Differential Notation and the Integral SDE

What a Stochastic Differential Equation Says

The notation $dX_t=b(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$ resembles a differential equation, but Brownian paths are not differentiable in the ordinary sense. In particular, $dB_t$ is not an ordinary infinitesimal increment to which the classical chain rule can be applied. The equation is instead shorthand for a statement about two integrals accumulated from an initial time. Making that statement explicit identifies the starting value, the time interval, the integrability required, and the sense in which the equality holds.

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 a finite horizon $T>0$. For the scalar equation below, suppose $X_t$ is an adapted continuous process, $b(t,X_t)$ is progressively measurable with $\int_0^T |b(s,X_s)|\,ds<\infty$ almost surely, and $\sigma(t,X_t)$ is progressively measurable with $\mathbb{E}\int_0^T \sigma(s,X_s)^2\,ds<\infty$. These assumptions make the drift integral a pathwise Lebesgue integral and the diffusion integral an Itô integral.

Definition: Under those assumptions, the Itô SDE $dX_t=b(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$, with initial condition $X_0=x$, means the integral equation $$ X_t=x+\int_0^t b(s,X_s)\,ds+\int_0^t \sigma(s,X_s)\,dB_s,\qquad 0\le t\le T. $$ The identity is required to hold for every $t\in[0,T]$ simultaneously, outside one $\mathbb{P}$-null set. Each stochastic integrand is adapted and progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning its expected integrated square is finite.

In this definition $x$ is a fixed, non-random starting value. The initial condition is part of the equation: putting $t=0$ gives $X_0=x$, since both integrals from $0$ to $0$ vanish. The equation can also be written from any $s$ to any later $t$ by subtracting the identities at those times. Its drift coefficient $b$ contributes through ordinary time integration; its diffusion coefficient $\sigma$ multiplies the Brownian noise in an Itô integral.

The Itô integral used here is defined for progressively measurable integrands $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. It is constructed by completing simple adapted integrands in the norm given by the Itô isometry, $\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds$. For simple integrands, the integral is a sum of past-measurable coefficients times Brownian increments; each increment has conditional mean zero given the past. Approximation in this norm therefore gives an integral with mean zero. The integral also has a continuous adapted version, which is the version used in the simultaneous identity above.

For comparison, a Stratonovich integral is written $\int_0^t H_s\circ dB_s$; it is not the same integral. In the scalar case, when $\sigma$ is differentiable in its state variable and the usual semimartingale integrability conditions hold, the equation $dX_t=b(t,X_t)\,dt+\sigma(t,X_t)\circ dB_t$ has Itô drift $b(t,X_t)+\tfrac12\sigma(t,X_t)\partial_x\sigma(t,X_t)$. Unless a circle is displayed, the differential notation in this tutorial means the Itô convention.

Integral Identities and Their Time-Interval Meaning

Proposition: Let $A_t=\int_0^t a_s\,ds$ and $M_t=\int_0^t H_s\,dB_s$, where $\int_0^T|a_s|\,ds<\infty$ almost surely and $H$ is progressively measurable with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. If $X_t=x+A_t+M_t$ holds simultaneously for $t\in[0,T]$ outside a null set, then, on that same event, for every $0\le s\le t\le T$, $$ X_t-X_s=\int_s^t a_u\,du+\int_s^t H_u\,dB_u. $$

Proof. Fix an outcome in the probability-one event on which the process identity holds for every time. Subtract the identity at $s$ from the identity at $t$. Additivity of the Lebesgue integral gives $A_t-A_s=\int_s^t a_u\,du$. The Itô integral is defined by limits of sums over time intervals and has the corresponding additivity property: $M_t-M_s=\int_s^t H_u\,dB_u$ almost surely. Taking the continuous versions of the integrals makes these equalities simultaneous in $s,t$. Combining the differences proves the stated increment equation. $\square$

The word “simultaneously” matters. An assertion that the equation holds almost surely for each fixed $t$ does not, by itself, say that a single null set works for all uncountably many times. In the integral formulation, both accumulated integrals have continuous versions, and the SDE identity is understood for those versions on the full interval. An equivalent way to specify the equation is to require the initial value and the increment relation on the interval, with the same continuous-version convention.

The formal multiplication rules $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ help organize Itô calculations, but they do not turn the SDE into an ordinary differential equation. In particular, $\sigma(t,X_t)\,dB_t$ is not the product of a coefficient and a classically differentiable noise path. The integral equation is the precise content of the notation.

Worked Integral Translations

Worked Example: Constant Drift and Diffusion

On $[0,T]$, consider $dX_t=2\,dt+3\,dB_t$ with $X_0=3$. Write the integral equation and check its mean and variance at $t=1$, where $T\ge1$.

The drift integrand is the constant process $2$, so $\int_0^t2\,ds=2t$. The diffusion integrand is the constant adapted process $3$, and $\mathbb{E}\int_0^T3^2\,ds=9T<\infty$. Thus the Itô integral is defined and the equation becomes

X_t=3+2t+3\int_0^t 1\,dB_s=3+2t+3B_t,\qquad 0\le t\le T.

