Why the Taylor Remainder Needs Its Own Estimate
For a smooth deterministic path, the second-order Taylor expansion has a remainder that is small compared with the square of the increment. Brownian increments are not uniformly small at a deterministic rate such as \(O(\Delta t)\). To use Taylor expansion along a Brownian partition, one therefore needs a bound that works for the actual increments of each path. The relevant control comes from uniform continuity of \(f''\) on a compact interval containing the path values.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix \(T>0\). For a partition \(\Pi=\{0=t_0<t_1<\cdots<t_n=T\}\), write \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\) and \(|\Pi|=\max_i(t_i-t_{i-1})\). The Brownian paths are continuous almost surely, so on each such path the values over \([0,T]\) are contained in some compact interval.
No stochastic integral is needed for the Taylor-remainder argument below. When an Ito integral \(\int_0^t H_s\,dB_s\) is used, its integrand is understood to be adapted and progressively measurable with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); this is the class \(\mathcal{L}^2\). The topic here is the ordinary, pathwise Taylor expansion of the increments \(B_{t_i}-B_{t_{i-1}}\), not a definition or proof of an Ito integral.
The Second-Order Expansion and Its Bound
Proof. The fundamental theorem of calculus, applied twice, gives \(f(x+h)=f(x)+f'(x)h+\int_0^h(h-u)f''(x+u)\,du\). Subtract \(\tfrac12 f''(x)h^2\), using \(\int_0^h(h-u)\,du=h^2/2\), to obtain the stated remainder. For \(h\ge0\), the definition of \(\omega_R\) gives \(|f''(x+u)-f''(x)|\le\omega_R(h)\) for \(0\le u\le h\). Therefore \(|r(x,h)|\le\omega_R(h)\int_0^h(h-u)\,du=\tfrac12\omega_R(h)h^2\). For \(h<0\), substitute \(v=-u\) in the oriented integral; the resulting integration is over an interval of length \(|h|\), the weight has absolute value at most \(|h|\), and the same bound is \(\tfrac12\omega_R(|h|)|h|^2\). Since \(|h|^2=h^2\), the estimate holds in both cases. \(\square\)
The bound uses only continuity of \(f''\), not a third derivative. If \(f''\) happens to be Lipschitz with constant \(L\), then \(\omega_R(\delta)\le L\delta\), giving the stronger familiar estimate \(|r(x,h)|\le L|h|^3/2\). In general, uniform continuity supplies a small factor \(\omega_R(|h|)\), while the square \(h^2\) records the scale of the second-order term.
Worked Example: A Cubic Function
Take \(f(x)=x^3\), \(x=1\), and \(h=\tfrac14\). Here \(f'(x)=3x^2\), \(f''(x)=6x\), and \(f''(1)=6\). The second-order Taylor polynomial at \(1\) is $$ f(1)+f'(1)h+\tfrac12 f''(1)h^2 =1+3\cdot\tfrac14+3\cdot\tfrac1{16} =\tfrac{31}{16}. $$ The exact endpoint value is \(f(\tfrac54)=\tfrac{125}{64}\), while \(\tfrac{31}{16}=\tfrac{124}{64}\). Thus \(r(1,\tfrac14)=\tfrac1{64}\).
Check: Expanding \((1+h)^3\) directly leaves the omitted term \(h^3\); at \(h=\tfrac14\), this is \(\tfrac1{64}\), matching the remainder. On the segment \([1,\tfrac54]\), the change in \(f''(z)=6z\) over distances at most \(\tfrac14\) is at most \(6\cdot\tfrac14=\tfrac32\). The theorem's bound is therefore \(\tfrac12\cdot\tfrac32\cdot(\tfrac14)^2=\tfrac3{64}\), and \(\tfrac1{64}\le\tfrac3{64}\).
