Time-Dependent Ito Formula for f(t,Wt)

Why Explicit Time Dependence Adds a Term

For a function of the Brownian value alone, Ito's formula contains a first-derivative stochastic term and a second-derivative correction. If the function also depends explicitly on time, a third contribution appears: the ordinary change in the time argument. The resulting formula separates these effects rather than treating the path $t\mapsto f(t,B_t)$ as differentiable.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. Fix $T>0$. For an integrand $H_s$, the Ito integral $\int_0^t H_s\,dB_s$ used below is defined for a progressively measurable process in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. A continuous adapted process is progressively measurable, so this condition applies to $H_s=f_x(s,B_s)$ whenever the stated square-integrability assumption holds.

Definition: Write $f_t$, $f_x$, and $f_{xx}$ for the time derivative, first space derivative, and second space derivative of $f(t,x)$. The notation $f\in C^{1,2}([0,T]\times\mathbb{R})$ means these derivatives exist and are continuous (with the time derivative interpreted one-sided at the time endpoints).

The Time-Dependent Ito Formula

Theorem: Suppose $f\in C^{1,2}([0,T]\times\mathbb{R})$ and $\mathbb{E}\int_0^T f_x(s,B_s)^2\,ds<\infty$. Then, for $0\le t\le T$, almost surely, $$ f(t,B_t)=f(0,B_0)+\int_0^t f_t(s,B_s)\,ds +\int_0^t f_x(s,B_s)\,dB_s +\frac12\int_0^t f_{xx}(s,B_s)\,ds. $$ The time integrals are finite pathwise because the derivatives are continuous and a Brownian path has compact range on $[0,T]$. The stochastic integrand is progressively measurable and in $\mathcal{L}^2$ by assumption. This is the standard time-dependent Ito formula; its complete partition-sum proof is given in the later course material on the complete proof of the time-dependent Ito formula.

In differential notation the same identity is $$ df(t,B_t)=f_t(t,B_t)\,dt+f_x(t,B_t)\,dB_t +\frac12 f_{xx}(t,B_t)\,dt. $$ The last term comes from Brownian quadratic variation, $[B]_t=t$, or in multiplication-table notation $dB_t\,dB_t=dt$. The other second-order products satisfy $dB_t\,dt=0$ and $dt\,dt=0$. Thus the time derivative is not a replacement for the Ito correction: both contribute to the coefficient of $dt$.

The integral against $dB_s$ here is an Ito integral, not a Stratonovich integral. For a continuous semimartingale integrand $H$, the standard conversion is $\int_0^t H_s\circ dB_s=\int_0^t H_s\,dB_s+\tfrac12[H,B]_t$, when these integrals are defined. In the Ito formula as stated, the correction is already written explicitly as $\tfrac12\int_0^t f_{xx}(s,B_s)\,ds$; one must not silently replace the Ito integral by a Stratonovich one.

Separating the Three Contributions

The formula can be read as a bookkeeping rule. The change in the time coordinate contributes $f_t\,dt$. The first-order response to the Brownian increment contributes $f_x\,dB_t$. Finally, the square of that increment contributes $\tfrac12 f_{xx}\,dt$, because $dB_t\,dB_t=dt$. This is why the time-dependent formula is not obtained by simply applying the ordinary chain rule to $f(t,B_t)$.

Differentiate in time.
Compute $f_t(t,x)$ while holding $x$ fixed.

Differentiate in space.
Compute $f_x(t,x)$ and $f_{xx}(t,x)$, then evaluate at $x=B_t$.

Assemble and integrate.
Use $f_t\,dt+f_x\,dB_t+\tfrac12 f_{xx}\,dt$, and verify that the stochastic integrand is in $\mathcal{L}^2$ on the interval considered.

Worked Example: A Time-Weighted Brownian Square

Apply the formula to $f(t,x)=tx^2$ on $0\le t\le1$. First compute the derivatives: $$ f_t(t,x)=x^2,\qquad f_x(t,x)=2tx,\qquad f_{xx}(t,x)=2t. $$ Consequently, $$ d(tB_t^2)=B_t^2\,dt+2tB_t\,dB_t+t\,dt =(B_t^2+t)\,dt+2tB_t\,dB_t. $$ Since $B_0=0$, integration gives $$ tB_t^2=\int_0^t(B_s^2+s)\,ds+\int_0^t2sB_s\,dB_s. $$

The stochastic integrand is in $\mathcal{L}^2$ on $[0,1]$: $$ \mathbb{E}\int_0^1(2sB_s)^2\,ds =4\int_0^1s^2\mathbb{E}[B_s^2]\,ds =4\int_0^1s^3\,ds=1. $$ Thus its Ito integral is square-integrable and has mean zero. At $t=1$, the expectation of the left side is $\mathbb{E}[B_1^2]=1$. On the right, the expected time integral is $$ \int_0^1(\mathbb{E}[B_s^2]+s)\,ds =\int_0^1 2s\,ds=1, $$ and the stochastic integral has expectation zero. The expectations agree, checking the formula independently.

Worked Example: A Time Factor Cancels the Ito Correction

Let $f(t,x)=e^{t/2}\cos x$ on $0\le t\le T$. Its derivatives are $$ f_t(t,x)=\tfrac12 e^{t/2}\cos x,\qquad f_x(t,x)=-e^{t/2}\sin x,\qquad f_{xx}(t,x)=-e^{t/2}\cos x. $$ The two $dt$ contributions cancel: $$ f_t(t,B_t)+\tfrac12 f_{xx}(t,B_t) =\tfrac12e^{t/2}\cos B_t-\tfrac12e^{t/2}\cos B_t=0. $$ Therefore $$ e^{t/2}\cos B_t=1-\int_0^t e^{s/2}\sin B_s\,dB_s. $$

The integrand is bounded in absolute value by $e^{T/2}$, so its squared expectation integrated over $[0,T]$ is at most $Te^T$; it is in $\mathcal{L}^2$. The Ito integral is therefore a square-integrable martingale, and the displayed process has constant conditional expectation equal to its current value: for $0\le s\le t$, $\mathbb{E}[e^{t/2}\cos B_t\mid\mathcal{F}_s]=e^{s/2}\cos B_s$. As an expectation check, $B_t$ has distribution $N(0,t)$, so $\mathbb{E}[\cos B_t]=e^{-t/2}$. Hence $\mathbb{E}[e^{t/2}\cos B_t]=1$, as the integral representation also implies.

