Two Limits from the Same Brownian Increments
In a fine partition sum, a Brownian increment contributes in two distinct ways. A first-order sum, with an adapted weight multiplying each increment, approaches an Itô integral. A second-order sum, with a weight multiplying each squared increment, approaches an ordinary time integral. The second limit arises because the conditional mean of a squared increment is the length of its time interval; the fluctuations around that mean cancel across the partition in mean square.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). In particular, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Fix \(T>0\). A partition \(\pi\) of \([0,T]\) is \(0=t_0<t_1<\cdots<t_n=T\), with mesh \(|\pi|=\max_i(t_i-t_{i-1})\). Write \(\Delta t_i=t_i-t_{i-1}\) and \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\).
We will use continuous adapted weights \(H\) and \(G\) satisfying \(\mathbb{E}[\sup_{0\le s\le T}|H_s|^2]<\infty\) and \(\mathbb{E}[\sup_{0\le s\le T}|G_s|^2]<\infty\), respectively. These assumptions ensure the needed integrability and that the left-endpoint approximations converge to their weights.
First-Order Sums Give the Itô Integral
Proof. Define the left-endpoint step process \(H^\pi_s=H_{t_{i-1}}\) for \(t_{i-1}\le s<t_i\), with its value at \(T\) immaterial to the integral. It is progressively measurable and belongs to \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^T(H^\pi_s)^2ds\le T\mathbb{E}[\sup_{s\le T}|H_s|^2]<\infty\). By the definition of the Itô integral for step processes, \(\int_0^T H^\pi_s\,dB_s=\sum_i H_{t_{i-1}}\Delta B_i\).
For almost every path, continuity on the compact interval implies that the maximum difference between \(H^\pi_s\) and \(H_s\) tends to zero as \(|\pi|\to0\). Also, \(\int_0^T|H^\pi_s-H_s|^2ds\le4T\sup_{s\le T}|H_s|^2\), whose expectation is finite. Dominated convergence therefore gives \(\mathbb{E}\int_0^T|H^\pi_s-H_s|^2ds\to0\). Apply the Itô isometry to the difference of the integrals: $$ \mathbb{E}\left[\left(\sum_iH_{t_{i-1}}\Delta B_i-\int_0^T H_s\,dB_s\right)^2\right] =\mathbb{E}\int_0^T|H^\pi_s-H_s|^2ds\longrightarrow0. $$ This is the claimed \(L^2\) convergence. \(\square\)
Worked Example: The Weight \(H_t=t\)
On \([0,1]\), take \(H_t=t\) and divide the interval into \(n\) equal subintervals. The left-endpoint sum is $$ S_n=\sum_{i=1}^n\frac{i-1}{n}\left(B_{i/n}-B_{(i-1)/n}\right). $$ The theorem applies because \(H\) is deterministic and continuous, and \(\sup_{0\le t\le1}|H_t|^2=1\). Thus \(S_n\to\int_0^1t\,dB_t\) in \(L^2\). The limiting integral has mean zero and, by the Itô isometry, variance \(\int_0^1t^2dt=1/3\).
For the two-interval partition, \(S_2=\tfrac12(B_1-B_{1/2})\). Its mean is zero and its variance is \(\tfrac14\cdot\tfrac12=\tfrac18\), since \(B_1-B_{1/2}\sim N(0,\tfrac12)\). The mean-square error can be computed directly by the isometry: $$ \mathbb{E}\left[\left(S_2-\int_0^1t\,dB_t\right)^2\right] =\int_0^{1/2}t^2dt+\int_{1/2}^1(t-\tfrac12)^2dt =\frac1{24}+\frac1{24}=\frac1{12}. $$ Check: the step weight is \(0\) on the first interval and \(\tfrac12\) on the second, so its squared distance from \(t\) has exactly the two integrals displayed. As the partition is refined, this distance tends to zero, as required by the theorem.
