Associativity of Semimartingale Integration

Why Associativity Matters

Many stochastic calculations produce an integral whose integrator is itself an integral. Associativity says that, when the relevant integrability conditions hold, the inner and outer integrations can be combined: integrating $K$ against $H\cdot X$ gives the same process as integrating the product $KH$ against $X$. The rule is useful for simplifying nested expressions, but it is not an ordinary algebraic cancellation: the integrals must exist, and the product integrand must be integrable.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ satisfying the usual conditions and carrying a Brownian motion $B_t$. In this tutorial, $X$ is a continuous semimartingale written $X=X_0+M+A$, where $M$ is a continuous local martingale with $M_0=0$, and $A$ is a continuous adapted finite-variation process with $A_0=0$. Integrands for general semimartingale integrals are predictable. For Brownian Itô integrals, we use progressively measurable integrands in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ on each stated finite horizon; these have predictable versions for the integrals used below.

Definition: For $X=X_0+M+A$, a predictable process $H$ is integrable with respect to $X$ on finite time intervals if, almost surely, $\int_0^T H_s^2\,d[M]_s<\infty$ and $\int_0^T |H_s|\,d|A|_s<\infty$ for every finite $T$, locally if these quantities become finite after stopping. Here $[M]$ is the quadratic variation of $M$, and $|A|$ is the total variation of $A$. The integral is $H\cdot X=H\cdot M+H\cdot A$.

The local version allows the displayed integrals to be finite only after an increasing sequence of stopping times tending to infinity. This formulation is sufficient for the theorem below; its conditions make explicit which parts of the decomposition control the martingale and finite-variation integrals. In particular, $[B]_t=t$, so a Brownian Itô integral $H\cdot B$ is locally defined when $\int_0^T H_s^2\,ds$ is locally finite. When this expectation is finite, $H$ belongs to $\mathcal{L}^2$.

The Associativity Theorem

Theorem: Let $X=X_0+M+A$ be a continuous semimartingale as above, and let $H$ be predictable and integrable with respect to $X$. Set $Y=H\cdot X$. Suppose $K$ is predictable and integrable with respect to $Y$, that is, locally $$ \int_0^T K_s^2\,d[H\cdot M]_s<\infty, \qquad \int_0^T |K_s|\,d|H\cdot A|_s<\infty. $$ Then $KH$ is integrable with respect to $X$, and, up to indistinguishability, $$ K\cdot(H\cdot X)=(KH)\cdot X. $$

In these conditions $H\cdot M$ is the local martingale part of $Y$, and $H\cdot A$ is its finite-variation part. The quadratic-variation formula for stochastic integrals gives $[H\cdot M]_t=\int_0^t H_s^2\,d[M]_s$. This is a pathwise identity of quadratic-variation processes; it is not the Itô isometry, which instead gives an identity of expectations for square-integrable integrals.

Proof. First consider the local martingale part. The quadratic-variation formula just stated yields $$ \int_0^T (K_sH_s)^2\,d[M]_s =\int_0^T K_s^2\,d[H\cdot M]_s. $$ Thus the assumed local finiteness for integration against the martingale part of $Y$ is exactly the local square-integrability condition for $KH$ against $M$. For the finite-variation part, the signed Stieltjes measure associated with $H\cdot A$ is $H_s\,dA_s$, whose total variation is $|H_s|\,d|A|_s$. Consequently, $$ \int_0^T |K_sH_s|\,d|A|_s =\int_0^T |K_s|\,d|H\cdot A|_s<\infty $$ locally. Both conditions for $KH$ to be integrable with respect to $X$ follow.

It remains to prove equality of the integrals. For the finite-variation part, integration is Lebesgue–Stieltjes integration against the signed measure $dA$. For a simple predictable $K$, splitting the time interval into its defining pieces gives $\int K\,d(H\cdot A)=\int KH\,dA$. General integrable $K$ follows by the defining extension of the Lebesgue integral, since $\int |K|\,d|H\cdot A|=\int |KH|\,d|A|$ is finite locally.

