Choosing Test Functions to Cancel the Ito Correction

Designing the Correction Rather Than Accepting It

In an Ito calculation, the second derivative of a test function determines the extra time-integral term. This makes test-function choice a useful design tool: prescribe the second derivative, integrate twice, and obtain a formula with a correction suited to the problem. The goal here is not to apply Ito's formula to a function chosen in advance, but to choose the function to produce a specified term or cancel one already present.

Throughout, 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, with increments independent of the past. Fix $T>0$. Every stochastic integral below is an Ito integral. Its integrand is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, each such integral is a square-integrable martingale, not merely a local martingale.

Definition: For a twice continuously differentiable function $f:\mathbb{R}\to\mathbb{R}$, the Ito correction in the formula for $f(B_t)$ is $\tfrac12 f''(B_t)\,dt$. The identity $dB_t\,dB_t=dt$, together with $dB_t\,dt=0$ and $dt\,dt=0$, accounts for this term; it is not part of the ordinary chain rule.

The Brownian Ito formula, proved earlier in the course, says that when $f\in C^2(\mathbb{R})$ and $f'(B_s)$ is progressively measurable with $\mathbb{E}\int_0^T f'(B_s)^2\,ds<\infty$,

f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds,\qquad 0\le t\le T.

The integral with respect to $B$ is an Ito integral, not a Stratonovich integral. Its martingale property follows from the Ito integral construction: for $s\le t$, $\mathbb{E}[\int_s^t H_u\,dB_u\mid\mathcal{F}_s]=0$, and the Ito isometry gives $\mathbb{E}[(\int_0^t H_u\,dB_u)^2]=\mathbb{E}\int_0^t H_u^2\,du<\infty$. These facts apply to the integrands in the examples after their square-integrability is checked.

Prescribing a State-Dependent Correction

For a spatial test function, prescribing $f''$ prescribes the correction as a function of the current Brownian value. The following result gives all such functions on the real line.

Proposition: Let $h:\mathbb{R}\to\mathbb{R}$ be continuous. The $C^2$ functions satisfying $f''(x)=2h(x)$ for every $x\in\mathbb{R}$ are exactly $$ f(x)=c_0+c_1x+2\int_0^x(x-u)h(u)\,du, $$ where $c_0,c_1\in\mathbb{R}$. For any such function, the Ito correction in $f(B_t)$ is $h(B_t)\,dt$.

Proof. Define $F(x)=2\int_0^x(x-u)h(u)\,du$, using the oriented integral convention when $x<0$. Differentiating by the fundamental theorem of calculus gives $F'(x)=2\int_0^x h(u)\,du$, because the boundary term $2(x-x)h(x)$ is zero. Differentiating again gives $F''(x)=2h(x)$. Thus every displayed $f$ has second derivative $2h$. Conversely, if $f''=2h$, then $g=f-F$ satisfies $g''=0$. The fundamental theorem of calculus implies $g'$ is constant, say $c_1$, and then $g(x)=c_0+c_1x$. Finally, $\tfrac12 f''(B_t)=h(B_t)$, which is the claimed correction coefficient. $\square$

The affine constants do not alter the correction, because their second derivatives vanish. They may be chosen to simplify the stochastic-integral term or to set a convenient initial value. To cancel an existing term $q(B_t)\,dt$, prescribe $h(x)=-q(x)$, or equivalently solve $f''(x)=-2q(x)$. The sign is important: the correction is one half of the second derivative.

Worked Example: Prescribing a Centered Quadratic Correction

Choose $f$ so that the correction is $(B_t^2-1)\,dt$, and write the resulting identity. Take $h(x)=x^2-1$. The condition is $f''(x)=2x^2-2$. Integrating once gives $f'(x)=\tfrac23x^3-2x+C$; integrating again gives $f(x)=\tfrac16x^4-x^2+Cx+D$. Choose $C=D=0$.

