Brownian Left and Right Sums Expose Riemann-Stieltjes Failure

Two Endpoint Choices, Two Limits

For a smooth integrator, left- and right-endpoint Riemann sums usually converge to the same integral. Brownian motion presents a sharp contrast: when the integrand and integrator are both the same Brownian path, the difference between these sums is the sum of squared increments. That sum does not vanish as the partition is refined. The resulting discrepancy shows that the ordinary tagged-partition Riemann–Stieltjes integral cannot be defined pathwise for Brownian motion integrated against itself.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ with respect to the filtration. Fix $T>0$. A deterministic partition of $[0,T]$ is $0=t_0<t_1<\cdots<t_n=T$; write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and $|\Pi|=\max_i\Delta t_i$ for its mesh.

Definition: The left-endpoint sum and right-endpoint sum for the integrand $B$ against the path $B$ on $[0,T]$ are $$ L_\Pi=\sum_{i=1}^n B_{t_{i-1}}\Delta B_i, \qquad R_\Pi=\sum_{i=1}^n B_{t_i}\Delta B_i. $$ The quadratic variation sum is $Q_\Pi=\sum_{i=1}^n(\Delta B_i)^2$. These are finite sums, not assumed Riemann–Stieltjes or stochastic integrals.

For each interval, $B_{t_i}=B_{t_{i-1}}+\Delta B_i$. Multiplying this identity by $\Delta B_i$ gives the basic relation between the two endpoint choices. Its consequence depends on the behavior of $Q_\Pi$ as the mesh tends to zero.

The Exact Finite-Partition Identity

Proposition: For every continuous path $x$ and every partition of $[0,T]$, if $\Delta x_i=x(t_i)-x(t_{i-1})$ and the endpoint sums are formed with integrand $x$ and integrator $x$, then $$ R_\Pi-L_\Pi=\sum_{i=1}^n(\Delta x_i)^2, \qquad L_\Pi=\frac12\left(x(T)^2-x(0)^2-\sum_{i=1}^n(\Delta x_i)^2\right), \qquad R_\Pi=\frac12\left(x(T)^2-x(0)^2+\sum_{i=1}^n(\Delta x_i)^2\right). $$

Proof. On interval $i$, the right-endpoint summand minus the left-endpoint summand is $x(t_i)\Delta x_i-x(t_{i-1})\Delta x_i=(x(t_i)-x(t_{i-1}))\Delta x_i=(\Delta x_i)^2$. Summing proves the first identity. Also, $x(t_i)^2-x(t_{i-1})^2=2x(t_{i-1})\Delta x_i+(\Delta x_i)^2$. Summing this equality over $i$ telescopes the left side to $x(T)^2-x(0)^2$, and gives the stated formula for $L_\Pi$. Replacing $x(t_{i-1})$ by $x(t_i)=x(t_{i-1})+\Delta x_i$ in each summand gives the formula for $R_\Pi$. These algebraic identities also cover zero increments and partitions with any finite number of intervals. $\square$

For Brownian motion, $B_0=0$, so the identities reduce to $$ L_\Pi=\frac12(B_T^2-Q_\Pi), \qquad R_\Pi=\frac12(B_T^2+Q_\Pi), \qquad R_\Pi-L_\Pi=Q_\Pi. $$ They hold path by path for every partition. No limiting argument is needed for these equalities.

Limits Along Deterministic Partitions

For Brownian motion, the increments over disjoint partition intervals are independent and $\Delta B_i$ has distribution $N(0,\Delta t_i)$. Thus $\mathbb{E}[(\Delta B_i)^2]=\Delta t_i$ and $\operatorname{Var}((\Delta B_i)^2)=2(\Delta t_i)^2$. Independence of the increments gives $\mathbb{E}[Q_\Pi]=T$ and $\operatorname{Var}(Q_\Pi)=2\sum_i(\Delta t_i)^2\le2T|\Pi|$. In particular, $Q_\Pi$ converges to $T$ in mean square as $|\Pi|\to0$.

Theorem: Along any sequence of deterministic partitions of $[0,T]$ whose mesh tends to zero, the endpoint sums satisfy $$ L_\Pi\longrightarrow\frac12(B_T^2-T), \qquad R_\Pi\longrightarrow\frac12(B_T^2+T) $$ in mean square. Their difference converges in mean square to $T$.