For the local martingale part, first take a simple predictable $K$, constant on each of finitely many time intervals with its value on an interval measurable at that interval’s left endpoint. On each interval, the increment of $K\cdot(H\cdot M)$ is the corresponding value of $K$ times the increment of $H\cdot M$. By the definition of $H\cdot M$, this equals the increment of $(KH)\cdot M$. For a general predictable $K$, localize so the relevant quadratic-variation integrals have finite expectation, then approximate $K$ by simple predictable processes in $L^2$ for the measure $d[H\cdot M]$. The Itô isometry makes the corresponding integrals converge in $L^2(\mathbb{P})$. The displayed quadratic-variation identity shows that the products of the approximants with $H$ converge in $L^2$ for $d[M]$ as well. The simple-integrand identity therefore passes to the limit. Removing the localization proves equality of the local martingale parts. Combining the two parts proves the claimed identity. $\square$

The proof separates the two mechanisms: the quadratic variation controls the local martingale integral, while total variation controls the finite-variation integral. In particular, associativity does not replace Itô multiplication rules: for Brownian motion, $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. Those rules are already reflected in the semimartingale integrals being combined.

Nested Brownian Integrals

Worked Example: Integrating Brownian Motion Against Its Own Integral

On a finite interval $[0,T]$, let $Y_t=\int_0^t B_s\,dB_s$. Compute $\int_0^T B_s\,dY_s$ using associativity, and check the result by another calculation.

Here $X=B$, $H_s=B_s$, and $K_s=B_s$. Brownian motion is adapted and continuous, hence progressively measurable. The needed integrability follows from $$ \mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=\frac{T^2}{2}<\infty, \qquad \mathbb{E}\int_0^T B_s^4\,ds=\int_0^T 3s^2\,ds=T^3<\infty. $$ The second equality uses the fourth moment of a centered normal variable with variance $s$. Thus both the inner integral and the outer Brownian integral are square-integrable martingales, and the theorem gives $$ \int_0^T B_s\,dY_s=\int_0^T B_s^2\,dB_s. $$

For an independent check, apply Itô’s formula to $f(x)=x^2$: $f'(x)=2x$, $f''(x)=2$, and the correction term is $\tfrac12 f''(B_s)\,ds=ds$. Hence $B_t^2=2\int_0^t B_s\,dB_s+t$, so $Y_t=(B_t^2-t)/2$ and $dY_t=B_t\,dB_t$, consistent with the definition. Apply Itô’s formula again to $f(x)=x^3$: $f'(x)=3x^2$, $f''(x)=6x$, and the correction term is $\tfrac12 f''(B_s)\,ds=3B_s\,ds$. It follows that $$ \int_0^T B_s^2\,dB_s=\frac{B_T^3}{3}-\int_0^T B_s\,ds. $$

Check: Both representations have mean zero. For the integral representation, the integrand $B_s^2$ is in $\mathcal{L}^2$, so the integral is a square-integrable martingale and has expectation zero. In the second representation, $\mathbb{E}[B_T^3]=0$ and $\mathbb{E}\int_0^T B_s\,ds=\int_0^T\mathbb{E}[B_s]\,ds=0$, giving the same answer.

Worked Example: Deterministic Weights in a Brownian Integral

For $T=1$, set $Y_t=\int_0^t s\,dB_s$ and compute $\int_0^1 s^2\,dY_s$, its mean and its variance.

The inner integrand $H_s=s$ and outer integrand $K_s=s^2$ are deterministic, adapted and progressively measurable. The theorem gives $$ \int_0^1 s^2\,dY_s=\int_0^1 s^3\,dB_s. $$ The product integrand is square-integrable because $\mathbb{E}\int_0^1 s^6\,ds=\tfrac17$. The integral is a square-integrable martingale, so its expectation is zero; its variance follows from the Itô isometry: $$ \mathbb{E}\left[\left(\int_0^1 s^3\,dB_s\right)^2\right] =\mathbb{E}\int_0^1 s^6\,ds =\left[\frac{s^7}{7}\right]_0^1 =\frac17. $$

Check: The nested-integral conditions give the same variance: $[Y]_t=\int_0^t s^2\,ds$, so $\mathbb{E}\int_0^1 s^4\,d[Y]_s=\int_0^1 s^4s^2\,ds=\tfrac17$. The bracket identity is the quadratic-variation formula for stochastic integrals; the expectation calculation is the separate application of the Itô isometry.

