Mixed Problems on the One-Dimensional Ito Formula

A Workflow for Mixed Ito Problems

Mixed Ito problems test more than differentiation. One must first decide whether the function depends on time, check that its derivatives are defined along the Brownian path, and then include the correction generated by quadratic variation. When a function has a singularity, the formula is applied only while the path remains in a region where that function is smooth.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. An integrand $H_s$ for an Ito integral $\int_0^t H_s\,dB_s$ is taken to be adapted and progressively measurable, with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ on each finite time interval under consideration. Such integrands form the class $\mathcal{L}^2$. Their Ito integrals are square-integrable martingales: in particular, $\mathbb{E}[\int_0^t H_s\,dB_s\mid\mathcal{F}_u]=\int_0^u H_s\,dB_s$ for $0\le u\le t$, and their expectations are zero.

Standard result: one-dimensional Ito formula. If $f\in C^2(\mathbb{R})$ and $f'(B_s)$ belongs to $\mathcal{L}^2$ on $[0,T]$, then $$ f(B_t)=f(B_0)+\int_0^t f'(B_s)\,dB_s+\frac12\int_0^t f''(B_s)\,ds,\qquad 0\le t\le T. $$ For $f\in C^{1,2}([0,T]\times\mathbb{R})$, with $f_x(s,B_s)\in\mathcal{L}^2$, the time-dependent version is $$ f(t,B_t)=f(0,B_0)+\int_0^t f_x(s,B_s)\,dB_s+ \int_0^t\left(f_t(s,B_s)+\frac12 f_{xx}(s,B_s)\right)ds. $$ These are standard results, proved in the earlier material on the one-dimensional Ito formula. If the integrability or smoothness conditions hold only after stopping, apply the formula to the stopped process on that region.

The practical distinction is whether the function has a time argument: use $f'(x)$, $f''(x)$ in the first formula, and add $f_t$ in the second. In both cases, the second spatial derivative contributes $\tfrac12 f_{xx}\,dt$, because $dB_t\,dB_t=dt$; the other multiplication rules are $dB_t\,dt=0$ and $dt\,dt=0$.

Polynomial Functions: Derivatives and a Moment Check

Proposition: For an integer $n\ge2$, if $B_0=0$, then $$ B_t^n=\int_0^t nB_s^{n-1}\,dB_s+\frac{n(n-1)}{2}\int_0^t B_s^{n-2}\,ds. $$ Both integrals are well-defined on each finite interval $[0,T]$.

Proof. Set $f(x)=x^n$. Then $f'(x)=nx^{n-1}$ and $f''(x)=n(n-1)x^{n-2}$. Brownian moments of every finite order are finite, so $\mathbb{E}\int_0^T n^2 B_s^{2n-2}\,ds =n^2\int_0^T\mathbb{E}[B_s^{2n-2}]\,ds<\infty$. Thus the stochastic integrand belongs to $\mathcal{L}^2$. Apply the time-independent Ito formula and use $B_0^n=0$ to obtain the claimed identity. The ordinary time integral is finite almost surely because Brownian paths are continuous on $[0,T]$. $\square$

Worked Example: A Quartic Polynomial with a Lower-Order Term

Apply Ito’s formula to $f(x)=x^4-3x^2$, and check its expected value by an independent calculation.

The derivatives are $f'(x)=4x^3-6x$ and $f''(x)=12x^2-6$. The Ito correction is explicitly $\tfrac12 f''(B_t)\,dt=(6B_t^2-3)\,dt$, so

d(B_t^4-3B_t^2)=(4B_t^3-6B_t)\,dB_t+(6B_t^2-3)\,dt.

Equivalently, integrating from $0$ to $t$, and using $B_0=0$, gives $$ B_t^4-3B_t^2=\int_0^t(4B_s^3-6B_s)\,dB_s+ \int_0^t(6B_s^2-3)\,ds. $$ The stochastic integrand is in $\mathcal{L}^2$: using the centered normal moments $\mathbb{E}[B_s^2]=s$, $\mathbb{E}[B_s^4]=3s^2$, and $\mathbb{E}[B_s^6]=15s^3$, its squared expectation is $16(15s^3)-48(3s^2)+36s=240s^3-144s^2+36s$, which is integrable on every finite interval. Therefore the stochastic integral is a true square-integrable martingale with mean zero.

