Statement of Ito's Formula for f(Wt)

The Formula and Its Two Integrals

For a differentiable path, the ordinary chain rule expresses the change in $f$ through its first derivative. Brownian paths are not differentiable, and their nonzero quadratic variation contributes an additional term involving the second derivative. Ito’s formula makes that correction precise. Here the process is standard Brownian motion itself; no time dependence or additional drift is included.

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 the filtration. For an interval $[0,T]$, an Ito integrand $H_s$ is progressively measurable and belongs to $\mathcal{L}^2$ when $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Progressive measurability ensures that the integrand at time $s$ uses information available up to that time.

Definition: For an adapted step process $H_s=\sum_{i=0}^{n-1}H_i\mathbf{1}_{(t_i,t_{i+1}]}(s)$, with each $H_i$ measurable with respect to $\mathcal{F}_{t_i}$, its Ito integral is $\int_0^T H_s\,dB_s=\sum_{i=0}^{n-1}H_i(B_{t_{i+1}}-B_{t_i})$. For a progressively measurable $H\in\mathcal{L}^2$, the Ito integral is the limit in $L^2(\mathbb{P})$ of such integrals. The ordinary integral $\int_0^T G_s\,ds$ is a Lebesgue integral in time; for a continuous sample path $G_s$, it is also the usual pathwise integral.

The defining isometry for the Ito integral is $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. In particular, the square-integrability condition ensures that the stochastic integral exists as an $L^2$ random variable. It also distinguishes the Ito integral from a Stratonovich integral: $dB_s$ below denotes Ito integration, not Stratonovich integration. In the shorthand multiplication rules, $dB_s\,dB_s=ds$, $dB_s\,ds=0$, and $ds\,ds=0$; the first rule is the source of the second-derivative term.

Theorem: Let $f:\mathbb{R}\to\mathbb{R}$ be twice continuously differentiable. Suppose that, for each finite $T$, the progressively measurable process $f'(B_s)$ satisfies $\mathbb{E}\int_0^T f'(B_s)^2\,ds<\infty$. Then, for every $t\in[0,T]$, almost surely, $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds. $$ The identity holds simultaneously for all $t\in[0,T]$ after choosing continuous versions of the processes. The first integral is an Ito integral in the class just defined; the second is an ordinary time integral, which is finite pathwise because $B_s$ is continuous and $f''$ is continuous. This is the standard Ito formula for Brownian motion; its proof is developed in a later part of the course.

Since $B_0=0$, the initial term is $f(0)$. The formula is stated with an integrability condition to ensure the stochastic integral is in $\mathcal{L}^2$. For a fixed finite $T$, functions with rapidly growing derivatives need not satisfy that condition automatically. For the polynomial and affine functions below, finiteness follows from the finite moments of the normal random variable $B_s$.

Why the Stochastic Integral Has Mean Zero

A useful way to check examples is to use the martingale property of a square-integrable Ito integral. The property includes both integrability and conditional expectation; it is not merely a formal claim that a stochastic integral “averages to zero.”

Proposition: If $H$ is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, then $M_t=\int_0^t H_s\,dB_s$ is a square-integrable martingale on $[0,T]$. In particular, $\mathbb{E}[M_t]=0$.

Proof. By the Ito isometry, $\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds<\infty$, so $M_t$ is integrable for each $t$. For an adapted step integrand, each increment of its integral over a step interval is a past-measurable coefficient multiplied by a Brownian increment independent of the information at the start of that interval and with conditional mean zero. Splitting an interval at a fixed time $s$, if necessary, therefore gives $\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s$ for $s\le t$. For general $H$, approximate in $\mathcal{L}^2$ by adapted step processes. The Ito isometry gives convergence of their integrals in $L^2$, and conditional expectation is a contraction in $L^2$; thus the conditional-expectation identity passes to the limit. Since $M_0=0$, taking expectations in that identity gives $\mathbb{E}[M_t]=0$. $\square$

Corollary: Under the hypotheses of Ito’s formula, if also $\mathbb{E}\int_0^T |f''(B_s)|\,ds<\infty$, then $$ \mathbb{E}[f(B_t)]=f(0)+\frac12\int_0^t\mathbb{E}[f''(B_s)]\,ds. $$

Proof. Take expectations in Ito’s formula. The stochastic integral has expectation zero by the Proposition. The assumed absolute integrability permits Fubini’s theorem to exchange expectation and the ordinary time integral, giving the displayed identity. $\square$

Worked Applications

Worked Example: The Square Function

Apply Ito’s formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$. The integrand is admissible on $[0,T]$, since $$ \mathbb{E}\int_0^T f'(B_s)^2\,ds =\int_0^T 4\mathbb{E}[B_s^2]\,ds =\int_0^T 4s\,ds =2T^2<\infty. $$ Substituting the function and both derivatives into the formula gives $$ B_t^2=0+\int_0^t 2B_s\,dB_s+\frac12\int_0^t 2\,ds =2\int_0^t B_s\,dB_s+t. $$

Check: The Ito integral has mean zero by the Proposition, so the right side has expectation $t$, matching $\mathbb{E}[B_t^2]=t$. Also, its stochastic term has variance $4\int_0^t\mathbb{E}[B_s^2]\,ds=4\int_0^t s\,ds=2t^2$, by the Ito isometry.

