Polynomial Tests of the Ito Correction

Two Polynomial Tests of Ito’s Formula

For a differentiable path $x(t)$, the ordinary chain rule gives $d(x^2)=2x\,dx$ and $d(x^3)=3x^2\,dx$. Brownian paths are not differentiable, and their nonzero quadratic variation contributes an additional term. The square and cube functions make the correction visible with minimal algebra: the square receives a constant time term, while the cube receives a term proportional to Brownian motion.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ relative to $(\mathcal{F}_t)$. In particular, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$ for $0\le s<t$. All stochastic integrals below are Ito integrals. Their integrands are progressively measurable and belong to $\mathcal{L}^2$ on each stated finite time interval: this means $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Under this condition, the Ito integral is a square-integrable martingale, with conditional mean-zero increments and the Ito isometry.

Definition: The Ito correction in the formula for $f(B_t)$ is the second-order term $\tfrac12\int_0^t f''(B_s)\,ds$. It arises because Brownian motion has quadratic variation $[B]_t=t$, summarized by $dB_t\,dB_t=dt$; the other differential products are $dB_t\,dt=0$ and $dt\,dt=0$.

We use the one-dimensional Ito formula, established in the earlier material on Ito’s formula for Brownian motion: for $f\in C^2(\mathbb{R})$, it gives $f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\tfrac12\int_0^t f''(B_s)\,ds$, provided the stochastic integral is defined. The applications below check that condition. The stochastic integrals are Ito, not Stratonovich integrals; in particular, the ordinary chain rule is not a rule for evaluating them.

The Square Identity

Proposition: For every finite $T\ge0$, almost surely for all $0\le t\le T$, $$ B_t^2=2\int_0^t B_s\,dB_s+t. $$ The integrand $B_s$ is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$.

Proof. Set $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$. Ito’s formula, including its second-order term, therefore gives $$ B_t^2=B_0^2+\int_0^t 2B_s\,dB_s+\frac12\int_0^t 2\,ds. $$ Since $B_0=0$, the last term is $t$, giving the identity. To check that the stochastic integral is defined on $[0,T]$, use $\mathbb{E}[B_s^2]=s$: $$ \mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=\frac{T^2}{2}<\infty. $$ The integrand is adapted with continuous paths, hence progressively measurable. The Ito integral is consequently square-integrable, and the asserted identity follows from Ito’s formula. $\square$

Worked Example: Checking the Square Identity at Time Two

Use the identity at $t=2$ to calculate the mean and variance of $B_2^2$, then check both against the normal distribution.

Write $I=\int_0^2 B_s\,dB_s$. The integrability calculation in the proposition applies, and the Ito integral is a square-integrable martingale. Thus $\mathbb{E}[I]=0$. By the Ito isometry, $$ \mathbb{E}[I^2]=\mathbb{E}\int_0^2 B_s^2\,ds=\int_0^2 s\,ds=\left.\frac{s^2}{2}\right|_0^2=2. $$ The square identity says $B_2^2=2I+2$. Therefore $$ \mathbb{E}[B_2^2]=2\mathbb{E}[I]+2=2, \qquad \operatorname{Var}(B_2^2)=\operatorname{Var}(2I+2)=4\operatorname{Var}(I)=4\cdot2=8. $$ Here $\operatorname{Var}(I)=\mathbb{E}[I^2]-(\mathbb{E}[I])^2=2$.

Check: $B_2\sim N(0,2)$. Its second moment is $2$, and its fourth moment is $3\cdot2^2=12$. Hence the variance of its square is $12-2^2=8$, agreeing with the identity-based calculation.

The Cube Identity

Proposition: For every finite $T\ge0$, almost surely for all $0\le t\le T$, $$ B_t^3=3\int_0^t B_s^2\,dB_s+3\int_0^t B_s\,ds. $$ The stochastic integrand $B_s^2$ is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$.

Proof. Set $f(x)=x^3$. Then $f'(x)=3x^2$ and $f''(x)=6x$. Ito’s formula gives the first-order integral and the explicit correction term: $$ B_t^3=B_0^3+\int_0^t 3B_s^2\,dB_s+\frac12\int_0^t 6B_s\,ds. $$ Since $B_0=0$, the correction is $3\int_0^t B_s\,ds$, proving the displayed identity. For the stochastic integral, Brownian normal moments give $\mathbb{E}[B_s^4]=3s^2$, so $$ \mathbb{E}\int_0^T (B_s^2)^2\,ds=\int_0^T 3s^2\,ds=T^3<\infty. $$ The process $B_s^2$ is adapted and continuous, hence progressively measurable. Thus the Ito integral exists in the square-integrable sense, as required. $\square$

Worked Example: Checking the Mean of the Cubic Identity

Apply the cube identity at $t=1$ and use it to compute $\mathbb{E}[B_1^3]$. Verify the result directly from the distribution of $B_1$.

The identity gives $$ B_1^3=3\int_0^1 B_s^2\,dB_s+3\int_0^1 B_s\,ds. $$ The stochastic integrand $B_s^2$ satisfies $\mathbb{E}\int_0^1 B_s^4\,ds=\int_0^1 3s^2\,ds=1<\infty$. Its Ito integral is a square-integrable martingale starting at zero, so its expectation at time $1$ is zero. For the ordinary time integral, Fubini’s theorem applies because $\int_0^1\mathbb{E}|B_s|\,ds\le\int_0^1\sqrt{\mathbb{E}[B_s^2]}\,ds=\int_0^1\sqrt{s}\,ds<\infty$. Since $\mathbb{E}[B_s]=0$, this integral also has mean zero. Consequently $\mathbb{E}[B_1^3]=0$.