A Mixed Brownian and Finite-Variation Example

Worked Example: A Brownian Integral Plus a Time Integral

Let $X_t=B_t+t$ on $[0,2]$, $H_s=s$, and $K_s=s$. Find $\int_0^2 K_s\,dY_s$, where $Y=H\cdot X$, and compute its mean and variance.

The semimartingale decomposition is $M=B$, $A_t=t$, so $$ Y_t=\int_0^t s\,dB_s+\int_0^t s\,ds =\int_0^t s\,dB_s+\frac{t^2}{2}. $$ The inner integrability requirements hold: $\int_0^2 s^2\,ds=8/3<\infty$ for the Brownian part, and $\int_0^2 s\,ds=2<\infty$ for the finite-variation part. For the outer integral, the corresponding conditions are $\int_0^2 s^4\,ds=32/5<\infty$ and $\int_0^2 s^2\,ds=8/3<\infty$. Associativity therefore gives $$ \int_0^2 s\,dY_s=\int_0^2 s^2\,dB_s+\int_0^2 s^2\,ds =\int_0^2 s^2\,dB_s+\frac83. $$

The Brownian term is a square-integrable martingale with mean zero. By the Itô isometry its variance is $$ \mathbb{E}\left[\left(\int_0^2 s^2\,dB_s\right)^2\right] =\int_0^2 s^4\,ds =\left[\frac{s^5}{5}\right]_0^2 =\frac{32}{5}. $$ Adding the deterministic constant $8/3$ gives mean $8/3$ and variance $32/5$. Check: the right-hand side separates into a zero-mean Brownian integral and a deterministic finite-variation integral, so its mean is exactly $0+8/3$, while adding a constant leaves the variance unchanged.

A Common Mistake: Using the Ordinary Chain Rule

A tempting but incorrect calculation is $\int_0^T B_s\,dB_s=B_T^2/2$, obtained by applying the ordinary chain rule to $B_t^2$. It omits Brownian quadratic variation. The correct Itô calculation has $f(x)=x^2$, $f'(x)=2x$, $f''(x)=2$, and the correction term $\tfrac12 f''(B_t)\,dt=dt$. Thus $$ B_T^2=2\int_0^T B_s\,dB_s+T, \qquad \int_0^T B_s\,dB_s=\frac{B_T^2-T}{2}. $$ The discrepancy is $T/2$, not zero. Associativity combines already-defined Itô semimartingale integrals; it does not license replacing them with ordinary integrals or dropping the correction associated with $dB_t\,dB_t=dt$.

The theorem also concerns Itô semimartingale integration, not a switch to Stratonovich integration. Stratonovich integrals use a different convention and correction term. Keep the chosen integral convention fixed throughout a calculation.

Takeaway: To use associativity, identify the martingale and finite-variation parts of the inner integral, verify outer integrability against both parts, and check that $KH$ satisfies the corresponding conditions for $X$. Then the two sides agree up to indistinguishability.

Check Your Understanding

For each question, state any integrability condition you use and distinguish a pathwise bracket identity from an expectation identity.

For $Y=H\cdot X$, which two local finiteness conditions ensure that $K$ is integrable with respect to $Y$?
If $X=B$, $H_s=s$, and $K_s=s^2$ on $[0,T]$, what product integrand appears on the right-hand side of the associativity identity?
Explain why $[H\cdot M]_t=\int_0^t H_s^2\,d[M]_s$ is not itself the Itô isometry.
For the mixed example with $X_t=B_t+t$ and $H_s=K_s=s$, identify the local martingale and finite-variation terms of the combined integral.
Use Itô’s formula for $f(x)=x^2$ to identify the term missing from the ordinary-chain-rule calculation of $\int_0^T B_s\,dB_s$.