Worked Scalar Transformations of Ito Processes

Transforming a Scalar Ito Process

A scalar Ito process combines a time-integrated drift with a stochastic integral driven by Brownian motion. Applying a nonlinear function to the process changes both coefficients: the diffusion is multiplied by the function's derivative, while the drift also receives a correction involving the second derivative. This tutorial works out that calculation for powers, logarithms, and trigonometric functions, with attention to the domains on which each transformation is valid.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. On a fixed finite interval $[0,T]$, suppose $X_t=X_0+\int_0^t\mu_s\,ds+\int_0^t\sigma_s\,dB_s$. Here $\mu_s$ and $\sigma_s$ are progressively measurable, $\int_0^T|\mu_s|\,ds<\infty$ almost surely, and $\mathbb{E}\int_0^T\sigma_s^2\,ds<\infty$. The stochastic integral is the Ito integral, with integrand $\sigma_s$ in $\mathcal{L}^2$. The drift coefficient is $\mu_s$; the diffusion coefficient is $\sigma_s$.

Scalar transformation rule: Let $I$ be an open interval and $f:I\to\mathbb{R}$ be twice continuously differentiable. If $X_t$ stays in $I$, the scalar Ito formula gives $$ f(X_t)=f(X_0)+\int_0^t f'(X_s)\mu_s\,ds+\int_0^t f'(X_s)\sigma_s\,dB_s+\frac12\int_0^t f''(X_s)\sigma_s^2\,ds. $$ For the displayed stochastic integral to be a true square-integrable Ito integral on $[0,T]$, require $\mathbb{E}\int_0^T(f'(X_s)\sigma_s)^2\,ds<\infty$. With weaker local integrability, the formula is understood after stopping and its stochastic-integral term is in general only a local martingale. The formula is the standard scalar Ito formula established in the earlier material on that result.

In differential notation, the correction comes from $dB_t\,dB_t=dt$, whereas $dB_t\,dt=0$ and $dt\,dt=0$. Thus the ordinary chain rule does not apply: the second derivative term remains even though the transformation is a function of $X_t$ alone.

Powers: Derivatives Determine Both Coefficients

Proposition: Suppose $X_t>0$ on the time interval under consideration and $p\in\mathbb{R}$. For $f(x)=x^p$, the transformed drift and diffusion are $$ \left(pX_t^{p-1}\mu_t+\frac12p(p-1)X_t^{p-2}\sigma_t^2\right)dt \quad\text{and}\quad pX_t^{p-1}\sigma_t\,dB_t, $$ respectively, whenever the integrals in the scalar transformation rule are defined.

Proof. On $(0,\infty)$, the function $f(x)=x^p$ is twice continuously differentiable, including when $p=0$ or $p=1$. Its derivatives are $f'(x)=p x^{p-1}$ and $f''(x)=p(p-1)x^{p-2}$. Substitution into the scalar Ito formula gives the drift integrand $pX_t^{p-1}\mu_t+\tfrac12p(p-1)X_t^{p-2}\sigma_t^2$ and stochastic-integral integrand $pX_t^{p-1}\sigma_t$, which proves the stated coefficients. For $p=0$, both coefficients vanish and $X_t^0=1$; for $p=1$, the correction vanishes and the original coefficients are recovered. $\square$

Worked Example: Cubing a Drift-Diffusion Process

Let $X_t=1+t+2B_t$ for $0\le t\le1$. Compute the Ito differential of $X_t^3$, then check its expectation at $t=1$. Here $\mu_t=1$, $\sigma_t=2$, and $X_t$ can take any real value, so use $f(x)=x^3$ on $I=\mathbb{R}$, rather than relying on a positive-domain power rule.

The derivatives are $f'(x)=3x^2$ and $f''(x)=6x$. The first-derivative drift contribution is $3X_t^2\,dt$. The second-derivative correction is $\tfrac12(6X_t)(2^2)\,dt=12X_t\,dt$. The stochastic-integral contribution is $(3X_t^2)(2)\,dB_t=6X_t^2\,dB_t$. Therefore

d(X_t^3)=(3X_t^2+12X_t)\,dt+6X_t^2\,dB_t.

