Reconstructing the Ito Formula Proof from Partition Sums

From Partition Sums to Ito's Formula

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carry a standard Brownian motion $B_t$ with respect to the filtration: its paths are continuous, and for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. Fix $T>0$, and take a partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, with mesh $|\Pi|=\max_i(t_{i+1}-t_i)$. Write $B_i=B_{t_i}$, $\Delta B_i=B_{i+1}-B_i$, and $\Delta t_i=t_{i+1}-t_i$, for $i=0,\ldots,n-1$.

The proof uses Taylor's formula on each interval. The first-order terms will converge to an Ito integral; the second-order terms survive because Brownian motion has quadratic variation $[B]_t=t$; and the Taylor remainders will vanish. The notation $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ summarizes the limiting contributions, but the proof below obtains the formula from sums.

Theorem: Suppose $f:\mathbb{R}\to\mathbb{R}$ is twice continuously differentiable and $f'$ and $f''$ are bounded. Then, for every $t\ge0$, almost surely, $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds. $$ Here the stochastic integral is the Ito integral: $H_s=f'(B_s)$ is adapted and progressively measurable, and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for every finite $T$.

The proof first establishes convergence at a fixed time $T$. Since $f'$ is bounded, the Ito integral is square-integrable: if $|f'|\le C$, then $\mathbb{E}\int_0^T f'(B_s)^2\,ds\le C^2T$. It is a martingale: the Ito isometry gives its integrability, and the conditional-mean-zero property follows first for adapted step integrands from independent Brownian increments, then for this integrand by approximation in $L^2$ and continuity of conditional expectation in $L^1$.

Controlling the Three Terms

We will use the quadratic-variation sum $Q_\Pi=\sum_{i=0}^{n-1}(\Delta B_i)^2$. For each interval, $\Delta B_i\sim N(0,\Delta t_i)$, so its second moment is $\Delta t_i$ and its fourth moment is $3(\Delta t_i)^2$. Independence of the increments therefore gives

\mathbb{E}[Q_\Pi]=T,\qquad \mathbb{E}\big[(Q_\Pi-T)^2\big] =\operatorname{Var}(Q_\Pi) =2\sum_{i=0}^{n-1}(\Delta t_i)^2 \le 2T|\Pi|.

In particular, $Q_\Pi\to T$ in $L^2(\mathbb{P})$ as the mesh tends to zero. Also, because Brownian paths are continuous, $\max_i|\Delta B_i|\to0$ almost surely along any sequence of partitions whose mesh tends to zero.

First-order sums. Define the step process $H^\Pi_s=f'(B_i)$ for $t_i\le s<t_{i+1}$. It is adapted and progressively measurable. Since $f'$ is bounded, $H^\Pi$ and $f'(B_s)$ belong to $\mathcal{L}^2$ on $[0,T]$. Path continuity and continuity of $f'$ imply $H^\Pi_s\to f'(B_s)$ for every $s<T$, almost surely. The bound $|H^\Pi_s-f'(B_s)|^2\le4C^2$ permits dominated convergence on $\Omega\times[0,T]$, giving convergence in the norm $\mathbb{E}\int_0^T|\cdot|^2ds$. By the Ito isometry,

\sum_{i=0}^{n-1} f'(B_i)\Delta B_i =\int_0^T H^\Pi_s\,dB_s \longrightarrow \int_0^T f'(B_s)\,dB_s \quad\text{in }L^2(\mathbb{P}).