Check: Taking expectations in the identity gives $\mathbb{E}[B_t^4-3B_t^2]=\int_0^t(6s-3)\,ds=3t^2-3t$. Directly, the same expectation is $\mathbb{E}[B_t^4]-3\mathbb{E}[B_t^2]=3t^2-3t$. The two calculations agree, including at $t=0$.

Time-Dependent Functions: Keep Both Drift Terms

For a function $f(t,x)$, its time derivative and its Ito correction are separate contributions. The coefficient of $dt$ is $f_t+\tfrac12 f_{xx}$, evaluated at $(t,B_t)$; it is not just $f_t$, nor just the second-derivative term.

Worked Example: A Decaying Trigonometric Function

Set $f(t,x)=e^{-t}\cos(2x)$. Find the Ito differential of $f(t,B_t)$ and verify its expectation.

All the required derivatives are $f_t(t,x)=-e^{-t}\cos(2x)$, $f_x(t,x)=-2e^{-t}\sin(2x)$, and $f_{xx}(t,x)=-4e^{-t}\cos(2x)$. Thus the correction is $\tfrac12 f_{xx}(t,B_t)\,dt=-2e^{-t}\cos(2B_t)\,dt$. Combining it with the time derivative yields

d(e^{-t}\cos(2B_t)) =-2e^{-t}\sin(2B_t)\,dB_t -3e^{-t}\cos(2B_t)\,dt.

On $[0,T]$, the stochastic integrand $H_s=-2e^{-s}\sin(2B_s)$ is adapted and progressively measurable, and $H_s^2\le4$. Hence $\mathbb{E}\int_0^T H_s^2\,ds\le4T<\infty$; its Ito integral is a true martingale with expectation zero.

Check: For a centered normal variable $B_t$ with variance $t$, $\mathbb{E}[\cos(2B_t)]=e^{-2t}$. Therefore $\mathbb{E}[e^{-t}\cos(2B_t)]=e^{-3t}$. The integrated Ito identity gives the same value: its expected drift is $-3\int_0^t e^{-s}\mathbb{E}[\cos(2B_s)]\,ds =-3\int_0^t e^{-3s}\,ds=e^{-3t}-1$. Adding the initial value $f(0,0)=1$ gives $e^{-3t}$, as required.

Stopping Before a Singular Boundary

A derivative calculation is useful only where the function is defined and sufficiently smooth. For instance, $\log(1+x)$ is smooth when $x>-1$, but its derivatives are singular at $x=-1$. Stopping the Brownian path at a safe distance from that boundary makes the formula legitimate on a specified finite interval.

Proposition: Let $a\in(0,1)$, let $\tau_a=\inf\{t\ge0:B_t\le-a\}$, with the infimum of the empty set defined as $+\infty$, and let $T<\infty$. Then, for $0\le t\le T$, $$ \log(1+B_{t\wedge\tau_a}) =\int_0^t\frac{\mathbf{1}_{\{s\le\tau_a\}}}{1+B_s}\,dB_s -\frac12\int_0^t\frac{\mathbf{1}_{\{s\le\tau_a\}}}{(1+B_s)^2}\,ds. $$ The stochastic integrand is understood to be zero after $\tau_a$.

Proof. The stopped path satisfies $B_{s\wedge\tau_a}\ge-a$, so its logarithm has argument at least $1-a>0$. For $f(x)=\log(1+x)$, the derivatives on $x>-1$ are $f'(x)=1/(1+x)$ and $f''(x)=-1/(1+x)^2$. Hence its correction is $-\tfrac12(1+B_s)^{-2}\,ds$. On the interval before stopping, $|f'(B_s)|\le(1-a)^{-1}$, so the stopped stochastic integrand has squared integral expectation at most $T(1-a)^{-2}$; it belongs to $\mathcal{L}^2$. Apply the stopped one-dimensional Ito formula, whose use follows by applying the standard formula to the stopped Brownian motion, to obtain the displayed identity. The endpoint causes no singularity because $1+B_{\tau_a}=1-a>0$ when $\tau_a<\infty$. $\square$

Worked Example: The Logarithm Stopped at a Safe Level

Take $a=\tfrac12$. Then $\tau_{1/2}=\inf\{t\ge0:B_t\le-\tfrac12\}$, and before and at this stopping time $1+B_{t\wedge\tau_{1/2}}\ge\tfrac12$. Compute the formula for $Y_t=\log(1+B_{t\wedge\tau_{1/2}})$.

