The Scalar Ito Formula for Time-Dependent Functions

From an Itô Process to a Transformed Process

A scalar Itô process combines a finite-variation drift with Brownian noise. Applying a smooth function to that process changes both coefficients: the new diffusion comes from the state derivative, while the new drift includes a second-derivative correction. For a function that also depends on time, its explicit time derivative contributes to the drift as well.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. A process is progressively measurable if, on each interval $[0,T]$, its restriction is measurable with respect to the product of the Borel sets on $[0,T]$ and $\mathcal{F}_T$, with the corresponding measurability adapted to each earlier time. The class $\mathcal{L}^2$ consists of progressively measurable $H$ such that $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for every finite $T$ under consideration. For $H\in\mathcal{L}^2$, the Itô integral $\int_0^t H_s\,dB_s$ is a square-integrable martingale.

Definition: A scalar Itô process on $[0,T]$ has the form $X_t=X_0+\int_0^t b_s\,ds+\int_0^t\sigma_s\,dB_s$, where $X_0$ is finite and $\mathcal{F}_0$-measurable, $b$ is progressively measurable with $\int_0^T|b_s|\,ds<\infty$ almost surely, and $\sigma$ is progressively measurable with $\int_0^T\sigma_s^2\,ds<\infty$ almost surely. The process $X$ is continuous and adapted; $b$ is its drift coefficient and $\sigma$ its diffusion coefficient.

In the common coefficient notation $b_s=b(s,X_s)$, $\sigma_s=\sigma(s,X_s)$, the coefficients are evaluated along the process. We use the differential shorthand $dX_t=b(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$. The multiplication rules are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. These encode quadratic variation, not ordinary products of infinitesimal real numbers.

The Time-Dependent Scalar Itô Formula

Theorem (scalar Itô formula): Suppose $X_t=X_0+\int_0^t b_s\,ds+\int_0^t\sigma_s\,dB_s$ satisfies the conditions above, and $f(t,x)$ has continuous partial derivatives $f_t,f_x,f_{xx}$ on an open set containing the time-state points $(t,X_t)$, $0\le t\le T$. Then, up to localization if necessary, $$ f(t,X_t)=f(0,X_0)+\int_0^t\left(f_t(s,X_s)+b_s f_x(s,X_s)+\frac12\sigma_s^2 f_{xx}(s,X_s)\right)ds+\int_0^t f_x(s,X_s)\sigma_s\,dB_s. $$ Here the stochastic integral is defined whenever its integrand is locally square-integrable. If $\mathbb{E}\int_0^T|f_x(s,X_s)\sigma_s|^2\,ds<\infty$, it is an $\mathcal{L}^2$ Itô integral on $[0,T]$. The ordinary drift integral must also be finite, locally or under a stronger global integrability assumption.

The notation $C^{1,2}$ means once continuously differentiable in time and twice continuously differentiable in the state variable, with these derivatives continuous. The formula is the standard time-dependent scalar Itô formula; its proof for a scalar Itô process is given in the subsequent treatment of that proof. When coefficients or derivatives are not globally bounded, the formula is read locally: stop the processes before they leave bounded regions and before the relevant integrals exceed finite bounds, apply the formula to the stopped processes, and then increase the stopping regions.

The transformed drift is $f_t+b f_x+\tfrac12\sigma^2f_{xx}$; the transformed diffusion is $f_x\sigma$. The explicit time dependence of $f$ appears only in the drift. The second-derivative term appears because the Brownian component has quadratic variation $dt$, unlike a differentiable path, for which the ordinary chain rule has no such correction.

Worked Example: A Time-Weighted Square

Let $X_t=1+2t+3B_t$, so $b=2$ and $\sigma=3$, and take $f(t,x)=tx^2$. Find the differential of $Y_t=f(t,X_t)$ and check its expectation.

The derivatives are $f_t(t,x)=x^2$, $f_x(t,x)=2tx$, and $f_{xx}(t,x)=2t$. The correction is $\tfrac12 f_{xx}\sigma^2=\tfrac12(2t)(9)=9t$. Therefore

dY_t=(X_t^2+4tX_t+9t)\,dt+6tX_t\,dB_t.

The stochastic integrand is in $\mathcal{L}^2$ on every finite interval: $\mathbb{E}[X_s^2]=1+13s+4s^2$, so $\mathbb{E}\int_0^T36s^2X_s^2\,ds<\infty$. Also $Y_0=0$, and the Itô integral has mean zero. Thus integrating the drift gives $\mathbb{E}[Y_t]=\int_0^t\mathbb{E}[X_s^2+4sX_s+9s]\,ds$. Here $\mathbb{E}[X_s]=1+2s$ and $\mathbb{E}[X_s^2]=(1+2s)^2+9s=1+13s+4s^2$. The expected drift is $1+26s+12s^2$, and hence $\mathbb{E}[Y_t]=t+13t^2+4t^3$.

Check: Directly, $\mathbb{E}[tX_t^2]=t(1+13t+4t^2)=t+13t^2+4t^3$, matching the formula.

Worked Example: A Zero-Drift Transform

For $X_t=B_t$, apply the formula to $f(t,x)=e^{t/2}\cos x$. Determine whether the result is a martingale on a fixed finite interval $[0,T]$.

