Partition Sums in the Proof of Ito's Formula

From One Increment to a Partition Sum

To organize the proof of Ito’s formula, first fix a finite time interval and split it into deterministic pieces. On each piece, compare the change in $f(B_t)$ with the first two Taylor terms at the left endpoint. Adding those comparisons over the partition gives an exact finite-sum identity. The limiting problem is then visible: the first-order sums correspond to a stochastic integral, the quadratic sums reflect Brownian quadratic variation, and the Taylor remainders must be controlled.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix $T>0$. A deterministic partition of $[0,T]$ is a finite set of times $0=t_0<t_1<\cdots<t_n=T$; write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and $|\Pi|=\max_i\Delta t_i$. For a function $f:\mathbb{R}\to\mathbb{R}$, set $X_i=f(B_{t_i})$. The basic fact is that $\sum_{i=1}^n(X_i-X_{i-1})=X_n-X_0$, regardless of the size or shape of the intervals.

Definition: For a twice continuously differentiable $f$, the first-order, quadratic, and remainder terms on the $i$-th interval are, respectively, $f'(B_{t_{i-1}})\Delta B_i$, $\tfrac12 f''(B_{t_{i-1}})(\Delta B_i)^2$, and the difference between $\Delta f(B)_i=f(B_{t_i})-f(B_{t_{i-1}})$ and those two terms.

The Exact Finite-Partition Decomposition

Proposition: For any finite deterministic partition and any function $f$, the increments telescope: $$ f(B_T)-f(B_0)=\sum_{i=1}^n\big(f(B_{t_i})-f(B_{t_{i-1}})\big). $$ If $f$ is twice continuously differentiable, define $$ R_i=f(B_{t_i})-f(B_{t_{i-1}})-f'(B_{t_{i-1}})\Delta B_i-\tfrac12 f''(B_{t_{i-1}})(\Delta B_i)^2. $$ Then the following identity is exact: $$ f(B_T)-f(B_0)=\sum_{i=1}^n f'(B_{t_{i-1}})\Delta B_i +\tfrac12\sum_{i=1}^n f''(B_{t_{i-1}})(\Delta B_i)^2 +\sum_{i=1}^n R_i. $$

Proof. In the first sum, each intermediate value $f(B_{t_i})$, for $1\le i<n$, occurs once with a plus sign and once with a minus sign. Only $f(B_{t_n})=f(B_T)$ and $-f(B_{t_0})=-f(B_0)$ remain. For the second identity, the definition of $R_i$ rearranges on each interval to $$ f(B_{t_i})-f(B_{t_{i-1}}) =f'(B_{t_{i-1}})\Delta B_i +\tfrac12 f''(B_{t_{i-1}})(\Delta B_i)^2+R_i. $$ Summing these equalities and applying the first telescoping identity proves the result. This is a finite algebraic identity; no limiting argument is needed. $\square$

The half in the quadratic term comes from the ordinary second-order Taylor coefficient. It is not inserted as an assumption about Brownian motion. Brownian motion matters when the partition is refined: its squared increments do not behave like the squared increments of a differentiable path.

What the Three Sums Represent

When $|\Pi|$ decreases, the first sum is a left-endpoint stochastic sum. To state precisely how it relates to an Ito integral, recall that an Ito integrand on $[0,T]$ belongs to $\mathcal{L}^2$ when it is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The Ito integral $\int_0^T H_s\,dB_s$ is defined for integrands in this class. If $f'$ is bounded, the process $H_s=f'(B_s)$ is progressively measurable and belongs to $\mathcal{L}^2$. The step process $$ H_s^{\Pi}=\sum_{i=1}^n f'(B_{t_{i-1}})\mathbf{1}_{(t_{i-1},t_i]}(s) $$ also belongs to $\mathcal{L}^2$: each coefficient is measurable with respect to $\mathcal{F}_{t_{i-1}}$, and boundedness gives $\mathbb{E}\int_0^T(H_s^{\Pi})^2\,ds<\infty$. By the definition of the Ito integral for elementary adapted step processes, $$ \int_0^T H_s^{\Pi}\,dB_s=\sum_{i=1}^n f'(B_{t_{i-1}})\Delta B_i. $$ Thus the finite first-order sum is exactly an Ito integral of a step integrand. Identifying the limit of these integrals requires a convergence argument; the finite identity itself does not supply one.

