Integration by Parts for Continuous Semimartingales

Why Products Need a Covariation Term

For differentiable functions $x$ and $y$, the product rule has two terms: $d(xy)=x\,dy+y\,dx$. For continuous semimartingales, increments are rough enough that multiplying two simultaneous increments leaves a contribution in the limit. That contribution is their quadratic covariation. Integration by parts therefore has a third term, and omitting it generally gives an incorrect formula.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. A continuous semimartingale $X$ is a continuous adapted process that can be written as $X_t=X_0+M_t+A_t$, where $M$ is a continuous local martingale and $A$ is a continuous finite-variation process. A local martingale is a process that becomes a martingale after stopping along a suitable sequence of stopping times; it need not itself be a true martingale.

Definition: For continuous semimartingales $X$ and $Y$, their quadratic covariation $[X,Y]$ is the continuous finite-variation process obtained as the limit in probability of the sums $\sum_i (X_{t_i}-X_{t_{i-1}})(Y_{t_i}-Y_{t_{i-1}})$ as the partition mesh tends to zero. It is the covariation of their local martingale parts; finite-variation parts contribute zero.

All integrals below are Ito semimartingale integrals unless specified otherwise. For a general continuous semimartingale integrator, a predictable, locally bounded integrand is integrable; a continuous adapted integrand is predictable, and continuous semimartingales are locally bounded after stopping. The notation $X\mathbin{\cdot}Y$ means $\int_0^t X_s\,dY_s$, with $X$ as integrand and $Y$ as integrator. When the integrator is $B$, a progressively measurable integrand $H$ belongs to $\mathcal{L}^2$ on $[0,T]$ if $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; its Ito integral is then square-integrable. Integrands that are only locally square-integrable are handled by stopping.

The Product Formula

Theorem: If $X$ and $Y$ are continuous semimartingales, then for all $t\ge0$, almost surely, $$ X_tY_t=X_0Y_0+\int_0^t X_s\,dY_s+\int_0^t Y_s\,dX_s+[X,Y]_t. $$

Proof. Fix $t$ and take partitions $0=t_0<\cdots<t_n=t$ whose mesh tends to zero. Write $\Delta_iX=X_{t_i}-X_{t_{i-1}}$ and $\Delta_iY=Y_{t_i}-Y_{t_{i-1}}$. The exact algebraic identity $$ X_{t_i}Y_{t_i}-X_{t_{i-1}}Y_{t_{i-1}} =X_{t_{i-1}}\Delta_iY+Y_{t_{i-1}}\Delta_iX+\Delta_iX\,\Delta_iY $$ holds for each interval, including when either increment is zero. Summing it gives the difference $X_tY_t-X_0Y_0$ on the left. By the construction of semimartingale integrals, the first two sums converge in probability to $\int_0^t X_s\,dY_s$ and $\int_0^t Y_s\,dX_s$, respectively. By the definition of quadratic covariation, the third sum converges in probability to $[X,Y]_t$. The left side is unchanged by refinement, so its value equals the sum of those three limits almost surely. Applying this identity at rational times and using continuity of the processes gives the formula simultaneously for all $t\ge0$, outside one null set. $\square$

The differential notation for the same identity is $d(X_tY_t)=X_t\,dY_t+Y_t\,dX_t+d[X,Y]_t$. In Brownian calculations, $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$; these are shorthand for the corresponding quadratic covariation rules. The product formula is an Ito formula, not the ordinary chain rule and not a Stratonovich product rule.

Worked Example: The Square of Brownian Motion

Apply the product formula with $X=Y=B$. Here $B_0=0$ and $[B,B]_t=t$, so $$ B_t^2=2\int_0^t B_s\,dB_s+t,\qquad \int_0^t B_s\,dB_s=\frac12(B_t^2-t). $$

The integrand $B_s$ is progressively measurable and belongs to $\mathcal{L}^2$ on every finite interval, since $\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty$. The Ito integral is therefore square-integrable. Its conditional expectation can also be checked directly: for $0\le r\le t$, independent Brownian increments give $\mathbb{E}[B_t^2\mid\mathcal{F}_r]=B_r^2+(t-r)$. Thus $$ \mathbb{E}\left[\frac12(B_t^2-t)\,\middle|\,\mathcal{F}_r\right] =\frac12(B_r^2-r). $$ This verifies the martingale property, rather than merely inferring it from a local martingale claim.

Check: taking expectations in the displayed integral formula gives $\mathbb{E}\int_0^t B_s\,dB_s=\tfrac12(\mathbb{E}[B_t^2]-t)=\tfrac12(t-t)=0$, consistent with the conditional-expectation calculation. The nonzero term $t$ is precisely the covariation correction.

Products with Drift

If one factor has finite variation, its covariation with any continuous semimartingale is zero. This recovers the ordinary product rule for a deterministic differentiable factor, while retaining a correction when both factors have Brownian variation.

Worked Example: A Product with Opposite Brownian Terms

Let $X_t=t+B_t$ and $Y_t=t-B_t$. Their differentials are $dX_t=dt+dB_t$ and $dY_t=dt-dB_t$. Their covariation is $[X,Y]_t=-t$: the two Brownian parts have coefficients $1$ and $-1$, and the finite-variation parts contribute zero.