The derivatives are $f_t=\tfrac12e^{t/2}\cos x$, $f_x=-e^{t/2}\sin x$, and $f_{xx}=-e^{t/2}\cos x$. With $b=0$ and $\sigma=1$, the drift is $\tfrac12e^{t/2}\cos X_t+\tfrac12(-e^{t/2}\cos X_t)=0$. Consequently, $$ Y_t=e^{t/2}\cos B_t=1-\int_0^t e^{s/2}\sin B_s\,dB_s. $$

On $[0,T]$, the integrand satisfies $\mathbb{E}\int_0^T e^s\sin^2(B_s)\,ds\le\int_0^T e^s\,ds<\infty$, so the integral is square-integrable. The initial value $Y_0=1$ is integrable. It follows that $Y$ is a martingale. For a direct conditional-expectation check, independent Brownian increments and the normal characteristic function give, for $s\le t$, $\mathbb{E}[\cos B_t\mid\mathcal{F}_s]=e^{-(t-s)/2}\cos B_s$. Multiplying by $e^{t/2}$ yields $\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_s$.

Check: At $s=0$, the conditional-expectation identity gives $\mathbb{E}[Y_t]=Y_0=1$. Independently, $\mathbb{E}[\cos B_t]=e^{-t/2}$, so $\mathbb{E}[e^{t/2}\cos B_t]=1$, as required.

Worked Example: A Time Derivative Cancels a Correction

Let $X_t=1+B_t$ and $f(t,x)=e^{-t}x^2$. Compute the transformed process and verify its expected value directly.

The derivatives are $f_t=-e^{-t}x^2$, $f_x=2e^{-t}x$, and $f_{xx}=2e^{-t}$. Here $b=0$, $\sigma=1$, and the correction is $\tfrac12f_{xx}=e^{-t}$. Thus

d\big(e^{-t}X_t^2\big)=e^{-t}(1-X_t^2)\,dt+2e^{-t}X_t\,dB_t.

The stochastic integrand is square-integrable on $[0,T]$, since $\mathbb{E}[4e^{-2s}X_s^2]=4e^{-2s}(1+s)$ is integrable there. Also $Y_0=1$. Since $\mathbb{E}[X_s^2]=1+s$, the expected drift is $-se^{-s}$. Integration gives $\mathbb{E}[Y_t]=1-\int_0^t se^{-s}\,ds=e^{-t}(1+t)$. Directly, $\mathbb{E}[e^{-t}X_t^2]=e^{-t}(1+t)$, verifying the result.

Two Consequences and Their Conditions

Corollary: If $g$ is continuously differentiable and $Y_t=g(t)$ is deterministic, then the time-dependent Itô formula reduces to $dg(t)=g'(t)\,dt$.

Proof. Set $X_t$ equal to a constant, so $b=0$ and $\sigma=0$, and set $f(t,x)=g(t)$. Then $f_t(t,x)=g'(t)$, $f_x(t,x)=0$, and $f_{xx}(t,x)=0$. Substitution in the formula leaves $g(t)=g(0)+\int_0^t g'(s)\,ds$, which is the ordinary fundamental theorem of calculus. $\square$

Proposition: Suppose the transformed drift $f_t(s,X_s)+b_s f_x(s,X_s)+\tfrac12\sigma_s^2f_{xx}(s,X_s)$ is zero for almost every $s$, almost surely. If $Y_0=f(0,X_0)$ is integrable and $H_s=f_x(s,X_s)\sigma_s$ satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, then $Y_t=f(t,X_t)$ is a martingale on $[0,T]$.

Proof. The formula gives $Y_t=Y_0+\int_0^t H_s\,dB_s$. The assumed square-integrability makes the Itô integral a square-integrable martingale, so it is integrable and satisfies $\mathbb{E}[\int_0^tH_s\,dB_s\mid\mathcal{F}_s]=\int_0^sH_u\,dB_u$ for $s\le t$. Since $Y_0$ is integrable and $\mathcal{F}_0$-measurable, $Y_t$ is integrable. Taking conditional expectations in the displayed representation gives $\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_0+\int_0^sH_u\,dB_u=Y_s$. This is precisely the martingale property. $\square$

The integrability of $Y_0$ is essential. A zero drift and a square-integrable stochastic integrand alone do not ensure that $Y_t$ is integrable: if $X_t=Z$ for a finite almost surely but non-integrable $\mathcal{F}_0$-measurable random variable $Z$, and $f(t,x)=x$, then the stochastic integrand is zero but $Y_t=Z$ is not a martingale.

Common Mistake: Omitting the Second-Derivative Term

Take $X_t=B_t$ and $f(t,x)=tx^2$. The derivatives are $f_t=x^2$, $f_x=2tx$, and $f_{xx}=2t$. The correct formula is $d(tB_t^2)=(B_t^2+t)\,dt+2tB_t\,dB_t$. A common wrong calculation drops the correction and writes $B_t^2\,dt+2tB_t\,dB_t$. Its expected drift is $\mathbb{E}[B_t^2]=t$, which would imply $\mathbb{E}[tB_t^2]=\int_0^t s\,ds=t^2/2$. But directly $\mathbb{E}[tB_t^2]=t^2$. Including the correction gives expected drift $t+t=2t$, whose integral is $t^2$, agreeing with the direct calculation.

This mismatch is the practical signal that the ordinary chain rule has been used where Brownian quadratic variation matters. The correct correction is $\tfrac12 f_{xx}\sigma^2\,dt$; it depends on the curvature of $f$ in the state variable and on the square of the diffusion coefficient, not on the drift coefficient.

Check Your Understanding

Use the formula with its stated hypotheses, and identify the time derivative and correction term separately.

For $X_t=2+B_t$ and $f(t,x)=t\sin x$, calculate the transformed drift and diffusion.
For $X_t=3t+2B_t$ and $f(t,x)=x^2$, find the drift and diffusion of $f(t,X_t)$.
For $X_t=B_t$ and $f(t,x)=e^{-t/2}\cos x$, determine whether the transformed drift vanishes.
State the integrability assumptions needed to conclude that a zero-drift transformed process is a martingale.
Explain why the ordinary chain rule gives the wrong expectation for $tB_t^2$.

The Scalar Ito Formula for Time-Dependent Functions

What You'll Learn