Complete Proof of Ito's Formula for f(Wt)

From Taylor Sums to Ito's Formula

The one-dimensional Ito formula is a chain rule for a Brownian path, but it is not the ordinary chain rule: the second derivative contributes because Brownian increments have nonvanishing quadratic variation. Here the formula is proved from Taylor expansion along deterministic partitions. The proof identifies the limits of the first- and second-order sums and shows that the accumulated Taylor remainders vanish.

Throughout, let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space carrying a standard Brownian motion $B_t$ with respect to this filtration. Thus, for $s<t$, $B_t-B_s$ has distribution $N(0,t-s)$ and is independent of $\mathcal{F}_s$. This filtration condition is needed for the conditional-mean and stochastic-integral arguments below. Fix a finite horizon $T>0$ and a deterministic partition $0=t_0<t_1<\cdots<t_n=T$; write $\Delta t_i=t_i-t_{i-1}$ and $\Delta B_i=B_{t_i}-B_{t_{i-1}}$.

Theorem: If $f:\mathbb{R}\to\mathbb{R}$ is twice continuously differentiable, then, for $0\le t\le T$, $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds. $$ The stochastic integral is the Ito integral. Since $f'$ need not be bounded, it is understood locally: stop the integrand when $B$ exits bounded intervals, define the stopped integrals in $\mathcal{L}^2$, and let the intervals increase. For each finite $T$, this defines the integral because Brownian paths are continuous and hence bounded on $[0,T]$, almost surely.

The Partition Argument

First prove the formula for $f\in C^2(\mathbb{R})$ with compact support. Then $f'$ and $f''$ are bounded, and $f''$ is uniformly continuous. The first-order sum will converge to the Ito integral. For this use the Ito isometry: for a progressively measurable integrand $H$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. Such integrals have mean zero: this holds for adapted step integrands by independence and zero mean of Brownian increments, and for $\mathcal{L}^2$ integrands by approximation in the isometry.

For each interval, Taylor's formula gives $f(B_{t_i})-f(B_{t_{i-1}})=f'(B_{t_{i-1}})\Delta B_i+\tfrac12 f''(B_{t_{i-1}})(\Delta B_i)^2+r_i$. Let $\omega(\delta)=\sup\{|f''(x)-f''(y)|:|x-y|\le\delta\}$. Uniform continuity gives $\omega(\delta)\to0$ as $\delta\to0$, and Taylor's integral remainder yields $|r_i|\le\tfrac12\omega(|\Delta B_i|)(\Delta B_i)^2$.

First-order sum.
On $[t_{i-1},t_i)$, set $H_s^n=f'(B_{t_{i-1}})$. This adapted step process is progressively measurable. Path continuity and continuity of $f'$ give $H_s^n\to f'(B_s)$ almost surely for almost every $s$. Boundedness of $f'$ and dominated convergence give $\mathbb{E}\int_0^T|H_s^n-f'(B_s)|^2\,ds\to0$. The Ito isometry therefore gives $\sum_i f'(B_{t_{i-1}})\Delta B_i\to\int_0^T f'(B_s)\,dB_s$ in $L^2(\mathbb{P})$.

Weighted quadratic sum.
Put $a_i=f''(B_{t_{i-1}})$. Split the sum into $\sum_i a_i\Delta t_i+\sum_i a_i((\Delta B_i)^2-\Delta t_i)$. The first term converges almost surely to $\int_0^T f''(B_s)\,ds$ by path continuity and Riemann-sum convergence. For the second term, each centered square has conditional mean zero given $\mathcal{F}_{t_{i-1}}$, and conditional variance $2(\Delta t_i)^2$, since a centered normal variable of variance $\Delta t_i$ has fourth moment $3(\Delta t_i)^2$. Distinct terms are orthogonal in $L^2$: condition the later term on the filtration just before its increment. Thus the second sum has mean zero and variance at most $2\|f''\|_\infty^2\sum_i(\Delta t_i)^2\le2\|f''\|_\infty^2T|\Pi|$, which tends to zero as the mesh $|\Pi|$ tends to zero.

Taylor remainders.
Let $m_\Pi=\max_i|\Delta B_i|$ and $Q_\Pi=\sum_i(\Delta B_i)^2$. Brownian path continuity implies $m_\Pi\to0$ almost surely as $|\Pi|\to0$. Also $\mathbb{E}[Q_\Pi]=T$ and $\operatorname{Var}(Q_\Pi)=2\sum_i(\Delta t_i)^2\le2T|\Pi|$, so $Q_\Pi\to T$ in $L^2$, in particular $Q_\Pi$ is bounded in probability. The remainder bound gives $|\sum_i r_i|\le\tfrac12\omega(m_\Pi)Q_\Pi\to0$ in probability.

Summing Taylor's identity telescopes the left side to $f(B_T)-f(B_0)$. The three limits just established show that the right side converges in probability to the two integrals in the theorem. A sequence that is identically $f(B_T)-f(B_0)$ can converge in probability only to that same random variable almost surely: if two probability limits differed with positive probability, their distance would exceed some positive threshold with positive probability, contradicting convergence in probability. This proves the formula at $T$ for compactly supported $f$. Applying the same argument at rational times and using the continuous versions of the integrals proves the process identity simultaneously for all $t\in[0,T]$.

