Symmetric Sums and the Stratonovich Integral

From Left-Endpoint Sums to Symmetric Sums

For an ordinary integral, left-endpoint and midpoint Riemann sums have the same limit when the integrand is sufficiently regular. With Brownian motion, the choice of sampling convention matters: the integrand can change on the same scale as the Brownian increment. Symmetric sums balance the two endpoints of each interval. Their limit is the Stratonovich integral, and the difference from the Itô integral is measured by quadratic covariation.

Throughout, 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 $T>0$. An Itô integrand $H$ on $[0,T]$ is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The Itô integral $\int_0^T H_s\,dB_s$ is defined for such integrands as an $L^2$ limit.

Take a deterministic partition $\Pi=\{0=t_0<t_1<\cdots<t_n=T\}$, with mesh $|\Pi|=\max_i(t_{i+1}-t_i)$. Write $\Delta B_i=B_{t_{i+1}}-B_{t_i}$ and $\Delta H_i=H_{t_{i+1}}-H_{t_i}$. The Itô sum uses the adapted value at the left endpoint. The symmetric sum averages the two endpoint values.

Definition: For a continuous adapted process $H$, its symmetric sum along $\Pi$ is $$ S_\Pi(H,B)=\sum_{i=0}^{n-1}\frac{H_{t_i}+H_{t_{i+1}}}{2}\,\Delta B_i. $$ When these sums converge in probability as $|\Pi|\to0$, their limit is the Stratonovich integral, denoted $\int_0^T H_s\circ dB_s$. In this tutorial we establish the limit for a class of continuous semimartingales; an arbitrary continuous adapted process need not have such a limit.

The endpoint average is the trapezoidal, or symmetric, rule. For a smooth deterministic integrand, the average of its endpoint values approximates its value at the interval midpoint. For a stochastic integrand, the average is used directly: an actual midpoint sample $H_{(t_i+t_{i+1})/2}$ is not in general the same as the endpoint average, nor is it an adapted left-endpoint sample for the entire interval.

The Covariation Correction

For continuous processes $H$ and $B$, quadratic covariation $[H,B]_T$, when it exists, is the limit in probability of $\sum_i\Delta H_i\Delta B_i$ as the mesh tends to zero. The elementary identity $$ \frac{H_{t_i}+H_{t_{i+1}}}{2}\Delta B_i =H_{t_i}\Delta B_i+\frac12\Delta H_i\Delta B_i $$ already indicates why covariation enters: the symmetric sum is the left-endpoint sum plus half the sum of the products of increments.

Theorem: Suppose $H$ has the representation $$ H_t=H_0+A_t+\int_0^t K_s\,dB_s,\qquad 0\le t\le T, $$ where $A$ is continuous, adapted, and has finite variation almost surely, $K$ is progressively measurable with $|K_s|\le C$ for a deterministic constant $C$, and $\mathbb{E}[\sup_{0\le s\le T}|H_s|^2]<\infty$. Then, along deterministic partitions whose mesh tends to zero, $$ S_\Pi(H,B)\longrightarrow \int_0^T H_s\,dB_s+\frac12\int_0^T K_s\,ds =\int_0^T H_s\,dB_s+\frac12[H,B]_T $$ in probability. Thus the symmetric sums define $\int_0^T H_s\circ dB_s$, and $[H,B]_T=\int_0^T K_s\,ds$.

Proof. First split each symmetric sum into its left-endpoint part and increment-product part: $$ S_\Pi(H,B)=\sum_i H_{t_i}\Delta B_i+\frac12\sum_i\Delta H_i\Delta B_i. $$ The left-endpoint step process $H^\Pi_s=H_{t_i}$ for $t_i\le s<t_{i+1}$ is progressively measurable. For almost every path, continuity of $H$ gives $H^\Pi_s\to H_s$ for almost every $s$. Also, $|H^\Pi_s-H_s|^2\le4\sup_{u\le T}|H_u|^2$, whose expectation is finite by hypothesis. Dominated convergence on $[0,T]\times\Omega$ gives $\mathbb{E}\int_0^T|H^\Pi_s-H_s|^2\,ds\to0$. The Itô isometry therefore gives $\sum_iH_{t_i}\Delta B_i=\int_0^T H^\Pi_s\,dB_s\to\int_0^T H_s\,dB_s$ in $L^2$.