Worked Example: The Cube Function

Take $f(x)=x^3$. Then $f'(x)=3x^2$ and $f''(x)=6x$. The required integrability holds because $$ \mathbb{E}\int_0^T f'(B_s)^2\,ds =9\int_0^T\mathbb{E}[B_s^4]\,ds =9\int_0^T 3s^2\,ds =9T^3<\infty. $$ Here the Brownian fourth moment is $\mathbb{E}[B_s^4]=3s^2$. Ito’s formula now gives $$ B_t^3=0+\int_0^t 3B_s^2\,dB_s+\frac12\int_0^t 6B_s\,ds =3\int_0^t B_s^2\,dB_s+3\int_0^t B_s\,ds. $$

Check: The first integral has mean zero. The ordinary integral has mean $3\int_0^t\mathbb{E}[B_s]\,ds=0$, so the right side has mean zero, in agreement with the symmetry of the centered normal variable $B_t$, which gives $\mathbb{E}[B_t^3]=0$.

Worked Example: An Affine Function

Let $f(x)=3x-2$. Its derivatives are $f'(x)=3$ and $f''(x)=0$; the integrability requirement is $\mathbb{E}\int_0^T 9\,ds=9T<\infty$. Thus $$ 3B_t-2=(3B_0-2)+\int_0^t 3\,dB_s+\frac12\int_0^t 0\,ds =-2+3B_t. $$

Check: From the defining step-integral sums, the integral of the constant $3$ is $3\sum_i(B_{t_{i+1}}-B_{t_i})=3B_t$, since the increments telescope and $B_0=0$. This agrees with the formula and with the ordinary chain rule: there is no second-derivative correction for an affine function.

The Correction Term and a Common Error

The term $\frac12\int_0^t f''(B_s)\,ds$ is the distinction between Ito’s formula and the ordinary chain rule. In Stratonovich notation the corresponding chain rule is written using $\circ dB_s$; for this integrand, the conversion is $\int_0^t f'(B_s)\circ dB_s=\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds$, when the integrals are defined. Thus the ordinary-looking chain rule belongs to the Stratonovich integral, not to the Ito integral. In particular, substituting an Ito integral into the ordinary chain rule silently drops the quadratic-variation correction.

Common mistake: Applying the ordinary chain rule to $f(x)=x^2$ can lead to the incorrect identity $B_t^2=2\int_0^t B_s\,dB_s$, omitting the $t$ term. Taking expectations already exposes the error for $t>0$: the left side has expectation $t$, whereas the right side has expectation zero by the martingale property. The second moments also disagree. Since $B_t$ is normal with variance $t$, $\mathbb{E}[(B_t^2)^2]=\mathbb{E}[B_t^4]=3t^2$. On the claimed right side, the Ito isometry gives $$ \mathbb{E}\left[\left(2\int_0^t B_s\,dB_s\right)^2\right] =4\int_0^t\mathbb{E}[B_s^2]\,ds =4\int_0^t s\,ds =2t^2. $$ The omitted term is exactly the correction in the square-function example.

For a general $f$, the first derivative controls the stochastic integral, while the second derivative controls the ordinary correction. These terms have different meanings and different integrability requirements: $f'(B_s)$ must be square-integrable in expectation over time for the Ito integral as stated, whereas the correction is a pathwise time integral and is absolutely integrable in expectation under the additional condition in the Corollary. The formula therefore gives both a precise chain rule and a framework for checking whether each term is well-defined.

Check Your Understanding

Use the formula and its stated hypotheses to answer each question.

For $f(x)=x^4$, write down $f'(x)$, $f''(x)$, and the resulting Ito identity.
For $f(x)=ax+b$, where $a,b$ are constants, why does the correction term vanish?
What integrability condition on $f'(B_s)$ ensures the Ito integral in the formula is square-integrable on $[0,T]$?
For the square-function example, compute the expectation of each term on the right side and compare their sum with $\mathbb{E}[B_t^2]$.
Explain why the identity $B_t^2=2\int_0^t B_s\,dB_s$ cannot hold for $t>0$, using either first moments or second moments.

Statement of Ito's Formula for f(Wt)

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