Weighted second-order sums. Put $g=f''$, and let $C_2$ be a bound for $|g|$. Decompose the weighted sum into a time-increment part and a centered part: $$ \sum_i g(B_i)(\Delta B_i)^2 =\sum_i g(B_i)\Delta t_i+ \sum_i g(B_i)\big((\Delta B_i)^2-\Delta t_i\big). $$ The centered summands have mean zero conditional on $\mathcal{F}_{t_i}$. For distinct indices they are orthogonal in $L^2$: condition on the information at the start of the later interval, which includes the earlier summand, and use the later summand's conditional mean zero. Thus $$ \mathbb{E}\left[\left(\sum_i g(B_i)\big((\Delta B_i)^2-\Delta t_i\big)\right)^2\right] =\sum_i\mathbb{E}[g(B_i)^2]\,2(\Delta t_i)^2 \le 2C_2^2T|\Pi|. $$ The centered part tends to zero in $L^2$. The other part is a left Riemann sum and converges almost surely to $\int_0^T g(B_s)\,ds$, because $s\mapsto g(B_s)$ is continuous. Hence the weighted second-order sums converge in probability to that integral.

Taylor remainders. Taylor's formula with a remainder bounded using the modulus of continuity gives $$ f(B_{i+1})=f(B_i)+f'(B_i)\Delta B_i+\frac12f''(B_i)(\Delta B_i)^2+r_i, \qquad |r_i|\le\frac12\omega_R\!\left(\max_j|\Delta B_j|\right)(\Delta B_i)^2, $$ whenever the endpoints lie in $[-R,R]$. Here $\omega_R(\delta)=\sup\{|f''(x)-f''(y)|:x,y\in[-R,R],\ |x-y|\le\delta\}$, which tends to zero as $\delta\to0$. For each continuous Brownian path on $[0,T]$, some finite $R$ contains all its values. On that path, $\max_i|\Delta B_i|\to0$, so $\sum_i|r_i|\le\frac12\omega_R(\max_i|\Delta B_i|)Q_\Pi\to0$ in probability: the modulus tends to zero almost surely and $Q_\Pi\to T$ in probability.

Assembling the Proof

Proof. Assume first that $f$ has the bounded derivatives in the theorem. Sum the Taylor identities over the partition. The left side telescopes to $f(B_T)-f(B_0)$. The first-order sum converges in $L^2$ to the Ito integral, the weighted second-order sum converges in probability to $\int_0^T f''(B_s)\,ds$, and the sum of remainders tends to zero in probability. Therefore the telescoping identity converges in probability to $$ f(B_T)-f(B_0)=\int_0^T f'(B_s)\,dB_s+\frac12\int_0^T f''(B_s)\,ds. $$ The left side does not depend on the partition. A sequence of random variables cannot converge in probability to two different limits, so the displayed identity holds almost surely.

For a general $f\in C^2(\mathbb{R})$, choose smooth cutoff functions $\chi_m$ that equal $1$ on $[-m,m]$ and vanish outside $[-m-1,m+1]$, and apply the proved result to $f_m=\chi_m f$. Each $f_m$ and its first two derivatives are bounded. Let $\tau_m=\inf\{t\ge0:|B_t|\ge m\}$. Up to $\tau_m$, $f_m,f_m'$, and $f_m''$ agree with $f,f'$, and $f''$. Consequently the identity for $f_m$, stopped at $\tau_m$, is the identity for $f$ up to that time. The integrals of the stopped integrands are well-defined Ito integrals: on each finite interval before $\tau_m$, the derivatives are bounded, so the integrands are adapted, progressively measurable, and square-integrable. Since continuous Brownian paths are bounded on every finite time interval, $\mathbb{P}(\tau_m>T)\to1$ as $m\to\infty$. The identity thus holds for $f$ on $[0,T]$ almost surely. In this general case the stochastic integral is understood locally, by these stopping times; if $\mathbb{E}\int_0^T f'(B_s)^2ds<\infty$, it is a square-integrable Ito integral on $[0,T]$. Applying the fixed-time identity at rational times and using continuity of both sides gives a single probability-one event on which it holds for all $t\ge0$. $\square$