Check: $B_1\sim N(0,1)$, a distribution symmetric about zero. Its third moment is zero, in agreement with the calculation from Ito’s identity.

A Conditional Cubic-Moment Formula

The cube identity also determines a conditional expectation. The result reflects the contribution of the Brownian increment’s variance: conditionally on the present, the expected cube at a later time is not just the cube of the present value.

Theorem: For $0\le s\le t<\infty$, $$ \mathbb{E}[B_t^3\mid\mathcal{F}_s]=B_s^3+3(t-s)B_s $$ almost surely.

Proof. Apply the cube identity at $t$ and at $s$, then subtract: $$ B_t^3-B_s^3=3\int_s^t B_u^2\,dB_u+3\int_s^t B_u\,du. $$ The first integral is an increment of a square-integrable Ito martingale: its integrand satisfies $\mathbb{E}\int_0^t B_u^4\,du=t^3<\infty$. Therefore its conditional expectation given $\mathcal{F}_s$ is zero. For the time integral, $\mathbb{E}\int_s^t|B_u|\,du<\infty$, as follows from $\mathbb{E}|B_u|\le\sqrt{u}$ on this finite interval. Conditional Fubini and $\mathbb{E}[B_u\mid\mathcal{F}_s]=B_s$ for $u\ge s$ give $$ \mathbb{E}\left[\int_s^t B_u\,du\mid\mathcal{F}_s\right] =\int_s^t\mathbb{E}[B_u\mid\mathcal{F}_s]\,du =\int_s^t B_s\,du=(t-s)B_s. $$ Taking conditional expectations in the subtracted identity proves the formula. When $s=t$, the stochastic and time integrals over $[s,t]$ vanish, and the formula reduces to $B_t^3=B_t^3$. $\square$

Worked Example: Conditional Cubic Moment from One Half to One

Specialize the theorem to $s=\tfrac12$ and $t=1$, and check the result by expanding the Brownian increment.

Substitution gives $$ \mathbb{E}[B_1^3\mid\mathcal{F}_{1/2}] =B_{1/2}^3+3\left(1-\frac12\right)B_{1/2} =B_{1/2}^3+\frac32B_{1/2}. $$ For an independent-increment check, write $B_1=B_{1/2}+Z$, where $Z=B_1-B_{1/2}$ is independent of $\mathcal{F}_{1/2}$, has mean zero, second moment $\tfrac12$, and third moment zero. Given $\mathcal{F}_{1/2}$, set $x=B_{1/2}$. Expanding the cube and taking the conditional expectation gives $$ \mathbb{E}[(x+Z)^3\mid\mathcal{F}_{1/2}] =x^3+3x^2\mathbb{E}[Z]+3x\mathbb{E}[Z^2]+\mathbb{E}[Z^3] =x^3+\frac32x. $$ This is exactly the formula obtained from Ito’s identity.

Check: The coefficient is $3(t-s)=3\cdot\tfrac12=\tfrac32$, not $3(t-s)^2$. The independent increment contributes through its second moment, which is $t-s=\tfrac12$.

What the Correction Changes

For the square function, the ordinary chain rule would suggest $B_t^2=2\int_0^t B_s\,dB_s$. The Ito identity instead has the additional $t$. For the cube, omitting the second derivative term would remove $3\int_0^t B_s\,ds$. Both omissions change the resulting identities, including their expectations and conditional expectations. In differential notation, the square calculation uses $d(B_t^2)=2B_t\,dB_t+dt$, while the cube calculation uses $d(B_t^3)=3B_t^2\,dB_t+3B_t\,dt$.

These are Ito-integral identities. They should not be confused with the Stratonovich chain rule, which uses the notation $\circ\,dB_t$ and has no extra second-derivative term in its chain rule. The integrals in this tutorial are all written with $dB_t$, not $\circ\,dB_t$. The distinction matters: replacing one integral convention with the other without accounting for the correction changes the calculation.

Common mistake: applying the ordinary chain rule to $B_t^2$ and concluding $B_t^2=2\int_0^t B_s\,dB_s$. This would make the expectation of $B_t^2$ equal to zero, because the integral is a square-integrable martingale starting at zero. But $\mathbb{E}[B_t^2]=t$. The corrected identity $B_t^2=2\int_0^t B_s\,dB_s+t$ gives $\mathbb{E}[B_t^2]=t$, as required.

Check Your Understanding

Use the displayed identities and state which integrability or conditional-expectation property justifies each stochastic calculation.

For $f(x)=x^2$, write $f'(x)$, $f''(x)$, and the Ito identity for $B_t^2$.
Verify that $B_s^2$ is an admissible Ito integrand on $[0,T]$ by computing $\mathbb{E}\int_0^T B_s^4\,ds$.
Use the square identity and the Ito isometry to compute $\operatorname{Var}(B_T^2)$.
Write the cube identity and identify its second-order correction term.
Use the conditional cubic-moment formula when $s=0$ to compute $\mathbb{E}[B_t^3]$.