Removing the Compact-Support Restriction

For general $f\in C^2(\mathbb{R})$, choose a smooth cutoff $\chi_R$ that equals one on $[-R,R]$ and vanishes outside $[-R-1,R+1]$, and set $f_R=\chi_R f$. The compact-support result applies to $f_R$. Define $\tau_R=\inf\{t\ge0:|B_t|\ge R\}$, with the infimum of the empty set taken as infinity. Up to $\tau_R$, $f_R(B_t)=f(B_t)$ and their first and second derivatives agree. Stopping the identity for $f_R$ at $\tau_R$ therefore gives the claimed identity for $f$ up to $\tau_R$. The stopped-integral equality follows first for adapted step integrands and then for their Ito-isometry limits. On every finite interval Brownian paths are bounded almost surely, so $\tau_R>T$ eventually as $R\to\infty$. The stopped identities consequently give the formula on $[0,T]$. This argument also explains why no global boundedness assumption on $f'$ or $f''$ is required. $\square$

Worked Examples and Checks

Worked Example: Taylor Terms for a Cubic Function

For $f(x)=x^3$, compute the Taylor increment at a point $x$ with increment $h$, and identify the terms that appear in the partition proof.

The derivatives are $f'(x)=3x^2$, $f''(x)=6x$, and $f'''(x)=6$. Direct expansion gives $(x+h)^3-x^3=3x^2h+3xh^2+h^3$. The first-order term is $f'(x)h=3x^2h$, and the quadratic term is $\tfrac12 f''(x)h^2=3xh^2$; the remainder is exactly $h^3$.

Check: substituting $x=1$ and $h=2$, the exact increment is $3^3-1^3=26$. The Taylor terms give $3(1)^2(2)=6$, $3(1)(2)^2=12$, and $2^3=8$, whose sum is $6+12+8=26$. The cubic remainder is consistent with the general remainder bound: on a bounded interval, its magnitude is controlled by the modulus of continuity of $f''$ times $h^2/2$; for a fixed large $h$, the bound need not be small, while along fine Brownian partitions the maximum increment tends to zero.

Worked Example: The First and Second Sums for the Square Function

Take $f(x)=x^2$. Compute the terms of the partition identity and check the correction in the resulting formula.

Here $f'(x)=2x$ and $f''(x)=2$, so Taylor expansion is exact: $B_{t_i}^2-B_{t_{i-1}}^2=2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2$. Summing over $i$ gives $B_T^2-B_0^2=2\sum_iB_{t_{i-1}}\Delta B_i+Q_\Pi$. The first sum converges to $\int_0^T2B_s\,dB_s$, and $Q_\Pi$ converges in probability to $T$. Since $B_0=0$, the formula is $B_T^2= \int_0^T2B_s\,dB_s+T$.

Check: the integrand $H_s=2B_s$ is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds=4\int_0^T\mathbb{E}[B_s^2]\,ds=4\int_0^T s\,ds=2T^2<\infty$. The Ito integral has mean zero by the step-integrand approximation described above. Taking expectations in the formula therefore gives $\mathbb{E}[B_T^2]=T$, agreeing with $B_T\sim N(0,T)$.

Worked Example: Convergence of the Quadratic-Variation Sum

On $[0,2]$, use $n$ equal intervals and calculate the mean and variance of $Q_\Pi$. Then evaluate them when $n=4$.

Each interval has length $2/n$. Since Brownian increments are independent centered normal variables, their squared increments have mean $2/n$ and variance $2(2/n)^2=8/n^2$. Thus $\mathbb{E}[Q_\Pi]=n(2/n)=2$ and $\operatorname{Var}(Q_\Pi)=n(8/n^2)=8/n$. For $n=4$, each interval has length $2/4=1/2$, so the mean is $4(1/2)=2$, and the variance is $4\cdot2(1/2)^2=4\cdot(1/2)=2$.

Check: the general variance bound is $2T|\Pi|=2\cdot2\cdot(2/n)=8/n$, exactly the computed variance for equal intervals. At $n=4$, the bound is $2\cdot2\cdot(1/2)=2$, again matching the direct calculation. As $n\to\infty$, the variance $8/n\to0$, confirming convergence to $T=2$ in $L^2$.

What the Proof Uses—and a Common Error

The correction term is not an extra feature inserted into Taylor's formula. It is the limit of the quadratic Taylor terms: the weighted squares approach the time integral of $f''(B_s)$, while the first-order terms approach the Ito integral. The continuity of $f''$, together with boundedness after localization, is what makes the remainder negligible. The formula is an identity of processes, not merely an equality of expectations.

Common mistake: treating a single squared increment as its time increment. It is not true that $(\Delta B_i)^2=\Delta t_i$ for each interval. For an interval of length $1/2$, $(\Delta B_i)^2$ is random, with mean $1/2$ and variance $2(1/2)^2=1/2$. The valid statement is that the sum of squared increments converges to elapsed time as the mesh tends to zero. Replacing every square by its interval length before taking a limit discards the random fluctuations at finite mesh and is not a valid partition identity.

Likewise, the ordinary chain rule cannot be substituted for this proof: it has no term corresponding to the nonzero limit of the squared-increment sum. The Ito formula applies here to a time-independent $C^2$ function of Brownian motion; the time-dependent formula requires an additional time derivative and is a separate statement.

Check Your Understanding

Use the hypotheses and the convergence steps in the proof to answer the following.

Why must Brownian increments be independent of $\mathcal{F}_{t_{i-1}}$, not merely have the correct normal distribution, in the weighted quadratic-sum argument?
For a deterministic partition, show that $\sum_i(\Delta t_i)^2\le T|\Pi|$, and explain where this bound enters the proof.
For $f(x)=x^2$, identify the first- and second-order Taylor terms and state the limit of the sum of the second-order terms.
Why does $m_\Pi\to0$ alone not prove that the total Taylor remainder tends to zero? What additional quantity is controlled in the proof?
Explain how a cutoff and the exit time $\tau_R$ allow the proof for compactly supported functions to yield the formula for an arbitrary $C^2$ function.