Stochastic Integration by Parts for Ito Processes

From the Product Rule to an Integral Identity

Ordinary integration by parts transfers a derivative from one factor to another. For Ito processes, the same idea works, but the product rule includes a covariation term. Integrating that rule gives an identity that can be rearranged to solve for a stochastic integral. The correction is not optional: it records the quadratic variation generated by the diffusion parts.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. All stochastic integrals below are Ito integrals. Their integrands are progressively measurable and belong to $\mathcal{L}^2$ on the time interval in question: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Here an Ito process has the form $X_t=X_0+\int_0^t a_s\,ds+\int_0^t b_s\,dB_s$, where $a$ is progressively measurable with $\mathbb{E}\int_0^T|a_s|\,ds<\infty$, and $b$ is progressively measurable and in $\mathcal{L}^2$. We impose the corresponding conditions on every process and integral used.

Definition: For $X_t=X_0+\int_0^t a_s\,ds+\int_0^t b_s\,dB_s$ and $Y_t=Y_0+\int_0^t c_s\,ds+\int_0^t d_s\,dB_s$, their covariation is $[X,Y]_t=\int_0^t b_s d_s\,ds$. This is the finite-variation correction in their Ito product rule.

For the integrated formula, assume additionally that $\mathbb{E}\int_0^T (X_s^2d_s^2+Y_s^2b_s^2)\,ds<\infty$ and $\mathbb{E}\int_0^T(|X_sc_s|+|Y_sa_s|)\,ds<\infty$. These conditions ensure that the stochastic integrals with integrands $X_sd_s$ and $Y_sb_s$ are in $\mathcal{L}^2$, and that the ordinary integrals are finite. Also, $\mathbb{E}\int_0^T|b_sd_s|\,ds<\infty$ follows from the square-integrability of $b,d$ and Cauchy–Schwarz.

The Integrated Ito Product Rule

Theorem: Under the assumptions above, for each $t\in[0,T]$, almost surely, $$ X_tY_t-X_0Y_0 =\int_0^t X_s\,dY_s+\int_0^t Y_s\,dX_s+[X,Y]_t. $$ Equivalently, in drift and diffusion coefficients, $$ X_tY_t-X_0Y_0 =\int_0^t(X_sc_s+Y_sa_s+b_sd_s)\,ds +\int_0^t(X_sd_s+Y_sb_s)\,dB_s. $$

Proof. The Ito product rule, established in the earlier material on Ito processes, states $d(X_tY_t)=X_t\,dY_t+Y_t\,dX_t+d[X,Y]_t$ for continuous Ito processes driven by the same Brownian motion. Integrate this differential identity from $0$ to $t$. The integral of the left side is $X_tY_t-X_0Y_0$, and the integral of the covariation differential is $[X,Y]_t$, since $[X,Y]_0=0$. Thus the first displayed identity follows. Substituting $dY_s=c_s\,ds+d_s\,dB_s$, $dX_s=a_s\,ds+b_s\,dB_s$, and $[X,Y]_t=\int_0^t b_sd_s\,ds$ gives the coefficient form. The stated integrability assumptions make every integral in these equalities well-defined. $\square$

This formula can be solved for either of the two Ito integrals. For example, isolating the integral against $dB_s$ in the coefficient form gives

\int_0^t(X_sd_s+Y_sb_s)\,dB_s =X_tY_t-X_0Y_0-\int_0^t(X_sc_s+Y_sa_s+b_sd_s)\,ds.

More often, one wants just one of the integrals appearing in the product rule. Rearranging the first form gives $\int_0^tX_s\,dY_s=X_tY_t-X_0Y_0-\int_0^tY_s\,dX_s-[X,Y]_t$. The correction has a plus sign in the product formula and therefore a minus sign when transferred to the other side.

Corollary: If $X,Y$ satisfy the theorem’s assumptions, then $$ \int_0^tX_s\,dY_s=X_tY_t-X_0Y_0-\int_0^tY_s\,dX_s-\int_0^t b_sd_s\,ds. $$ The integral $\int Y_s\,dX_s$ is itself the sum $\int Y_sa_s\,ds+\int Y_sb_s\,dB_s$.