It remains to identify the increment-product sum. The finite-variation part satisfies $$ \left|\sum_i\Delta A_i\Delta B_i\right| \le \max_i|\Delta B_i|\,\operatorname{Var}_{[0,T]}(A), $$ where $\operatorname{Var}_{[0,T]}(A)$ is the total variation of $A$ on $[0,T]$. Brownian paths are continuous, so their maximum increment over a partition tends to zero almost surely as the mesh tends to zero. The displayed bound consequently tends to zero almost surely.

Set $M_t=\int_0^tK_s\,dB_s$. On each interval $[t_i,t_{i+1}]$, Itô's product formula applied to $f(x,y)=xy$, with $f_x=y$, $f_y=x$, $f_{xx}=f_{yy}=0$, and $f_{xy}=1$, gives $$ \Delta M_i\Delta B_i =\int_{t_i}^{t_{i+1}}(B_s-B_{t_i})K_s\,dB_s +\int_{t_i}^{t_{i+1}}(M_s-M_{t_i})\,dB_s +\int_{t_i}^{t_{i+1}}K_s\,ds. $$ The last term is the covariation contribution: the mixed derivative term in the product formula is $d[M,B]_s=K_s\,ds$. To see that the first two terms vanish after summation, let $\delta_i=t_{i+1}-t_i$. The Itô isometry and $|K_s|\le C$ give $$ \mathbb{E}\left[\left(\sum_i\int_{t_i}^{t_{i+1}}(B_s-B_{t_i})K_s\,dB_s\right)^2\right] \le C^2\sum_i\int_{t_i}^{t_{i+1}}(s-t_i)\,ds \le \frac{C^2T|\Pi|}{2}. $$ For the other sum, the Itô isometry gives $\mathbb{E}|M_s-M_{t_i}|^2=\mathbb{E}\int_{t_i}^sK_u^2\,du\le C^2(s-t_i)$. Applying the same bound, $$ \mathbb{E}\left[\left(\sum_i\int_{t_i}^{t_{i+1}}(M_s-M_{t_i})\,dB_s\right)^2\right] \le\frac{C^2T|\Pi|}{2}. $$ Both sums tend to zero in $L^2$. Hence $\sum_i\Delta H_i\Delta B_i\to\int_0^T K_s\,ds$ in probability, using also the vanishing finite-variation contribution. Combining this with the left-sum limit proves the asserted formula and identifies $[H,B]_T=\int_0^T K_s\,ds$. $\square$

The theorem uses an explicit integrability assumption on the supremum of $H$ to justify convergence of the left sums. In the wider theory, the symmetric-integral relation extends to continuous semimartingales by localization and standard semimartingale convergence results. That extension does not make the theorem's stated hypotheses unnecessary for the proof just given.

Worked Examples

Worked Example: A Constant Integrand

Let $H_t=c$, where $c$ is a fixed real number. Every symmetric sum telescopes: $$ S_\Pi(c,B)=\sum_i c(B_{t_{i+1}}-B_{t_i})=c(B_T-B_0)=cB_T. $$ The Itô integral is also $cB_T$, since the constant integrand is progressively measurable and $\mathbb{E}\int_0^T c^2\,ds=c^2T<\infty$. Here $K=0$, so $[H,B]_T=0$, and the theorem predicts no correction.

Check: substituting the values into the covariation formula gives $\int_0^T c\,dB_s+\tfrac12\int_0^T0\,ds=cB_T+0$, exactly the telescoping sum.

