What Makes a Process an Ito Process?
An Ito process combines two different kinds of accumulation. Its drift accumulates through ordinary time integration, while its diffusion accumulates through an Ito integral against Brownian motion. The distinction matters: the diffusion term is random even when its coefficient is constant, and its stochastic integral has martingale properties that the drift term need not have.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) adapted to the filtration, with increments after time \(s\) independent of \(\mathcal{F}_s\). Fix a finite horizon \(T>0\). A process is progressively measurable if, for each \(t\), its restriction to \([0,t]\times\Omega\) is measurable with respect to the product of the Borel sets on \([0,t]\) and \(\mathcal{F}_t\). This condition ensures that its value at time \(s\) uses information available by time \(s\).
The time-integrability condition implies \(\int_0^t|\mu_s|\,ds<\infty\) almost surely. The Ito integral is defined for the stated \(\mathcal{L}^2\) integrand and has a continuous version. Thus the displayed representation defines a continuous adapted process. The coefficient \(\mu_s\) is called the drift; \(\sigma_s\) is the diffusion coefficient. When \(\mu_s=\mu(s,X_s)\) and \(\sigma_s=\sigma(s,X_s)\), the common differential notation is $$ dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t. $$ This notation abbreviates the integral representation; it is not an ordinary differential equation.
The Integrability Conditions and the Martingale Part
The \(\mathcal{L}^2\) condition is more than a technical convenience: it makes the Ito integral square-integrable and gives the Ito isometry. The weaker condition on \(\mu\) makes its time integral integrable, so the expectation of \(X_t\) is defined. For every stochastic integral below, the integrand is progressively measurable and satisfies the stated \(\mathcal{L}^2\) condition.
Proof. First consider a simple adapted integrand \(H\) on \([s,t]\), constant on finitely many time intervals and with each interval’s value measurable at its left endpoint. Each term of \(\int_s^t H_u\,dB_u\) is a past-measurable coefficient multiplied by a Brownian increment independent of the information at the start of that interval and with mean zero. Taking conditional expectations successively over the intervals gives \(\mathbb{E}[\int_s^tH_u\,dB_u\mid\mathcal{F}_s]=0\). For general progressively measurable \(H\in\mathcal{L}^2\), choose simple adapted integrands converging to \(H\) in \(\mathcal{L}^2\). By the Ito isometry their integrals converge in \(L^2(\mathbb{P})\), hence in \(L^1(\mathbb{P})\); conditional expectation is continuous in \(L^1\). Therefore the same zero conditional expectation holds for \(\int_s^tH_u\,dB_u\). Apply this fact to \(H=\sigma\), and split both integrals defining \(X_t\) at \(s\). The part up to \(s\) is \(X_s\), while the stochastic integral from \(s\) to \(t\) has conditional expectation zero. The drift integral is integrable because \(\mathbb{E}\int_0^T|\mu_u|\,du<\infty\), so its conditional expectation exists. This proves the formula. If \(\mu=0\) almost everywhere, the formula reduces to \(\mathbb{E}[X_t\mid\mathcal{F}_s]=X_s\); \(X_t\) is integrable by the assumptions and the Ito isometry. Hence \(X\) is a martingale. \(\square\)
More generally, the stochastic integral \(M_t=\int_0^t\sigma_s\,dB_s\) is a square-integrable martingale: it is integrable by the Ito isometry, and the conditional-expectation argument in the proof gives \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). The full Ito process is not automatically a martingale; its drift term must also be accounted for.
Worked Examples: Identifying the Coefficients
Worked Example: Brownian Motion with Constant Drift
Let \(X_t=x+2t+3B_t\), where \(x\) is a fixed real number. Identify its coefficients, check the integrability requirements on \([0,T]\), and find its mean and variance.
The representation is $$ X_t=x+\int_0^t 2\,ds+\int_0^t 3\,dB_s. $$ The drift integrability is \(\mathbb{E}\int_0^T|2|\,ds=2T<\infty\), and the diffusion integrability is \(\mathbb{E}\int_0^T3^2\,ds=9T<\infty\). Thus the drift is \(2\) and the diffusion coefficient is \(3\). Since the Ito integral has mean zero, \(\mathbb{E}[X_t]=x+2t\). The Ito isometry gives \(\operatorname{Var}(X_t)=\mathbb{E}[(3B_t)^2]=9t\).
Check: \(B_t\sim N(0,t)\), so \(x+2t+3B_t\) has mean \(x+2t\) and variance \(3^2t=9t\), agreeing with the integral calculation.
Worked Example: Deterministic Time-Varying Coefficients
Consider \(Y_t=\int_0^t(1+s)\,ds+\int_0^t s\,dB_s\). Verify the coefficient conditions and calculate the mean and variance at \(t=2\).
The drift is the deterministic, hence progressively measurable, function \(\mu_s=1+s\). Its absolute integral on \([0,2]\) is $$ \int_0^2(1+s)\,ds=\left[s+\frac{s^2}{2}\right]_0^2=4. $$ The diffusion is \(\sigma_s=s\), and \(\mathbb{E}\int_0^2s^2\,ds=8/3<\infty\). The mean of the stochastic integral is zero, so \(\mathbb{E}[Y_2]=4\). By the Ito isometry, the stochastic integral has second moment \(8/3\). Its mean is zero, so that second moment is also its variance; the deterministic drift contributes no variance. Therefore \(\operatorname{Var}(Y_2)=8/3\).