The second sum weights the squared increments by the curvature $f''$ at the left endpoint. Brownian quadratic variation is expressed by $dB_t\,dB_t=dt$; in a partition proof, this notation points to the role of the squared-increment sum, not to an equality for each individual increment. The remainders $R_i$ record precisely what the two displayed Taylor terms do not capture. Their total must be handled rather than discarded just because each individual interval is short.

Partition term	Finite-sum form	Role in the decomposition
First order	$\sum_i f'(B_{t_{i-1}})\Delta B_i$	Left-endpoint stochastic sum
Quadratic	$\tfrac12\sum_i f''(B_{t_{i-1}})(\Delta B_i)^2$	Curvature weighted by squared increments
Remainder	$\sum_i R_i$	Difference between the exact increment and the two Taylor terms

Worked Partition Decompositions

Worked Example: The Square Function on Two Intervals

Let $f(x)=x^2$, and consider a path whose values at $t_0,t_1,t_2$ are $0,1,3$. Compute each part of the decomposition for the partition $t_0<t_1<t_2$.

Here $f'(x)=2x$, $f''(x)=2$, and the increments are $\Delta B_1=1-0=1$ and $\Delta B_2=3-1=2$. The first-order terms are $2(0)(1)=0$ and $2(1)(2)=4$, so their sum is $4$. The quadratic terms are $\tfrac12(2)(1)^2=1$ and $\tfrac12(2)(2)^2=4$, so their sum is $5$. For a quadratic function the Taylor expansion is exact, giving $R_1=R_2=0$. Therefore the right side is $4+5=9$.

Check: directly, $f(3)-f(0)=3^2-0^2=9$. The increment-by-increment changes are $1^2-0^2=1$ and $3^2-1^2=8$; they sum to $1+8=9$, as required.

Worked Example: A Cubic Function and Nonzero Remainders

Let $f(x)=x^3$, and take consecutive path values $-1,1,2$. Compute the sums and remainders. These values illustrate the finite algebraic identity; they are not a claim about a typical Brownian path.

For $f(x)=x^3$, $f'(x)=3x^2$ and $\tfrac12 f''(x)=3x$. On the first interval, $x=-1$ and $\Delta B=2$. The first-order term is $3(-1)^2(2)=6$; the quadratic term is $3(-1)(2)^2=-12$. The exact function increment is $1^3-(-1)^3=2$, so $R_1=2-6-(-12)=8$.

On the second interval, $x=1$ and $\Delta B=1$. The first-order term is $3(1)^2(1)=3$; the quadratic term is $3(1)(1)^2=3$. The exact function increment is $2^3-1^3=7$, so $R_2=7-3-3=1$. The total first-order sum is $6+3=9$; the quadratic sum is $-12+3=-9$; and the remainder sum is $8+1=9$. Together they give $9-9+9=9$.

Check: the endpoint change is $f(2)-f(-1)=8-(-1)=9$. Also, expanding $(x+\Delta B)^3$ gives the exact remainder $(\Delta B)^3$: its values here are $2^3=8$ and $1^3=1$, agreeing with $R_1$ and $R_2$.

Worked Example: Trigonometric Values on a Partition

Let $f(x)=\sin x$, with path values $0,\tfrac{\pi}{2},\pi$. Compute the three sums. This bounded-derivative function also fits the global integrability condition stated above for its first-order integrand.