For the expectation check, $X_t$ is normal with mean $1+t$ and variance $4t$. Hence $\mathbb{E}[X_t]=1+t$ and $\mathbb{E}[X_t^2]=(1+t)^2+4t$. The stochastic-integral integrand $6X_t^2$ is square-integrable on $[0,1]$, since $\mathbb{E}[X_t^4]$ is finite and integrable there. Its Ito integral is therefore a true martingale with mean zero. Taking expectations in the differential and integrating gives $\mathbb{E}[X_1^3]=1+\int_0^1\{3[(1+t)^2+4t]+12(1+t)\}\,dt$. The integrand simplifies to $15+30t+3t^2$, whose integral is $15+15+1=31$. Thus $\mathbb{E}[X_1^3]=32$.

Independent check: $X_1=2+2B_1$. Since $\mathbb{E}[B_1]=0$, $\mathbb{E}[B_1^2]=1$, and $\mathbb{E}[B_1^3]=0$, expanding gives $\mathbb{E}[(2+2B_1)^3]=8+24\mathbb{E}[B_1]+24\mathbb{E}[B_1^2]+8\mathbb{E}[B_1^3]=32$, as above.

Logarithms: Positivity and Stopping

The logarithm is defined only for positive arguments. Consequently, applying it to a diffusion requires either that positivity is known on the time interval or that the calculation is restricted to a region before the process reaches zero. A stopping time can enforce such a region; stopping does not remove the Ito correction.

Logarithm rule: For a positive scalar Ito process $X_t$, take $f(x)=\log x$. Since $f'(x)=1/x$ and $f''(x)=-1/x^2$, its transformed drift and diffusion are $$ \left(\frac{\mu_t}{X_t}-\frac{\sigma_t^2}{2X_t^2}\right)dt \quad\text{and}\quad \frac{\sigma_t}{X_t}\,dB_t. $$ This statement applies while $X_t>0$; integrability of the displayed coefficients is required, or the formula must be localized.

Worked Example: Logarithm Before a Boundary

Set $X_t=4+t+B_t$, and let $\tau=\inf\{t\ge0:X_t\le1\}$, with $\inf\varnothing=\infty$. Compute the differential of $Y_t=\log(X_{t\wedge\tau})$. Because $X_0=4$ and paths are continuous, the process stopped at $\tau$ stays at least $1$. In stopped-process notation, $dX_{t\wedge\tau}=\mathbf{1}_{\{t<\tau\}}dt+\mathbf{1}_{\{t<\tau\}}dB_t$. The stochastic-integral integrand $\mathbf{1}_{\{t<\tau\}}$ is progressively measurable and has expected integrated square at most $T$ on every finite horizon.

For $f(x)=\log x$, $f'(x)=1/x$ and $f''(x)=-1/x^2$. Before stopping, $\mu_t=1$ and $\sigma_t=1$, so the first-derivative drift is $1/X_t$, the correction is $-1/(2X_t^2)$, and the diffusion is $1/X_t$. Including the stopping indicator,

dY_t=\mathbf{1}_{\{t<\tau\}}\left(\frac{1}{X_t}-\frac{1}{2X_t^2}\right)dt +\mathbf{1}_{\{t<\tau\}}\frac{1}{X_t}\,dB_t.

The stopped stochastic-integral integrand is bounded in absolute value by $1$, so its Ito integral is square-integrable and is a true martingale. This also justifies taking its expectation as zero on any finite horizon.

Independent check: Apply Ito's formula to $e^{Y_t}$. Its derivatives with respect to $y$ are $e^y$ and $e^y$. Before stopping, $e^{Y_t}=X_t$, and the drift in $dY_t$ is $1/X_t-1/(2X_t^2)$, with diffusion $1/X_t$. The transformed drift is $X_t(1/X_t-1/(2X_t^2))+\tfrac12X_t(1/X_t)^2=1-\tfrac1{2X_t}+\tfrac1{2X_t}=1$; the diffusion is $X_t(1/X_t)=1$. These are precisely the drift and diffusion of $X_t$ before stopping. After $\tau$, both stopped processes are constant.