For this $f$, $f'(x)=\tfrac23x^3-2x$ and $f''(x)=2x^2-2$. Ito's formula therefore gives

\frac16B_t^4-B_t^2 =\int_0^t\left(\frac23B_s^3-2B_s\right)\,dB_s +\int_0^t(B_s^2-1)\,ds.

The stochastic integrand is progressively measurable because $B$ is adapted and continuous. Its square is integrable: using the Brownian moments $\mathbb{E}[B_s^2]=s$, $\mathbb{E}[B_s^4]=3s^2$, and $\mathbb{E}[B_s^6]=15s^3$,

\mathbb{E}\int_0^T\left(\frac23B_s^3-2B_s\right)^2ds =\int_0^T\left(\frac49(15s^3)-\frac83(3s^2)+4s\right)ds =\int_0^T\left(\frac{20}{3}s^3-8s^2+4s\right)ds<\infty.

Check: the expected left side is $\tfrac16(3t^2)-t=\tfrac12t^2-t$. The expected time integral on the right is $\int_0^t(\mathbb{E}[B_s^2]-1)\,ds=\int_0^t(s-1)\,ds=\tfrac12t^2-t$. The stochastic integral has mean zero, as its integrand is in $\mathcal{L}^2$, so both sides have the same expectation.

Prescribing a Deterministic Time Correction

A spatial function alone has correction $\tfrac12 f''(B_t)\,dt$, which generally depends on $B_t$. To obtain a prescribed deterministic function of time, use a time-dependent test function. The time-dependent Ito formula, also established earlier, applies to $F\in C^{1,2}$ when $F_x(s,B_s)$ is progressively measurable and square-integrable. It reads

F(t,B_t)=F(0,B_0)+\int_0^t F_t(s,B_s)\,ds+\int_0^t F_x(s,B_s)\,dB_s+\frac12\int_0^t F_{xx}(s,B_s)\,ds.

In addition to the second-derivative correction, this formula contains the time derivative $F_t$. A successful design must account for both terms.

Proposition: Let $r\in C^1([0,T])$. The functions of the form $F(t,x)=r(t)x^2+a(t)x+b(t)$, with $a,b\in C^1([0,T])$, satisfy $F_{xx}(t,x)=2r(t)$. Their Ito correction is $r(t)\,dt$. In particular, for any prescribed continuous deterministic correction coefficient $r(t)$, this construction produces that coefficient; the time-derivative term remains $F_t(t,B_t)\,dt$.

Proof. Differentiating the stated function with respect to $x$ gives $F_x(t,x)=2r(t)x+a(t)$ and $F_{xx}(t,x)=2r(t)$. Hence the Ito correction is $\tfrac12F_{xx}(t,B_t)\,dt=r(t)\,dt$. Conversely, for each fixed $t$, any twice differentiable function of $x$ whose second derivative is $2r(t)$ differs from $r(t)x^2$ by an affine function of $x$. Writing its two affine coefficients as $a(t)$ and $b(t)$ gives the displayed form, and $C^1$ regularity of these coefficients ensures the required time derivative exists. $\square$

Worked Example: Prescribing and Cancelling a Time Correction

Construct a test function with correction $t\,dt$, and arrange that its time derivative cancels the resulting $-t\,dt$ term. Start with $F(t,x)=tx^2$, for which $F_{xx}=2t$ and the correction is $t\,dt$. Its time derivative is $F_t=x^2$, so this choice does not produce the desired cancellation. Instead take $F(t,x)=tx^2-\tfrac12t^2$.

The partial derivatives are $F_t(t,x)=x^2-t$, $F_x(t,x)=2tx$, and $F_{xx}(t,x)=2t$. The time-dependent Ito formula gives

d\left(tB_t^2-\frac12t^2\right) =(B_t^2-t)\,dt+2tB_t\,dB_t+\frac12(2t)\,dt =B_t^2\,dt+2tB_t\,dB_t.