Worked Checks of the Formula

Worked Example: The Square Function

Take $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$. The general formula applies by localization. The integrand $2B_s$ is square-integrable on $[0,T]$, since $$ \mathbb{E}\int_0^T(2B_s)^2\,ds =4\int_0^T s\,ds=2T^2. $$ Thus $$ B_T^2=2\int_0^T B_s\,dB_s+T. $$ At the partition level, the exact algebraic identity is $$ B_T^2=2\sum_i B_i\Delta B_i+Q_\Pi. $$ For $T=1$ and $n=4$ equal intervals, each $\Delta t_i=\tfrac14$, so $\mathbb{E}[Q_\Pi]=4\cdot\tfrac14=1$ and $\operatorname{Var}(Q_\Pi)=2\cdot4\cdot(\tfrac14)^2=\tfrac12$. As $n$ increases, the variance for $n$ equal intervals on $[0,1]$ is $2n(1/n)^2=2/n$, which tends to zero. Check: taking expectations in the formula gives $\mathbb{E}[B_T^2]=T$, because the Ito integral has mean zero. This agrees with the variance of $B_T$.

Worked Example: The Exponential Function

For $f(x)=e^x$, all three derivatives needed are $f(x)=f'(x)=f''(x)=e^x$. The formula gives $$ e^{B_T}=1+\int_0^T e^{B_s}\,dB_s+\frac12\int_0^T e^{B_s}\,ds. $$ The Ito integral is square-integrable because $$ \mathbb{E}\int_0^T e^{2B_s}\,ds =\int_0^T e^{2s}\,ds =\frac{e^{2T}-1}{2}<\infty. $$ Here we used the normal moment formula $\mathbb{E}[e^{aB_s}]=e^{a^2s/2}$. Its integrand is adapted and progressively measurable, so the integral is a martingale with mean zero. Taking expectations gives $$ \mathbb{E}[e^{B_T}] =1+\frac12\int_0^T e^{s/2}\,ds =1+(e^{T/2}-1)=e^{T/2}. $$ Check: direct use of $B_T\sim N(0,T)$ gives $\mathbb{E}[e^{B_T}]=e^{T/2}$, the same result.

Worked Example: The Cosine Function

Let $f(x)=\cos x$, so $f'(x)=-\sin x$, $f''(x)=-\cos x$, and $f(0)=1$. The derivatives are bounded, and the formula is $$ \cos(B_T)=1-\int_0^T\sin(B_s)\,dB_s-\frac12\int_0^T\cos(B_s)\,ds. $$ The stochastic integrand is adapted and progressively measurable, and $\mathbb{E}\int_0^T\sin^2(B_s)\,ds\le T$, so the Ito integral is a square-integrable martingale with mean zero. The characteristic function of $N(0,s)$ gives $\mathbb{E}[\cos(B_s)]=e^{-s/2}$. Therefore $$ \mathbb{E}[\cos(B_T)] =1-\frac12\int_0^T e^{-s/2}\,ds =1-(1-e^{-T/2})=e^{-T/2}. $$ Check: directly, $\mathbb{E}[\cos(B_T)]=\operatorname{Re}\mathbb{E}[e^{iB_T}]=\operatorname{Re}(e^{-T/2})=e^{-T/2}$.

Common Pitfall: Dropping the Correction

The ordinary chain rule would give the incorrect calculation $B_T^2=2\int_0^T B_s\,dB_s$. The partition identity shows exactly what is missing: $B_T^2=2\sum_i B_i\Delta B_i+\sum_i(\Delta B_i)^2$, and the last sum converges to $T$, not to zero. The correct result is $B_T^2=2\int_0^T B_s\,dB_s+T$. In differential notation, the extra term comes from $dB_t\,dB_t=dt$; the rules $dB_t\,dt=0$ and $dt\,dt=0$ express that the mixed and purely time-increment second-order sums vanish.

Check Your Understanding

Use the partition proof and the stated hypotheses to answer the following questions.

For $f(x)=x^3$, calculate $f'$ and $f''$, and write the Ito formula for $f(B_T)$.
Why does the centered weighted sum $\sum_i f''(B_i)((\Delta B_i)^2-\Delta t_i)$ have variance tending to zero when $f''$ is bounded?
Which step in the proof uses continuity of the Brownian paths to control the Taylor remainder?
For $f(x)=e^{2x}$, check whether the Ito integral in the formula is square-integrable on a finite interval $[0,T]$.