At $t=1$, this gives $X_1=5+3B_1$. Since $B_1$ has mean $0$ and variance $1$, $\mathbb{E}[X_1]=5+3\cdot0=5$ and $\operatorname{Var}(X_1)=3^2\cdot1=9$. The integral equation also gives $X_0=3+0+0=3$, as required. The variance check confirms that the diffusion coefficient $3$ scales Brownian variance by its square.

Worked Example: Time-Dependent Integrands

On $[0,2]$, translate $dX_t=t\,dt+2t\,dB_t$, $X_0=1$, and compute the mean and variance at $t=2$.

The drift integral is $\int_0^t s\,ds=t^2/2$. For the diffusion integral the integrand $H_s=2s$ is deterministic, hence progressively measurable, and its integrability is checked directly: $\mathbb{E}\int_0^2(2s)^2\,ds=4\int_0^2s^2\,ds=4(8/3)=32/3<\infty$. Therefore the integral equation is

X_t=1+\frac{t^2}{2}+\int_0^t2s\,dB_s,\qquad 0\le t\le2.

The stochastic integral has mean zero, as follows from the simple-integrand construction and its $L^2$ extension. Its variance is its expected integrated square by the Itô isometry. At $t=2$, the deterministic part is $1+2^2/2=3$, so $\mathbb{E}[X_2]=3$, and

\operatorname{Var}(X_2)=\mathbb{E}\left[\left(\int_0^2 2s\,dB_s\right)^2\right] =\int_0^2 4s^2\,ds =4\left[\frac{s^3}{3}\right]_0^2 =\frac{32}{3}.

As a check, the drift has no variance because it is deterministic, while the diffusion variance is exactly the integral of $(2s)^2$. Also, setting $t=0$ in the displayed equation returns $X_0=1$.

Worked Example: Checking a Candidate for a Driftless SDE

Consider $dX_t=X_t\,dB_t$, $X_0=1$, on $[0,T]$. Verify that $X_t=e^{B_t-t/2}$ satisfies its integral equation.

Set $f(t,x)=e^{x-t/2}$. Its derivatives are $f_t=-\tfrac12e^{x-t/2}$, $f_x=e^{x-t/2}$, and $f_{xx}=e^{x-t/2}$. Itô's formula, including its second-derivative correction, gives

df(t,B_t)=\left(f_t+\frac12f_{xx}\right)dt+f_x\,dB_t =\left(-\frac12e^{B_t-t/2}+\frac12e^{B_t-t/2}\right)dt+e^{B_t-t/2}\,dB_t =X_t\,dB_t.

Integrating from $0$ to $t$, and using $f(0,B_0)=e^0=1$, yields the required identity $X_t=1+\int_0^t X_s\,dB_s$. The integrand is progressively measurable because $X$ is continuous and adapted. It is in $\mathcal{L}^2$ on every finite interval: the normal moment formula gives $\mathbb{E}[X_s^2]=\mathbb{E}[e^{2B_s-s}]=e^{-s}e^{2s}=e^s$, and $\int_0^T e^s\,ds=e^T-1<\infty$.

As an independent check, the integral equation implies $\mathbb{E}[X_t]=1$, since its Itô integral has mean zero. The candidate gives the same answer: $\mathbb{E}[e^{B_t-t/2}]=e^{-t/2}e^{t/2}=1$. The initial value is $X_0=e^0=1$.

A Common Differential-Notation Pitfall

A frequent error is to apply the ordinary chain rule to $dX_t=X_t\,dB_t$ and propose $X_t=e^{B_t}$. The proposed process does not satisfy that equation. For $g(t,x)=e^x$, the derivatives are $g_t=0$, $g_x=e^x$, and $g_{xx}=e^x$. Itô's formula gives $d(e^{B_t})=\tfrac12e^{B_t}\,dt+e^{B_t}\,dB_t$. The nonzero $\tfrac12e^{B_t}\,dt$ correction shows that this candidate instead has drift $\tfrac12X_t$.

The correction is not optional: it arises because $dB_t\,dB_t=dt$, rather than from an ordinary derivative of a Brownian path. The corrected candidate $e^{B_t-t/2}$ was verified above and has no drift term. In integral language, the distinction is between $1+\int_0^t e^{B_s}\,dB_s$ and $1+\int_0^t e^{B_s-s/2}\,dB_s$; these are different equations with different integrands.

When reading or writing an SDE, check four items before manipulating it: the initial condition; the interval on which the identity is asserted; whether the stochastic integral is Itô or Stratonovich; and whether the integrands meet the stated measurability and integrability conditions. Those checks prevent differential shorthand from hiding the actual mathematical claim.

Check Your Understanding

For each question, state the interval and initial condition when writing an integral equation, and distinguish the drift integral from the Itô integral.

Write the integral equation for $dX_t=(1+t)\,dt+2\,dB_t$, $X_0=0$, on $[0,T]$.
For $H_s=s$ on $[0,3]$, compute $\mathbb{E}\int_0^3H_s^2\,ds$ and state whether $H\in\mathcal{L}^2$.
Explain why identities holding almost surely separately at each fixed time are not the same statement as one identity holding simultaneously for all times outside one null set.
Apply Itô's formula to $e^{B_t}$. Which drift term shows that it does not solve $dX_t=X_t\,dB_t$?
State the integral equation corresponding to $dX_t=b(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$ with $X_0=x$, and give the required square-integrability condition on the diffusion integrand.

Differential Notation and the Integral SDE

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