Worked Example: The Exponential Function
For \(f(x)=e^x\), expand at \(x=0\) with \(h=\tfrac12\). Since \(f(0)=f'(0)=f''(0)=1\), the second-order approximation is \(1+\tfrac12+\tfrac18=\tfrac{13}{8}\). The exact value at the endpoint is \(e^{1/2}=\sqrt e\), so the remainder is \(\sqrt e-\tfrac{13}{8}\).
Here \(f''(u)=e^u\). The integral expression for the remainder gives $$ r(0,\tfrac12)=\int_0^{1/2}(\tfrac12-u)(e^u-1)\,du. $$ For \(0\le u\le\tfrac12\), \(0\le e^u-1\le\sqrt e-1\). Hence $$ 0\le r(0,\tfrac12)\le(\sqrt e-1)\int_0^{1/2}(\tfrac12-u)\,du =\frac{\sqrt e-1}{8}. $$ Check: evaluating the Taylor formula gives \(r=\sqrt e-\tfrac{13}{8}\), while evaluating the integral gives the same value: \(\int_0^{1/2}(\tfrac12-u)e^u\,du-\int_0^{1/2}(\tfrac12-u)\,du=\sqrt e-\tfrac32-\tfrac18=\sqrt e-\tfrac{13}{8}\). The displayed integral bound also verifies that this remainder is nonnegative and no larger than \((\sqrt e-1)/8\).
Summing Remainders Along a Partition
For each interval of a Brownian partition, set \(x=B_{t_{i-1}}\) and \(h=\Delta B_i\). On a path that stays in \([-R,R]\), every segment from \(B_{t_{i-1}}\) to \(B_{t_i}\) lies in that interval. Applying the theorem separately on each segment gives remainders \(r_i\) satisfying $$ \left|\sum_{i=1}^n r_i\right| \le \sum_{i=1}^n |r_i| \le \tfrac12\omega_R\!\left(\max_i|\Delta B_i|\right)\sum_{i=1}^n(\Delta B_i)^2. $$ This inequality is the useful estimate: the modulus is evaluated at the largest Brownian increment, while the sum of squared increments controls the total second-order scale.
Proof. Define \(R_\Pi=\sum_i r(B_{t_{i-1}},\Delta B_i)\), \(m_\Pi=\max_i|\Delta B_i|\), and \(Q_\Pi=\sum_i(\Delta B_i)^2\). Brownian paths are continuous almost surely. Each such path is uniformly continuous on \([0,T]\), so \(m_\Pi\to0\) almost surely as \(|\Pi|\to0\). The quadratic-variation result for Brownian motion gives \(Q_\Pi\to T\) in probability; in particular, \(Q_\Pi\) is bounded in probability. These facts about Brownian paths and quadratic variation are the standard results established in the earlier material on Brownian quadratic variation.
Fix \(\varepsilon>0\) and an arbitrary probability tolerance \(\eta>0\). Since \(\sup_{0\le t\le T}|B_t|\) is finite almost surely, choose \(R\) large enough that the event \(A_R=\{\sup_{0\le t\le T}|B_t|\le R\}\) has probability at least \(1-\eta\). On \(A_R\), the summed bound applies. Given \(K>0\), it yields $$ \mathbb{P}(|R_\Pi|>\varepsilon) \le \mathbb{P}(A_R^c)+\mathbb{P}(Q_\Pi>K) +\mathbb{P}\!\left(\tfrac12\omega_R(m_\Pi)K>\varepsilon\right). $$ Because \(Q_\Pi\to T\) in probability, \(K\) can be chosen so that the second probability is arbitrarily small for all sufficiently fine partitions. For this fixed \(R\) and \(K\), \(m_\Pi\to0\) almost surely and \(\omega_R(\delta)\to0\) as \(\delta\downarrow0\); hence the last probability tends to zero. The first probability is at most \(\eta\), and \(\eta\) was arbitrary. Thus \(\mathbb{P}(|R_\Pi|>\varepsilon)\to0\), which is convergence in probability. The argument also covers the empty-sum case \(T=0\), where the remainder is zero; here the proposition was stated for \(T>0\). \(\square\)
This proposition controls only the accumulated Taylor remainder. It does not, by itself, establish convergence of the sums involving \(f'(B_{t_{i-1}})\Delta B_i\) or \(f''(B_{t_{i-1}})(\Delta B_i)^2\). Those terms require their own convergence arguments. Keeping these tasks separate prevents the Taylor estimate from being mistaken for a complete proof of Ito's formula.