Worked Example: The Brownian Integrand

Take $H_t=B_t$. Brownian motion is a continuous semimartingale with $A=0$ and $K=1$; on a finite interval it also satisfies $\mathbb{E}[\sup_{s\le T}|B_s|^2]<\infty$. Each symmetric sum is exactly $$ S_\Pi(B,B)=\frac12\sum_i(B_{t_i}+B_{t_{i+1}})(B_{t_{i+1}}-B_{t_i}) =\frac12\sum_i(B_{t_{i+1}}^2-B_{t_i}^2)=\frac12 B_T^2. $$ Thus $\int_0^T B_s\circ dB_s=\tfrac12B_T^2$. The theorem says the Itô integral is $\tfrac12B_T^2-\tfrac12T$, because $[B,B]_T=T$.

Check: adding back half the covariation gives $\tfrac12(B_T^2-T)+\tfrac12T=\tfrac12B_T^2$, agreeing with the exact telescoping sum. Its expectation is $T/2$, since $\mathbb{E}[B_T^2]=T$; the Itô integral has expectation zero, also consistent with $\mathbb{E}[\tfrac12(B_T^2-T)]=0$.

Worked Example: A Time Trend Added to Brownian Motion

Let $H_t=t+B_t$ on $[0,2]$. This has $A_t=t$, $K=1$, and finite expected squared supremum. The theorem gives $$ \int_0^2(t+B_t)\circ dB_t =\int_0^2(t+B_t)\,dB_t+1. $$ To make the terms explicit, the Itô integral splits into $\int_0^2t\,dB_t+\int_0^2B_t\,dB_t$. For the deterministic part, summation by parts on left sums and passage to the limit gives $\int_0^2t\,dB_t=2B_2-\int_0^2B_s\,ds$. The second part, as in the preceding example, is $\tfrac12(B_2^2-2)$. Therefore $$ \int_0^2(t+B_t)\circ dB_t =2B_2-\int_0^2B_s\,ds+\frac12B_2^2. $$

Check: the correction is half of $[t+B,B]_2=2$, hence $1$, as displayed. The random variable $2B_2-\int_0^2B_s\,ds$ has mean zero: both terms have mean zero. Since $\mathbb{E}[B_2^2]=2$, the symmetric integral has mean $1$, matching the expectation of its Itô integral plus the correction.

Why the Convention Matters

The conversion formula is useful because the two integrals encode different sampling conventions. An Itô integral uses adapted left-endpoint values and has the martingale properties developed for that integral. A Stratonovich integral uses symmetric sums and, for the processes in the theorem, differs by half the covariation. The quadratic-variation rule $dB_t\,dB_t=dt$ is the reason that the Brownian integrand in the second example produces a nonzero correction. The other multiplication rules are $dB_t\,dt=0$ and $dt\,dt=0$; they do not replace the covariation term.

Common pitfall: dropping the increment product. A tempting but incorrect calculation is to identify the symmetric sum with the left-endpoint sum because the endpoint average “approaches” the left value. For $H=B$, that would give $\int_0^T B_s\circ dB_s=\int_0^T B_s\,dB_s=\tfrac12(B_T^2-T)$, omitting half the covariation. The correction is not negligible: the average-minus-left difference on interval $i$ is $\tfrac12\Delta B_i$, and multiplying by $\Delta B_i$ gives $\tfrac12(\Delta B_i)^2$. Summing these terms converges to $T/2$, not zero.

Nor does continuity alone guarantee a Stratonovich integral. The definition requires convergence of the symmetric sums; the theorem supplies it for a stated semimartingale class. For more general continuous semimartingales, the standard extension uses localization and the semimartingale covariation theorem. It should not be inferred for every continuous adapted process merely from the endpoint-average notation.

Check Your Understanding

Use the increment identity and the covariation relation to check each answer.

For $H_t=3$, compute the symmetric sum on any partition of $[0,T]$ and identify its covariation with $B$.
For $H_t=B_t$, explain why the sum $\sum_i\Delta H_i\Delta B_i$ converges to $T$.
If $H_t=t+B_t$ on $[0,T]$, what is $[H,B]_T$, and what correction is added to the Itô integral?
Why is an actual midpoint value $H_{(t_i+t_{i+1})/2}$ not automatically the same as the symmetric endpoint average?
Which integrability hypothesis in the theorem justifies convergence of the left-endpoint sums in $L^2$?

Symmetric Sums and the Stratonovich Integral

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