Proof. The partition intervals have total length $T$, so $\sum_i\Delta t_i=T$. Since $\Delta t_i\le|\Pi|$, we have $\sum_i(\Delta t_i)^2\le|\Pi|\sum_i\Delta t_i=T|\Pi|$. Using the variance formula and the mean of $Q_\Pi$, $$ \mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)=2\sum_i(\Delta t_i)^2\le2T|\Pi|\longrightarrow0. $$ The finite-partition identities show that $L_\Pi-\tfrac12(B_T^2-T)=-\tfrac12(Q_\Pi-T)$ and $R_\Pi-\tfrac12(B_T^2+T)=\tfrac12(Q_\Pi-T)$. The mean squares of both errors are therefore $\tfrac14\mathbb{E}[(Q_\Pi-T)^2]$, which tend to zero. Subtracting the two identities gives $R_\Pi-L_\Pi=Q_\Pi$, whose mean-square limit is $T$. $\square$

The limits are not equal: their difference is the positive elapsed time $T$. They are not two versions of a single ordinary integral obtained by changing endpoint convention. The endpoint choice changes the limit by a nonvanishing quantity.

Worked Example: Four Equal Intervals on [0, 1]

Take $t_i=i/4$ for $i=0,1,2,3,4$. Then $\Delta t_i=1/4$, and the sums are $L_\Pi=B_0(B_{1/4}-B_0)+B_{1/4}(B_{1/2}-B_{1/4})+B_{1/2}(B_{3/4}-B_{1/2})+B_{3/4}(B_1-B_{3/4})$ and $R_\Pi=B_{1/4}(B_{1/4}-B_0)+B_{1/2}(B_{1/2}-B_{1/4})+B_{3/4}(B_{3/4}-B_{1/2})+B_1(B_1-B_{3/4})$. The first term of $L_\Pi$ is zero since $B_0=0$. For each interval, the right summand is the left summand plus the squared increment. Thus $R_\Pi-L_\Pi=(B_{1/4}-B_0)^2+(B_{1/2}-B_{1/4})^2+(B_{3/4}-B_{1/2})^2+(B_1-B_{3/4})^2=Q_\Pi$.

Here $\mathbb{E}[Q_\Pi]=4(1/4)=1$, and $\operatorname{Var}(Q_\Pi)=2\cdot4\cdot(1/4)^2=1/2$. The mean-square error from replacing $Q_\Pi$ by $1$ is therefore $1/2$. The exact sums are $L_\Pi=\tfrac12(B_1^2-Q_\Pi)$ and $R_\Pi=\tfrac12(B_1^2+Q_\Pi)$. Check: subtracting these expressions gives $Q_\Pi$, as the interval-by-interval calculation did. Replacing $Q_\Pi$ by its limit $1$ gives a difference of $1$, the elapsed time on $[0,1]$.

Worked Example: A Nonuniform Partition of [0, 2]

Use the partition $0<\tfrac12<2$. Its increments are $\Delta B_1=B_{1/2}$ and $\Delta B_2=B_2-B_{1/2}$. The left sum is $L_\Pi=B_0B_{1/2}+B_{1/2}(B_2-B_{1/2})=B_{1/2}(B_2-B_{1/2})$. The right sum is $R_\Pi=B_{1/2}B_{1/2}+B_2(B_2-B_{1/2})=B_{1/2}^2+B_2(B_2-B_{1/2})$. Subtracting gives $R_\Pi-L_\Pi=B_{1/2}^2+(B_2-B_{1/2})^2=Q_\Pi$.

The same identity gives $L_\Pi=\tfrac12(B_2^2-Q_\Pi)$ and $R_\Pi=\tfrac12(B_2^2+Q_\Pi)$. The interval lengths are $1/2$ and $3/2$. Hence $\mathbb{E}[Q_\Pi]=1/2+3/2=2$, and $\operatorname{Var}(Q_\Pi)=2((1/2)^2+(3/2)^2)=2(1/4+9/4)=5$. Check: the mean of the difference $R_\Pi-L_\Pi$ is $2$, exactly the total elapsed time. The nonuniform mesh affects the variance of this finite-partition difference, but as the mesh tends to zero the difference converges in mean square to $2$.

Why This Rules Out a Riemann–Stieltjes Integral

For continuous functions $f$ and $g$, a tagged Riemann–Stieltjes sum on a partition is $\sum_i f(\xi_i)(g(t_i)-g(t_{i-1}))$, where each tag $\xi_i$ lies in $[t_{i-1},t_i]$. A Riemann–Stieltjes integral exists in the tagged-partition sense if these sums approach one common finite limit as the mesh tends to zero, independently of the partitions and tags. In particular, the left- and right-endpoint choices must have the same limit along any one sequence of partitions.