We have $f'(x)=\cos x$ and $\tfrac12 f''(x)=-\tfrac12\sin x$. Both increments equal $\tfrac{\pi}{2}$. On the first interval, the first-order term is $\cos(0)\tfrac{\pi}{2}=\tfrac{\pi}{2}$, and the quadratic term is $-\tfrac12\sin(0)(\tfrac{\pi}{2})^2=0$. The exact change is $\sin(\tfrac{\pi}{2})-\sin(0)=1$, so $R_1=1-\tfrac{\pi}{2}$.

On the second interval, the first-order term is $\cos(\tfrac{\pi}{2})\tfrac{\pi}{2}=0$, and the quadratic term is $-\tfrac12\sin(\tfrac{\pi}{2})(\tfrac{\pi}{2})^2=-\tfrac{\pi^2}{8}$. The exact change is $\sin(\pi)-\sin(\tfrac{\pi}{2})=-1$, so $R_2=-1-(-\tfrac{\pi^2}{8})=\tfrac{\pi^2}{8}-1$.

The sums are therefore $\tfrac{\pi}{2}$, $-\tfrac{\pi^2}{8}$, and $\tfrac{\pi^2}{8}-\tfrac{\pi}{2}$. Their total is $0$. Check: the endpoint change is $\sin(\pi)-\sin(0)=0$, and the calculated total simplifies to $0$ by cancellation.

Why the Remainder Is Kept

The exact identity is useful because it puts all unaccounted-for behavior in one explicit term. For an interval beginning at $x$ with increment $h$, twice continuous differentiability gives the integral form $$ f(x+h)=f(x)+f'(x)h+h^2\int_0^1(1-u)f''(x+uh)\,du. $$ Subtracting $\tfrac12 f''(x)h^2$ shows that the Taylor remainder in the decomposition is $$ r(x,h)=h^2\int_0^1(1-u)\big(f''(x+uh)-f''(x)\big)\,du. $$ Indeed, $\int_0^1(1-u)\,du=\tfrac12$. This expression explains why continuity of $f''$ matters: when $h$ is small, the values of $f''$ between $x$ and $x+h$ are close to $f''(x)$. To conclude that the sum of these remainders vanishes, however, one must control the sum across all partition intervals. Smallness of each separate remainder alone is not that conclusion.

Common mistake: treating $f(B_{t_i})-f(B_{t_{i-1}})$ as only $f'(B_{t_{i-1}})\Delta B_i$ drops the quadratic term. For $f(x)=x^2$, $x=0$, and $h=1$, that incorrect calculation predicts $f(1)-f(0)=2(0)(1)=0$. The actual change is $1^2-0^2=1$. The correction is $\tfrac12 f''(0)h^2=\tfrac12(2)(1)^2=1$, which restores the exact value. On Brownian partitions, squared increments cannot be omitted on the grounds that each increment is small; their aggregate contribution is central to the formula.

Check Your Understanding

For each question, distinguish the exact finite-partition identity from any assertion about a limit as the mesh tends to zero.

Write the three-term decomposition for a partition with exactly two intervals, using the definition of $R_i$.
For $f(x)=x^2$, explain why the remainder is zero for every interval, whatever its left endpoint and increment.
For $f(x)=x^3$, show directly from expansion that the remainder on an interval with increment $h$ is $h^3$.
State the progressive-measurability and square-integrability conditions required for an Ito integrand on $[0,T]$.
Why does the fact that each interval has a small increment not, by itself, prove that the sum of all Taylor remainders tends to zero?

Partition term	Finite-sum form	Role in the decomposition
First order	\(\sum_i f'(B_{t_{i-1}})\Delta B_i\)	Left-endpoint stochastic sum
Quadratic	\(\tfrac12\sum_i f''(B_{t_{i-1}})(\Delta B_i)^2\)	Curvature weighted by squared increments
Remainder	\(\sum_i R_i\)	Difference between the exact increment and the two Taylor terms