Check: for any \(t\in[0,2]\), the mean is \(t+t^2/2\) and the variance is \(t^3/3\). Substituting \(t=2\) gives \(2+2=4\) and \(2^3/3=8/3\), as calculated.
Worked Example: Geometric Brownian Motion
Let \(S_t=s\exp(t/2+B_t)\), with \(s>0\). Show that this is an Ito process and identify its drift and diffusion coefficients.
Use the time-dependent Ito formula, with \(f(t,x)=s\exp(t/2+x)\). Its derivatives are \(f_t=\tfrac12 f\), \(f_x=f\), and \(f_{xx}=f\). Since \(dB_t\,dB_t=dt\), the formula, including its correction term, gives $$ dS_t=\left(f_t+\frac12 f_{xx}\right)(t,B_t)\,dt+f_x(t,B_t)\,dB_t =S_t\,dt+S_t\,dB_t. $$ Equivalently, \(S_t=s+\int_0^t S_u\,du+\int_0^t S_u\,dB_u\). The integrands are progressively measurable because \(S\) is continuous and adapted. For each finite \(T\), the normal moment-generating function gives $$ \mathbb{E}[S_t^2]=s^2e^t\mathbb{E}[e^{2B_t}] =s^2e^te^{2t}=s^2e^{3t}, \qquad \mathbb{E}\int_0^T S_t^2\,dt=s^2\int_0^T e^{3t}\,dt<\infty. $$ Also \(\mathbb{E}\int_0^T|S_t|\,dt<\infty\), since \(\mathbb{E}[S_t]=s e^t\) and \(s\int_0^T e^t\,dt<\infty\). The drift and diffusion coefficients are therefore both \(S_t\), and the required conditions hold.
Check: the process definition implies \(\mathbb{E}[S_t]=s+\int_0^t\mathbb{E}[S_u]\,du\); directly, the normal moment-generating function gives \(\mathbb{E}[S_t]=s e^{t/2}e^{t/2}=s e^t\). This agrees with the drift calculation. In general, \(s\exp((a-\sigma^2/2)t+\sigma B_t)\) has drift \(aS_t\) and diffusion \(\sigma S_t\), provided the corresponding finite-horizon integrability conditions hold.
Notation, Stratonovich Integrals, and Common Pitfalls
The Ito integral and the Stratonovich integral are different notions. For a continuous semimartingale integrand \(H\) for which the covariation exists, the Stratonovich convention is related to the Ito convention by \(\int_0^t H_s\circ dB_s=\int_0^t H_s\,dB_s+\tfrac12[H,B]_t\). In particular, the Stratonovich chain rule does not remove the Ito correction from an Ito calculation. For example, \(d(e^{B_t})=e^{B_t}\,dB_t+\tfrac12e^{B_t}\,dt\); omitting the \(dt\) term would confuse the Ito and Stratonovich rules. In differential shorthand, the multiplication rules include \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); these encode limiting stochastic integral calculations, not ordinary algebraic identities between finite increments.
A common mistake is to identify the drift with the whole conditional mean change, even when the coefficients are random. For a random drift, the conditional mean change from \(s\) to \(t\) is \(\mathbb{E}[\int_s^t\mu_u\,du\mid\mathcal{F}_s]\), not necessarily a deterministic number. Another mistake is to omit the integrability check: a formal expression \(\int_0^t\sigma_s\,dB_s\) is not covered by the square-integrable Ito integral construction unless \(\sigma\) is progressively measurable and \(\mathbb{E}\int_0^T\sigma_s^2\,ds<\infty\).
An Ito-process representation on a given filtered space uses that space’s Brownian motion and filtration. This is a strong construction; a weak solution instead allows the probability space and Brownian motion to be chosen as part of the solution. Existence of a process with specified state-dependent coefficients is a separate question from defining an Ito process once its drift and diffusion are given. Conditions often used to establish existence and uniqueness include Lipschitz continuity in the state variable and linear growth: for some constant \(C\), the coefficients satisfy \(|\mu(t,x)-\mu(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le C|x-y|\) and \(|\mu(t,x)|+|\sigma(t,x)|\le C(1+|x|)\). These conditions are not required for the definition above; they are sufficient conditions used in particular existence and uniqueness theorems.
Check Your Understanding
Use the definition and conditional-expectation formula to organize each answer.
- For \(X_t=1+\int_0^t(2-s)\,ds+\int_0^t(1+s)\,dB_s\), verify the two integrability conditions on any finite horizon \(T\).
- If \(\mu=0\) and \(\sigma\in\mathcal{L}^2\), which conditions ensure that \(X_t=X_0+\int_0^t\sigma_s\,dB_s\) is a martingale?
- For \(Z_t=\int_0^t s\,dB_s\), calculate \(\mathbb{E}[Z_1]\) and \(\operatorname{Var}(Z_1)\).
- Apply the time-dependent Ito formula to \(f(t,x)=e^{2t+x}\), displaying \(f_t\), \(f_x\), and \(f_{xx}\), and identify the drift and diffusion of \(f(t,B_t)\).
- Explain why square-integrability of \(\sigma\), rather than just integrability of \(|\sigma|\), is the condition used for the Ito integral in this definition.