Proof. Subtract $\int_0^tY_s\,dX_s+[X,Y]_t$ from both sides of the theorem’s first identity, and then use the definition $[X,Y]_t=\int_0^t b_sd_s\,ds$. The assumptions of the theorem ensure that both sides are defined. $\square$

Worked Applications

Worked Example: The Integral of Brownian Motion Against Itself

Compute $\int_0^2 B_s\,dB_s$ using integration by parts, and check the result. Here $X=Y=B$, so both drift coefficients are zero and both diffusion coefficients are $1$. The product rule gives $B_2^2-B_0^2=2\int_0^2B_s\,dB_s+[B,B]_2$. Since $B_0=0$ and $[B,B]_2=2$,

\int_0^2 B_s\,dB_s=\frac12(B_2^2-2).

For an independent check, apply Ito’s formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$, so the correction is $\tfrac12 f''(B_s)\,ds=ds$. Thus $d(B_s^2)=2B_s\,dB_s+ds$; integrating from $0$ to $2$ gives the same expression.

The Ito integral is square-integrable because $\mathbb{E}\int_0^2B_s^2\,ds=\int_0^2s\,ds=2<\infty$. Its mean is zero, as follows from the martingale property of square-integrable Ito integrals: for simple adapted integrands, independent mean-zero Brownian increments give zero conditional mean; passage to an $\mathcal{L}^2$ limit preserves that conditional expectation. The answer has mean $\tfrac12(\mathbb{E}[B_2^2]-2)=\tfrac12(2-2)=0$. Its variance also checks: $B_2\sim N(0,2)$, so $\mathbb{E}[B_2^4]=3(2)^2=12$ and $\operatorname{Var}(B_2^2)=12-2^2=8$. Therefore $\operatorname{Var}(\tfrac12(B_2^2-2))=8/4=2$, matching the Ito isometry value $\mathbb{E}\int_0^2B_s^2\,ds=2$.

Worked Example: A Deterministic Factor Times Brownian Motion

Express $\int_0^1s\,dB_s$ in terms of $B_1$ and an ordinary time integral, then verify its mean and variance. Take $X_s=s$ and $Y_s=B_s$. Their coefficients are $a_s=1,b_s=0,c_s=0,d_s=1$, and their covariation is zero. The product rule, with $X_0Y_0=0$, yields

\int_0^1s\,dB_s=B_1-\int_0^1B_s\,ds.

The stochastic integral is in $\mathcal{L}^2$, since $\mathbb{E}\int_0^1s^2\,ds=1/3$. Its mean is zero. To check the right side, $\mathbb{E}[B_1]=0$ and $\mathbb{E}[\int_0^1B_s\,ds]=\int_0^1\mathbb{E}[B_s]\,ds=0$, where interchange is justified by $\int_0^1\mathbb{E}|B_s|\,ds<\infty$.

For the variance check, $\operatorname{Var}(B_1)=1$. Also, using $\mathbb{E}[B_sB_u]=\min(s,u)$,

\operatorname{Var}\left(\int_0^1B_s\,ds\right) =\int_0^1\int_0^1\min(s,u)\,ds\,du =2\int_0^1\int_0^u s\,ds\,du =2\int_0^1\frac{u^2}{2}\,du =\frac13.

The covariance is $\operatorname{Cov}(B_1,\int_0^1B_s\,ds)=\int_0^1\mathbb{E}[B_1B_s]\,ds=\int_0^1s\,ds=\tfrac12$. Hence the variance of the difference is $1+\tfrac13-2(\tfrac12)=\tfrac13$. This agrees with the Ito isometry, which gives $\mathbb{E}\int_0^1s^2\,ds=\tfrac13$.

