reversed-increment process. For example, it starts at \(B_T\), not zero, and its covariance for \(s\le t\) is \(\mathbb{E}[B_{T-s}B_{T-t}]=T-t\), not \(s\).
Common mistake. A tempting but incorrect calculation is to assert that \(\widetilde R_t=B_{T-t}\) is Brownian because it reads Brownian values backward. On \([0,T]\), \(\widetilde R_0=B_T\), which has variance \(T\), whereas Brownian motion must start at zero. Its covariance is \(T-\max(s,t)\), not \(\min(s,t)\). The correction is to reverse the increments: \(R_t=B_T-B_{T-t}\). Then \(R_0=0\), and the covariance calculation gives \(\min(s,t)\).
For time inversion, the covariance formula at positive times is not by itself a path-continuity argument at zero. The multiplier \(t\) and the fact that \(B_u/u\to0\) almost surely are both relevant. Once continuity is established, the covariance and Gaussian properties give the Brownian finite-dimensional laws and the independent-increment property.
Check Your Understanding
Use the covariance identities and the definitions to answer these questions.
- For the reversed process on \([0,6]\), compute \(\mathbb{E}[R_2R_5]\) and \(\operatorname{Var}(R_5-R_2)\).
- For \(0<s\le t\), derive \(\mathbb{E}[I_sI_t]\) directly from \(I_u=uB_{1/u}\).
- Explain why zero covariance between a Gaussian increment and every finite collection of past values gives independence from the natural past sigma-field.
- For the times \(1,3,5\), compute the covariance matrix of \((I_1,I_3,I_5)\).
- Identify the initial-value and covariance errors in claiming that \(B_{T-t}\) itself is Brownian motion in reversed time.