Worked Example: An Exponential with Explicit Time Dependence

Take $f(t,x)=te^x$ on $0\le t\le1$. The derivatives are $$ f_t(t,x)=e^x,\qquad f_x(t,x)=te^x,\qquad f_{xx}(t,x)=te^x. $$ Thus $$ d(t e^{B_t})=e^{B_t}\,dt+t e^{B_t}\,dB_t+\tfrac12 t e^{B_t}\,dt =e^{B_t}(1+\tfrac12t)\,dt+t e^{B_t}\,dB_t. $$ At time zero $f(0,B_0)=0$. The integrand is square-integrable because $$ \mathbb{E}\int_0^1s^2e^{2B_s}\,ds =\int_0^1s^2e^{2s}\,ds<\infty, $$ where the equality uses the normal moment formula $\mathbb{E}[e^{aB_s}]=e^{a^2s/2}$. The Ito integral is consequently a square-integrable martingale with mean zero.

At $t=1$, the expectation of the left side is $\mathbb{E}[e^{B_1}]=e^{1/2}$. The expected drift integral on the right is $$ \int_0^1\mathbb{E}[e^{B_s}](1+\tfrac12s)\,ds =\int_0^1e^{s/2}(1+\tfrac12s)\,ds =\left[s e^{s/2}\right]_0^1=e^{1/2}. $$ The expected stochastic integral is zero, so the expectation check agrees exactly.

Two Consequences of the Formula

Proposition: Suppose the theorem's assumptions hold and $f_t(t,B_t)+\tfrac12f_{xx}(t,B_t)=0$ for all $t\in[0,T]$, almost surely. Then $f(t,B_t)$ is a martingale on $[0,T]$.

Proof. With the stated drift equal to zero, the formula gives $f(t,B_t)=f(0,B_0)+I_t$, where $I_t=\int_0^t f_x(s,B_s)\,dB_s$. Since the integrand is in $\mathcal{L}^2$, the Ito isometry gives $\mathbb{E}[I_t^2]=\mathbb{E}\int_0^t f_x(s,B_s)^2\,ds<\infty$, so $I_t$ is integrable. The Ito integral of a progressively measurable $\mathcal{L}^2$ integrand is a martingale: its construction from adapted step integrands gives $\mathbb{E}[I_t-I_s\mid\mathcal{F}_s]=0$, and the identity passes to the $L^2$ limit by continuity of conditional expectation in $L^2$. Therefore $\mathbb{E}[f(t,B_t)\mid\mathcal{F}_s]=f(s,B_s)$ for $s\le t$; the process is integrable as a constant plus $I_t$. $\square$

Corollary: If the theorem's assumptions hold and $f_t+\tfrac12f_{xx}=0$ everywhere on $[0,T]\times\mathbb{R}$, then $\mathbb{E}[f(t,B_t)]=f(0,0)$ for $0\le t\le T$.

Proof. The hypothesis makes the drift vanish along every Brownian path, so the preceding proposition applies. Its martingale conditional-expectation identity with $s=0$, followed by taking expectations, gives $\mathbb{E}[f(t,B_t)]=\mathbb{E}[f(0,B_0)]$. Since $B_0=0$ almost surely, the right side is $f(0,0)$. $\square$

Common Pitfall: Dropping the Time Derivative

For $f(t,x)=tx$, the ordinary product $tB_t$ is a useful test. Its derivatives are $f_t=x$, $f_x=t$, and $f_{xx}=0$, so the correct formula is $$ d(tB_t)=B_t\,dt+t\,dB_t. $$ A common incorrect calculation keeps only the space derivative and writes $d(tB_t)=t\,dB_t$. That omits $B_t\,dt$, the change caused by the explicit time argument. Integrating the incorrect expression would give $tB_t=\int_0^t s\,dB_s$, whereas the correct identity is $tB_t=\int_0^t B_s\,ds+\int_0^t s\,dB_s$. The omitted term is generally a nonzero random variable; it is not an Ito correction, since $f_{xx}=0$.

For any application, keep the three contributions visible until the derivatives have been evaluated. In particular, a zero drift in the resulting formula may arise from cancellation between $f_t$ and $\tfrac12f_{xx}$, as in the trigonometric example; it does not mean either term can be discarded before calculation.

Check Your Understanding

For each question, assume Brownian motion is defined on the filtered probability space described above and check that any stochastic integrand you use is in $\mathcal{L}^2$ on the stated finite interval.

Apply the time-dependent Ito formula to $f(t,x)=t^2x$, writing all three derivatives and the integrated identity.
For $f(t,x)=e^{-t}\cos x$, calculate $f_t+\tfrac12f_{xx}$. Does the formula imply a martingale in this case?
Apply the formula to $f(t,x)=x^3$. Identify the drift correction and state which Brownian moment verifies that the stochastic integrand is square-integrable on a finite interval.
For $f(t,x)=t e^x$, explain why the expected stochastic integral at time $1$ is zero and compute the expected drift integral.
In one sentence each, distinguish the contribution $f_t\,dt$ from the correction $\tfrac12f_{xx}\,dt$.