To obtain an almost-sure contradiction, use dyadic partitions $t_i=iT2^{-n}$. For these partitions, $\operatorname{Var}(Q_{\Pi_n})=2T^2 2^{-n}$. For every $\varepsilon>0$, Chebyshev's inequality gives $\mathbb{P}(|Q_{\Pi_n}-T|>\varepsilon)\le 2T^2 2^{-n}/\varepsilon^2$. For fixed $\varepsilon$, the right side is summable over $n$. The Borel–Cantelli lemma implies that only finitely many of the events $\{|Q_{\Pi_n}-T|>\varepsilon\}$ occur, almost surely. Applying this for $\varepsilon=1,1/2,1/3,\ldots$ shows $Q_{\Pi_n}\to T$ almost surely.

Corollary: For each fixed $T>0$, with probability one, the left- and right-endpoint sums for $B$ against $B$ along dyadic partitions converge to different limits. Consequently, with probability one the pathwise tagged Riemann–Stieltjes integral of $B$ against itself on $[0,T]$ does not exist.

Proof. On the probability-one event where $Q_{\Pi_n}\to T$, the exact identities give $L_{\Pi_n}\to\tfrac12(B_T^2-T)$ and $R_{\Pi_n}\to\tfrac12(B_T^2+T)$. Their difference tends to $T$, which is strictly positive by the assumption $T>0$. Both endpoint choices are valid tagged sums, and their limits differ. Therefore the tagged sums cannot all converge to one common limit, as the Riemann–Stieltjes definition requires. $\square$

This conclusion is pathwise and does not claim that every possible integration procedure fails. It says that the ordinary Riemann–Stieltjes integral, which requires tag-independent convergence, is unavailable for this pair of paths. The left-endpoint sums are important in stochastic calculus, but treating their limit as an ordinary Riemann–Stieltjes integral would erase the distinction established here. No Ito or Stratonovich integral is being defined in this calculation.

Worked Example: Endpoint Sums for a Smooth Path

For contrast, take the deterministic path $x(t)=t$ on $[0,1]$, with integrand also $x$, and divide the interval into $n$ equal pieces. The left sum is $\sum_{i=1}^n\frac{i-1}{n}\frac1n=\frac1{n^2}\sum_{i=1}^n(i-1)=\frac{n(n-1)}{2n^2}=\frac{n-1}{2n}$. The right sum is $\sum_{i=1}^n\frac{i}{n}\frac1n=\frac1{n^2}\sum_{i=1}^n i=\frac{n(n+1)}{2n^2}=\frac{n+1}{2n}$. Their difference is $1/n$, which tends to zero. For $n=4$, the sums are $3/8$ and $5/8$, whose difference is $2/8=1/4$, matching $1/n$. Both converge to $1/2$, consistent with the ordinary integral of $t$ over $[0,1]$.

Check: the finite-partition identity gives the same difference directly: each increment is $1/n$, so the sum of squared increments is $n(1/n)^2=1/n$. Unlike Brownian motion, this quantity vanishes as the mesh tends to zero.

Common Mistake: Dropping the Squared Increments

A tempting but incorrect calculation is to write $B_{t_i}\Delta B_i\approx B_{t_{i-1}}\Delta B_i$ because the endpoints become close as the mesh shrinks, and then conclude that the two sums have the same limit. The exact difference is not controlled by the largest increment alone: all squared increments must be added. On $[0,T]$, that total approaches $T$, not zero.

For example, along the four-interval partition of $[0,1]$, the difference is $(\Delta B_1)^2+(\Delta B_2)^2+(\Delta B_3)^2+(\Delta B_4)^2=Q_\Pi$, not zero. Its expectation is $1$, and its variance is $1/2$. As the partition is refined, the mean-square error from $1$ tends to zero. The correction is to retain the full squared-increment sum and use its quadratic-variation limit; small individual increments do not imply that their squared sum is small.

Check Your Understanding

Use the finite-partition identities and the quadratic-variation limit to answer each question.

For any partition of $[0,T]$, derive $R_\Pi-L_\Pi=Q_\Pi$ directly from the summands.
For equal intervals on $[0,T]$, compute $\operatorname{Var}(Q_\Pi)$ when there are $n$ intervals, and state its limit as $n\to\infty$.
Write the mean-square limits of $L_\Pi$ and $R_\Pi$, and calculate their difference.
Why is it enough to compare left and right sums along the same dyadic sequence to disprove the tagged Riemann–Stieltjes integral?
For $x(t)=t$ on $[0,1]$, what are the left and right sums on $n$ equal intervals, and why do they have the same limit?