For $f(x)=\log(1+x)$, $f'(x)=1/(1+x)$, $f''(x)=-1/(1+x)^2$, and the correction is $-1/(2(1+x)^2)\,dt$. Therefore $$ Y_t=\int_0^t\frac{\mathbf{1}_{\{s\le\tau_{1/2}\}}}{1+B_s}\,dB_s -\frac12\int_0^t\frac{\mathbf{1}_{\{s\le\tau_{1/2}\}}}{(1+B_s)^2}\,ds. $$ The stochastic integrand has absolute value at most $2$, so its expected squared integral on $[0,T]$ is at most $4T$. Its Ito integral is thus a square-integrable martingale, not merely a local martingale.

Check: Apply Ito’s formula once more to $g(y)=e^y$, for which $g'(y)=e^y$, $g''(y)=e^y$, and the correction is $\tfrac12 e^{Y_t}\,d[Y]_t$. Here the Brownian coefficient of $Y$ is $\mathbf{1}_{\{t\le\tau_{1/2}\}}/(1+B_t)$, so $d[Y]_t=\mathbf{1}_{\{t\le\tau_{1/2}\}}(1+B_t)^{-2}\,dt$. The two drift contributions cancel: $$ e^{Y_t}\left(-\frac{\mathbf{1}_{\{t\le\tau_{1/2}\}}}{2(1+B_t)^2}\right)dt+ \frac12e^{Y_t}\frac{\mathbf{1}_{\{t\le\tau_{1/2}\}}}{(1+B_t)^2}dt=0. $$ The remaining differential is $d(e^{Y_t})=\mathbf{1}_{\{t\le\tau_{1/2}\}}\,dB_t$, consistent with $e^{Y_t}=1+B_{t\wedge\tau_{1/2}}$. This check confirms both the sign and the size of the correction.

A Reliable Problem-Solving Checklist

Check the domain.
Identify where the function is smooth. If a singular boundary may be reached, stop inside the smooth region before differentiating along the path.

Differentiate explicitly.
Write $f'$ and $f''$, or $f_t$, $f_x$, and $f_{xx}$. The correction is always $\tfrac12 f''\,dt$ or $\tfrac12 f_{xx}\,dt$.

Check the integral.
Verify that the stochastic integrand is adapted, progressively measurable, and square-integrable; if only local conditions hold, use a suitable stopping argument.

Test the answer.
Where possible, compare expectations, substitute the proposed transformation into another Ito calculation, or test the initial value and a special case.

Common mistake. Treating Brownian motion as differentiable and writing $d f(B_t)=f'(B_t)\,dB_t$ drops the quadratic-variation term. For $f(x)=x^2$, that incorrect calculation gives $d(B_t^2)=2B_t\,dB_t$. The correct derivatives are $f'(x)=2x$ and $f''(x)=2$, so the correction is $\tfrac12 f''(B_t)\,dt=dt$, and $d(B_t^2)=2B_t\,dB_t+dt$. Integrating and taking expectations gives $\mathbb{E}[B_t^2]=t$; the incorrect rule would instead give zero, since the square-integrable Ito integral has mean zero. The missing term is therefore essential, not a notational refinement.

Check Your Understanding

For each problem, identify the required derivatives and state how you would check the result.

Apply Ito’s formula to $f(x)=x^3$, writing the stochastic and time-integral terms explicitly.
For $f(t,x)=e^{-2t}\sin(x)$, compute $f_t$, $f_x$, $f_{xx}$, and the full $dt$ coefficient.
Why must $\log(1+B_t)$ be stopped before $B_t$ reaches $-1$? Give a stopping level that keeps its derivatives bounded.
For a proposed Ito identity with stochastic integrand $H_s$, what condition makes its Ito integral a square-integrable martingale on $[0,T]$?
Explain how taking expectations can reveal an omitted second-derivative correction.