Worked Example: Isolating a Weighted Brownian Integral

Let $Y_t=tB_t$, and find a formula for $J=\int_0^2sB_s\,dB_s$ in terms of $B_2$ and an ordinary integral. First apply Ito’s formula to $f(t,x)=tx$. Its derivatives are $f_t=x$, $f_x=t$, and $f_{xx}=0$, so the correction $\tfrac12f_{xx}\,dt$ is zero. Therefore $dY_t=B_t\,dt+t\,dB_t$.

Now use the product rule with $X_t=B_t$ and $Y_t=tB_t$. The diffusion coefficients are $1$ and $t$, so $[B,Y]_2=\int_0^2s\,ds=2$. Also $Y_0=0$, and the two integrals on the right are $\int_0^2B_s\,dY_s=\int_0^2B_s^2\,ds+J$ and $\int_0^2Y_s\,dB_s=J$. Thus

2B_2^2=\int_0^2B_s^2\,ds+2J+2, \qquad J=B_2^2-1-\frac12\int_0^2B_s^2\,ds.

Check by applying Ito’s formula directly to $g(t,x)=tx^2$. Here $g_t=x^2$, $g_x=2tx$, and $g_{xx}=2t$, so the correction is $\tfrac12g_{xx}\,dt=t\,dt$. Consequently $d(tB_t^2)=B_t^2\,dt+2tB_t\,dB_t+t\,dt$. Integrating to $2$ gives $2B_2^2=\int_0^2B_s^2\,ds+2J+2$, exactly the identity above.

The integral defining $J$ is square-integrable because $\mathbb{E}\int_0^2s^2B_s^2\,ds=\int_0^2s^3\,ds=4<\infty$, and its mean is zero by the square-integrable Ito-integral martingale property described above. The formula also has mean zero: $\mathbb{E}[B_2^2]=2$ and $\mathbb{E}\int_0^2B_s^2\,ds=\int_0^2s\,ds=2$, so its right side has expectation $2-1-\tfrac12(2)=0$.

Why the Correction Matters

All integrals in the formulas here are Ito integrals. In differential shorthand, their correction follows $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. In particular, the product of two diffusion terms contributes to covariation; drift-diffusion and drift-drift products do not. This is different from ordinary integration by parts, where there is no quadratic-variation term. Stratonovich integrals use a different convention and have a different conversion correction; none of the displayed Ito identities silently replaces an Ito integral with a Stratonovich integral.

A common error is to apply ordinary integration by parts to $B_t$ and $tB_t$, writing $2B_2^2=\int_0^2B_s\,d(tB_s)+\int_0^2tB_s\,dB_s$. This omits the covariation term $2$. The correct formula is $2B_2^2=\int_0^2B_s\,d(tB_s)+\int_0^2tB_s\,dB_s+2$. The value $2$ is $[B,tB]_2=\int_0^2s\,ds$, not an arbitrary adjustment. Omitting it would contradict the direct Ito calculation of $d(tB_t^2)$.

The product rule also does not imply that a product of martingales is automatically a martingale. The formula is an identity of processes; to establish a martingale property for a particular term requires checking integrability and conditional expectations. For example, the Ito integrals in the examples are square-integrable martingales because their integrands satisfy the stated $\mathcal{L}^2$ condition. No such conclusion follows merely from seeing a product such as $X_tY_t$.

Check Your Understanding

Use the integrated Ito product rule and state which covariation term is present in each calculation.

For $X_t=B_t$ and $Y_t=2B_t$, express $\int_0^T B_s\,dY_s$ using $B_T$ and $T$.
Apply the product rule to $X_t=t^2$ and $Y_t=B_t$ to express $\int_0^Tt^2\,dB_t$ using $B_T$ and an ordinary integral.
For $Y_t=tB_t$, compute $[B,Y]_T$ and explain why it is not zero.
Use Ito’s formula for $f(t,x)=tx^2$ to identify the correction term, and explain how it affects an expression for $\int_0^TtB_t\,dB_t$.
What integrability condition ensures that $\int_0^T X_sd_s\,dB_s$ is a square-integrable Ito integral?

Stochastic Integration by Parts for Ito Processes

What You'll Learn