Trigonometric Functions: Bounded Derivatives, Nonzero Correction

Sine and cosine have bounded derivatives, so their transformed stochastic-integral integrands are often square-integrable whenever the original diffusion is. Boundedness does not eliminate the correction: the second derivative of sine is minus sine, and that of cosine is minus cosine.

Worked Example: Sine of a Drifted Brownian Motion

Let $X_t=t+2B_t$ on $0\le t\le1$, and find the Ito differential of $\sin(X_t)$. Here $\mu_t=1$, $\sigma_t=2$. For $f(x)=\sin x$, $f'(x)=\cos x$ and $f''(x)=-\sin x$. The first-derivative drift is $\cos(X_t)$, the correction is $\tfrac12[-\sin(X_t)](2^2)=-2\sin(X_t)$, and the stochastic-integral term is $2\cos(X_t)\,dB_t$. Thus

d\sin(X_t)=\big(\cos(X_t)-2\sin(X_t)\big)\,dt+2\cos(X_t)\,dB_t.

The stochastic-integral integrand $2\cos(X_t)$ is bounded by $2$, so its Ito integral is square-integrable on $[0,1]$, a true martingale, and has expectation zero.

Independent check: A normal-variable calculation gives $\mathbb{E}[\sin(t+2B_t)]=e^{-2t}\sin t$. Differentiating this expression gives $e^{-2t}(\cos t-2\sin t)$. Also, $\mathbb{E}[\cos(t+2B_t)]=e^{-2t}\cos t$ and $\mathbb{E}[\sin(t+2B_t)]=e^{-2t}\sin t$. Therefore the expected drift in the displayed Ito differential is $e^{-2t}(\cos t-2\sin t)$, exactly the derivative of the independently computed expectation. At $t=0$, both expectations are $0$.

Why the Correction Matters

The transformation rule is useful because it gives the coefficients of a new scalar Ito process directly, without guessing which terms will remain after transformation. The drift changes by $\tfrac12 f''(X_t)\sigma_t^2$; the diffusion changes by the factor $f'(X_t)$. The correction depends on the square of the diffusion coefficient, not on the drift coefficient. In particular, a nonlinear function can create drift even when the original process has zero drift.

One must also distinguish the stochastic-integral term from the full transformed process. When its integrand belongs to $\mathcal{L}^2$, the Ito integral is a square-integrable martingale: its conditional expectation at a later time equals its current value. Under only local integrability, the corresponding term is generally a local martingale, and one cannot conclude that its expectation is constant. The drift integral is a finite-variation term, not a martingale merely because it appears in the same differential.

Common mistake: For the logarithm example, the ordinary chain rule would give $d\log X_t=(1/X_t)\,dt+(1/X_t)\,dB_t$ before stopping. It omits the correction $-1/(2X_t^2)\,dt$, because it treats $(dB_t)^2$ as zero. The Ito multiplication rules instead give $(dB_t)^2=dt$, $dB_t\,dt=0$, and $dt\,dt=0$; with $f''(x)=-1/x^2$ and $\sigma_t=1$, the missing term is $\tfrac12f''(X_t)\sigma_t^2dt=-1/(2X_t^2)dt$. The exponential check above confirms that this term is necessary to recover the original stopped process.

Check Your Understanding

For each question, state any domain or integrability condition needed for the calculation.

If $dX_t=\mu_t\,dt+\sigma_t\,dB_t$ and $X_t>0$, what are the drift and diffusion coefficients of $X_t^p$ for a fixed real $p$?
For $dX_t=2\,dt+3\,dB_t$ with $X_t>0$, compute the drift and diffusion coefficients of $\log X_t$.
Apply Ito's formula to $\cos(X_t)$ when $dX_t=\mu_t\,dt+\sigma_t\,dB_t$, writing out both derivatives and the correction term.
Why does square-integrability of the transformed stochastic-integral integrand permit taking its expectation to be zero?
In the sine example, identify the term that would be missing under the ordinary chain rule and explain which multiplication rule produces it.