Worked Example: Adding Remainders Across Two Increments
Use \(f(x)=\sin x\) and the two successive increments \(0\to\tfrac14\to0\). This is a deterministic calculation of the Taylor remainders for those path values. For the first increment, \(x=0\), \(h=\tfrac14\), and \(f'(0)=1,\ f''(0)=0\), so $$ r_1=\sin(\tfrac14)-\tfrac14. $$ For the second, \(x=\tfrac14\), \(h=-\tfrac14\), \(f'(x)=\cos(\tfrac14)\), and \(f''(x)=-\sin(\tfrac14)\). Substituting all terms gives $$ r_2=0-\left(\sin(\tfrac14)-\tfrac14\cos(\tfrac14)-\tfrac1{32}\sin(\tfrac14)\right) =-\sin(\tfrac14)+\tfrac14\cos(\tfrac14)+\tfrac1{32}\sin(\tfrac14). $$ Adding, $$ r_1+r_2=-\tfrac14+\tfrac14\cos(\tfrac14)+\tfrac1{32}\sin(\tfrac14). $$ Check: Since \(f''(z)=-\sin z\) and its derivative \(-\cos z\) has absolute value at most \(1\), the mean value theorem gives \(|f''(z)-f''(w)|\le|z-w|\). Each remainder is therefore bounded by \(\tfrac12(\tfrac14)(\tfrac14)^2=\tfrac1{128}\), so \(|r_1+r_2|\le|r_1|+|r_2|\le\tfrac1{64}\). The exact expression above is the sum obtained by substituting the two Taylor remainders, and it obeys this bound. The endpoint values agree, \(f(0)-f(0)=0\); the total remainder compensates exactly for the sum of the two first- and second-order Taylor terms.
Common Pitfall: Assuming a Third Derivative
A frequent shortcut is to write a remainder as \(f'''(\xi)h^3/6\) without checking that \(f'''\) exists and is bounded on the relevant interval. That form of Taylor's theorem is not available for every \(C^2\) function. The integral remainder above requires only continuity of \(f''\), and its bound uses the modulus of continuity on a compact interval.
For example, the incorrect calculation for a general \(C^2\) function would be to assert \(|r(x,h)|\le C|h|^3\) for some constant \(C\) merely because \(f''\) is continuous. Continuity on a compact interval gives \(\omega_R(|h|)\to0\), but does not assert \(\omega_R(|h|)\le L|h|\). The valid conclusion is $$ |r(x,h)|\le\tfrac12\omega_R(|h|)h^2. $$ A cubic bound follows if one additionally proves that \(f''\) is Lipschitz; it must not be assumed from continuity alone.
The compact interval is also essential to the uniform estimate. A Brownian path is bounded on a fixed finite time interval almost surely, but the bound can depend on the path. Localization to \(A_R\), followed by taking \(R\) large, handles this dependence without claiming a single compact interval contains every Brownian path with probability one.
Check Your Understanding
Use the integral remainder and the compact-interval modulus bound in your answers.
- For \(f(x)=x^3\), expand at \(x=0\) with increment \(h\), and compute the exact second-order remainder.
- Explain why uniform continuity of \(f''\) on \([-R,R]\) implies \(\omega_R(\delta)\to0\), and state where compactness is used.
- For a partition, derive the bound on \(\left|\sum_i r_i\right|\) in terms of \(\omega_R(\max_i|\Delta B_i|)\) and \(\sum_i(\Delta B_i)^2\).
- Why does convergence of the summed Taylor remainders not prove convergence of the first-order sums?
- What additional condition on \(f''\) gives a cubic remainder bound, and what constant appears when \(f''\) is Lipschitz with constant \(L\)?