The integrals in the product formula are well-defined: integrals against $dt$ are ordinary finite-variation integrals, and the Brownian integrands $s$ and $B_s$ are progressively measurable and square-integrable on each finite interval. Expanding each term gives $$ \int_0^t X_s\,dY_s =\frac{t^2}{2}+\int_0^t B_s\,ds-\int_0^t s\,dB_s-\int_0^t B_s\,dB_s, $$ $$ \int_0^t Y_s\,dX_s =\frac{t^2}{2}-\int_0^t B_s\,ds+\int_0^t s\,dB_s-\int_0^t B_s\,dB_s. $$ Adding cancels the two ordinary integrals involving $B_s\,ds$ and the two integrals involving $s\,dB_s$. The remaining sum is $t^2-2\int_0^t B_s\,dB_s$. Including $X_0Y_0=0$ and $[X,Y]_t=-t$ gives $$ X_tY_t=t^2-2\int_0^t B_s\,dB_s-t=t^2-B_t^2, $$ where the last equality uses the square identity from the preceding example.

Check: directly, $(t+B_t)(t-B_t)=t^2-B_t^2$. The product-formula calculation agrees exactly; in particular, the $-t$ covariation term is required.

Reciprocals of Nonvanishing Processes

Suppose $X$ is a continuous semimartingale and $X_t\ne0$ throughout the time interval under consideration. The reciprocal is then another continuous semimartingale. Its formula includes a positive covariation correction, even though the first derivative of $x\mapsto1/x$ is negative.

Proposition: Let $X$ be a continuous semimartingale with $X_0\ne0$, and fix $T>0$. On the event that $X_t\ne0$ for every $t\in[0,T]$, for $0\le t\le T$, $$ \frac{1}{X_t}=\frac{1}{X_0}-\int_0^t \frac{1}{X_s^2}\,dX_s+\int_0^t \frac{1}{X_s^3}\,d[X]_s. $$

Proof. On the stated event, continuity and compactness imply $\inf_{0\le s\le T}|X_s|>0$. Thus $x\mapsto1/x$ is twice continuously differentiable on a neighborhood of the range of $X$ over this interval. Applying Ito's formula to $f(x)=1/x$, its derivatives are $f'(x)=-x^{-2}$ and $f''(x)=2x^{-3}$. The Ito correction is explicitly $\tfrac12 f''(X_s)\,d[X]_s=X_s^{-3}\,d[X]_s$, giving $$ d(1/X_s)=-X_s^{-2}\,dX_s+X_s^{-3}\,d[X]_s. $$ Integrating from $0$ to $t$ yields the identity. For a formal localization, stop when $|X|$ first reaches $1/n$ or $n$, apply Ito's formula while $X$ stays in that compact subset of $\mathbb{R}\setminus\{0\}$, and let $n$ increase. On the event specified in the proposition, the path on $[0,T]$ is bounded and bounded away from zero, so these stopping times eventually exceed $T$. $\square$

Worked Example: Reciprocal of a Stopped Brownian Shift

Set $X_t=2+B_t$ and $\tau=\inf\{t\ge0:|2+B_t|\le1\}$. Before and at $\tau$, $X$ stays away from zero. For $t\le\tau$, its quadratic variation is $[X]_t=t$, and the reciprocal formula gives $$ \frac{1}{2+B_t} =\frac12-\int_0^t\frac{1}{(2+B_s)^2}\,dB_s +\int_0^t\frac{1}{(2+B_s)^3}\,ds. $$

The Brownian integrand is progressively measurable and bounded in absolute value by $1$ before $\tau$; on $[0,T]$, its squared integral has expectation at most $T$. It is therefore in $\mathcal{L}^2$ when stopped at $\tau$. To verify the correction and its sign, write $R_t=1/X_t$. The reciprocal formula gives $dR_t=-X_t^{-2}dB_t+X_t^{-3}dt$. The product formula then yields $$ d(X_tR_t)=X_t\,dR_t+R_t\,dX_t+d[X,R]_t =-X_t^{-1}dB_t+X_t^{-2}dt+X_t^{-1}dB_t-X_t^{-2}dt=0, $$ because $[X,R]_t=-\int_0^t X_s^{-2}\,ds$. Since $X_0R_0=1$, this checks $X_tR_t=1$ up to $\tau$, as required.

A Common Error: Dropping Covariation

A frequent mistake is to use only the two integral terms in the product formula. For $B_t^2$, that would give $B_t^2=2\int_0^t B_s\,dB_s$, which contradicts the correctly computed identity $2\int_0^t B_s\,dB_s=B_t^2-t$ whenever $t>0$. The missing term is $[B,B]_t=t$. In the reciprocal calculation, dropping the Ito correction would likewise make the proposed differential $dR=-X^{-2}dB$; the product check would leave the nonzero remainder $-X^{-2}dt$, so $XR$ would not remain constant.

The product rule is useful because it converts products of semimartingales into integrals and covariations, and it provides a direct consistency check for derived identities. The reciprocal formula is restricted to intervals on which the process does not hit zero; at a zero, the function $1/x$ is not defined and the argument cannot be extended through that time. Ito integrals and Stratonovich integrals must also not be confused: the integrals denoted here by $dX$ are Ito integrals, and the covariation correction is part of that convention.

Check Your Understanding

For each question, specify the relevant covariation or nonvanishing condition before applying a formula.

If $X$ has finite variation and $Y$ is a continuous semimartingale, what is $[X,Y]$, and how does the product formula simplify?
Apply the product formula to $B_t$ and $t$, and identify why there is no covariation correction.
For a nonvanishing continuous semimartingale $X$, state the first and second derivatives used to obtain the formula for $1/X_t$.
Explain why the Brownian integrand in the stopped reciprocal example is square-integrable on every finite time interval.
What incorrect identity results from omitting $[B,B]_t$ when computing $B_t^2$?