Time Increments and Mixed Terms in the Proof

Separating Time and Space Increments

In a partition proof for a function of time and Brownian motion, the increment from \(f(t_{i-1},B_{t_{i-1}})\) to \(f(t_i,B_{t_i})\) changes both arguments at once. Expanding both changes together can obscure which terms survive refinement. The useful first step is to separate the time change, the space change, and the interaction between them. The pure time second-order remainder and the mixed time-space increment both vanish after summation as the mesh tends to zero. The spatial second-order contribution is different: its behavior is tied to Brownian quadratic variation and is not discarded by these estimates.

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 \((\mathcal{F}_t)\). Fix \(T>0\). For a deterministic partition \(\Pi=\{0=t_0<t_1<\cdots<t_n=T\}\), write \(\Delta t_i=t_i-t_{i-1}\), \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\), and \(|\Pi|=\max_i\Delta t_i\). Consider a function \(f(t,x)\) whose partial derivatives \(f_t,f_{tt},f_x,f_{xx},f_{tx}\) exist and are continuous on a neighborhood of \([0,T]\times\mathbb{R}\). Brownian paths are continuous almost surely, so the estimates below apply on every continuous path, and hence almost surely for \(B\).

For each interval, the exact decomposition is

The first bracket changes time while holding space fixed; the second changes space while holding time fixed. The final term accounts exactly for the interaction of the two changes. No stochastic integral is used in this decomposition or in the estimates that follow.

The Pure Time Remainder

Proof. Fix the path \(x\). Its range is bounded, so choose \(R\) with \(|x(t)|\le R\) for all \(t\in[0,T]\). Continuity of \(f_{tt}\) on the compact rectangle \([0,T]\times[-R,R]\) gives a finite bound \(M_{tt}=\sup|f_{tt}|\) there. Taylor's theorem in the time variable, with space held at \(x(t_{i-1})\), gives \(|A_i|\le \tfrac12 M_{tt}(\Delta t_i)^2\). Since \((\Delta t_i)^2\le|\Pi|\Delta t_i\), summing yields \(\sum_i|A_i|\le\tfrac12M_{tt}|\Pi|\sum_i\Delta t_i=\tfrac12M_{tt}T|\Pi|\), which tends to zero. \(\square\)

This is a pathwise estimate: its constant \(M_{tt}\) may depend on the range of the path, but it is finite for each continuous path. For Brownian motion, continuity holds with probability one. The argument does not require independence of increments or a moment bound on their sizes; it uses only the shrinking time mesh and boundedness of \(f_{tt}\) over the path's range.

The Mixed Time-Space Increment

Proof. Fix a continuous path \(x\), and choose \(R\) so that its range lies in \([-R,R]\). The continuous derivative \(f_{tx}\) is bounded on \([0,T]\times[-R,R]\); write \(M_{tx}=\sup|f_{tx}|\) there. Applying the fundamental theorem of calculus first in space and then in time gives the exact identity \(C_i=\int_{x(t_{i-1})}^{x(t_i)}\int_{t_{i-1}}^{t_i} f_{tx}(s,y)\,ds\,dy\), where an integral with reversed endpoints has its usual signed meaning. Therefore \(|C_i|\le M_{tx}\Delta t_i|x(t_i)-x(t_{i-1})|\). Let \(\omega_x(\delta)=\sup\{|x(t)-x(s)|:s,t\in[0,T],\,|t-s|\le\delta\}\). Uniform continuity of \(x\) on \([0,T]\) implies \(\omega_x(\delta)\to0\) as \(\delta\to0\). Hence \(\sum_i|C_i|\le M_{tx}\omega_x(|\Pi|)\sum_i\Delta t_i=M_{tx}T\omega_x(|\Pi|)\to0\). \(\square\)

The same argument applies to Brownian paths almost surely. In particular, no assertion that a Brownian increment is bounded by a fixed constant times \(\Delta t_i\) is needed. Instead, continuity controls the largest increment across intervals of length at most the mesh: \(\max_i|\Delta B_i|\le\omega_B(|\Pi|)\to0\) almost surely. The time factors \(\Delta t_i\) then make the total mixed contribution vanish.

How the Bounds Enter a Partition Expansion

Combining the exact decomposition with one-variable Taylor expansions makes the role of these estimates explicit. At the left endpoint of the interval, expand the time bracket to first order and the space bracket to second order. The resulting identity has the form

Here \(A_i\) is the pure time remainder defined above, \(C_i\) is the mixed rectangular increment, and \(R_i\) is the remainder in the spatial second-order Taylor expansion. The two propositions prove that \(\sum_i A_i\to0\) and \(\sum_i C_i\to0\) almost surely as the mesh tends to zero. They do not prove that the spatial remainders vanish, nor do they remove the spatial second-order term. Those require the separate analysis of spatial increments and their quadratic variation.

This distinction is the point of separating the increments. The time variable has finite variation along the partition, so its squared increments have total size at most \(T|\Pi|\). For a mixed term, continuity supplies a vanishing maximum space increment, while the time increments sum to \(T\). Brownian squared space increments, by contrast, have a nonzero limiting sum. In the differential shorthand used in stochastic calculus, the discarded types correspond to \(dt\,dt=0\) and \(dt\,dB_t=0\); the surviving quadratic contribution corresponds to \(dB_t\,dB_t=dt\). These are limit rules for partition sums, not ordinary algebraic identities for individual increments.

Common Pitfall: Dropping the Time Factor

A mistaken calculation can make a mixed sum appear not to vanish: replacing \(\sum_i\Delta t_i\Delta B_i\) by \(\sum_i\Delta B_i=B_T\) silently drops the factor \(\Delta t_i\). The unweighted Brownian increments telescope to \(B_T\), which generally is not zero; the mixed sum is a different quantity. On the equal partition of \([0,1]\), the correct sum for \(f(t,x)=tx\) is \(B_1/n\), not \(B_1\). The correction is not that Brownian increments themselves vanish after summation, but that the mixed increments carry time weights whose total contribution is controlled by continuity and the shrinking mesh.

Another tempting but unnecessary shortcut is to treat every \(|\Delta B_i|\) as if it were bounded by a deterministic constant times \(\sqrt{\Delta t_i}\). Such a bound is not supplied by path continuity and is not needed here. The proof uses the valid pathwise statement \(\max_i|\Delta B_i|\to0\) as the mesh shrinks, together with \(\sum_i\Delta t_i=T\). This is enough to establish the mixed-term limit without an unproved estimate on the size of each Brownian increment.