Thus the correction $t\,dt$ combines with the $-t\,dt$ from $F_t$ to cancel. The stochastic integrand $2sB_s$ is progressively measurable, and $\mathbb{E}\int_0^T(2sB_s)^2ds=4\int_0^T s^3ds=T^4<\infty$.

Check: after integration, the left side is $tB_t^2-\tfrac12t^2$, whose expectation is $t\mathbb{E}[B_t^2]-\tfrac12t^2=t^2-\tfrac12t^2=\tfrac12t^2$. The right side has expectation $\int_0^t\mathbb{E}[B_s^2]\,ds=\int_0^t s\,ds=\tfrac12t^2$, since the Ito integral is square-integrable and has mean zero.

Canceling a State-Dependent Term in an Expression

Cancellation can also be used to remove a time integral already present in a Brownian expression. The design step is to match the correction to the negative of that term. The following example keeps the sign visible and checks the resulting identity independently.

Worked Example: Removing a Brownian Time Integral

Find a function $f$ for which the correction cancels $B_t\,dt$. The required equation is $\tfrac12 f''(x)=-x$, so $f''(x)=-2x$. Integrating twice and choosing zero affine constants gives $f'(x)=-x^2$ and $f(x)=-\tfrac13x^3$.

Here $f'(x)=-x^2$ and $f''(x)=-2x$. Ito's formula gives

-\frac13B_t^3 =-\int_0^t B_s^2\,dB_s-\int_0^t B_s\,ds.

Rearranging adds the ordinary time integral to both sides:

-\frac13B_t^3+\int_0^t B_s\,ds=-\int_0^t B_s^2\,dB_s.

The integrand $B_s^2$ is progressively measurable and $\mathbb{E}\int_0^T B_s^4\,ds=\int_0^T3s^2ds=T^3<\infty$. Consequently the right side is a square-integrable martingale with mean zero. The time integral on the left is an ordinary Lebesgue integral, not an Ito integral.

Check: the expectation of the left side is $-\tfrac13\mathbb{E}[B_t^3]+\int_0^t\mathbb{E}[B_s]\,ds=0$, since centered normal random variables have zero odd moments. This agrees with the mean of the square-integrable Ito integral on the right. The sign also follows directly from the correction: $\tfrac12(-2B_t)\,dt=-B_t\,dt$.

Common Pitfalls and Scope

A common mistake is to use $f''$, rather than $\tfrac12f''$, as the correction coefficient. For example, if the target correction is $h(B_t)\,dt$, the equation to solve is $f''=2h$, not $f''=h$. With $f''=h$, Ito's formula produces only $\tfrac12h(B_t)\,dt$.

A second mistake is to treat a time-dependent design as if it were spatial only. In the example above, $F_{xx}=2t$ does give correction $t\,dt$, but $F_t$ contributes as well. Omitting $F_t$ would incorrectly leave a $-t\,dt$ term uncancelled. The derivatives $F_t$, $F_x$, and $F_{xx}$ must all be computed before deciding what cancels.

Finally, Ito's correction is specific to the Ito integral convention. The ordinary chain rule omits it, and a Stratonovich integral uses a different chain-rule formula. The identities here use $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ in the Ito sense. There is no change of measure or SDE solution involved: the probability measure throughout is $\mathbb{P}$, and $B_t$ is the Brownian motion on the stated filtered space.

Check Your Understanding

For each question, keep the factor one half in the Ito correction and account for any time derivative.

Find a $C^2$ function whose correction in $f(B_t)$ is $(3B_t-2)\,dt$.
If $f''(x)=4\cos(x)$, what is the correction coefficient in Ito's formula for $f(B_t)$?
For $F(t,x)=r(t)x^2+a(t)x+b(t)$, compute $F_t$, $F_x$, and the Ito correction.
What second-derivative condition cancels a term $q(B_t)\,dt$ already present in an expression?
Why does the condition $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ matter when claiming that $\int_0^t H_s\,dB_s$ is a true martingale?