Second-Order Sums Give an Integral Against Time
Proof. Separate the sum into its centered fluctuation and its conditional-mean part: $$ \sum_iG_{t_{i-1}}(\Delta B_i)^2 = \sum_iG_{t_{i-1}}\big((\Delta B_i)^2-\Delta t_i\big) +\sum_iG_{t_{i-1}}\Delta t_i. $$ The second sum is a left-endpoint Riemann sum. For a continuous path, its difference from \(\int_0^T G_sds\) is bounded in absolute value by \(T\omega_G(|\pi|)\), where \(\omega_G(\delta)=\sup\{|G_t-G_s|:s,t\in[0,T],\,|t-s|\le\delta\}\). This modulus tends to zero almost surely. The difference is also bounded by \(2T\sup_{s\le T}|G_s|\), which is square-integrable. Dominated convergence shows that the Riemann-sum error tends to zero in \(L^2\).
It remains to control the centered sum. Set \(\xi_i=(\Delta B_i)^2-\Delta t_i\). Independence of \(\Delta B_i\) from \(\mathcal{F}_{t_{i-1}}\), together with the second and fourth moments of a centered normal variable, gives $$ \mathbb{E}[\xi_i\mid\mathcal{F}_{t_{i-1}}]=0,\qquad \mathbb{E}[\xi_i^2\mid\mathcal{F}_{t_{i-1}}]=2(\Delta t_i)^2. $$ For \(i<j\), \(G_{t_{i-1}}\xi_i\) is \(\mathcal{F}_{t_{j-1}}\)-measurable, whereas \(\mathbb{E}[G_{t_{j-1}}\xi_j\mid\mathcal{F}_{t_{j-1}}]=0\). Consequently the distinct terms in the centered sum have zero pairwise expectations. Its second moment is therefore $$ \mathbb{E}\left[\left(\sum_iG_{t_{i-1}}\xi_i\right)^2\right] =\sum_i\mathbb{E}[G_{t_{i-1}}^2\xi_i^2] =2\sum_i\mathbb{E}[G_{t_{i-1}}^2](\Delta t_i)^2 \le2|\pi|T\,\mathbb{E}\left[\sup_{s\le T}|G_s|^2\right]. $$ The bound tends to zero. Both parts of the decomposition thus converge in \(L^2\) to the asserted limit. \(\square\)
The conditional-mean calculation explains the distinction between the two limits: \(\mathbb{E}[(\Delta B_i)^2\mid\mathcal{F}_{t_{i-1}}]=\Delta t_i\), while \(\mathbb{E}[\Delta B_i\mid\mathcal{F}_{t_{i-1}}]=0\). In shorthand, \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\) encode rules for limits of sums; they are not term-by-term identities for finite increments. The second-order theorem concerns a weighted sum of squared Brownian increments, not a Stratonovich integral or the ordinary chain rule.
Worked Example: A Deterministic Weight \(G_t=t\)
On \([0,1]\), let \(G_t=t\), and use \(n\) equal intervals. The weighted squared-increment sum is $$ Q_n=\sum_{i=1}^n\frac{i-1}{n}\left(B_{i/n}-B_{(i-1)/n}\right)^2. $$ The theorem gives \(Q_n\to\int_0^1t\,dt=\tfrac12\) in \(L^2\). Its expectation and variance can also be calculated at each finite \(n\). Each increment is independent of the others, with mean zero and variance \(1/n\); its squared value has mean \(1/n\) and variance \(2/n^2\). Hence $$ \mathbb{E}[Q_n]=\frac1n\sum_{i=1}^n\frac{i-1}{n}=\frac{n-1}{2n}, \qquad \operatorname{Var}(Q_n)=\frac{2}{n^2}\sum_{i=1}^n\left(\frac{i-1}{n}\right)^2 =\frac{(n-1)(2n-1)}{3n^3}. $$ The expectation tends to \(\tfrac12\) and the variance tends to zero.
For \(n=2\), the sum is \(Q_2=\tfrac12(B_1-B_{1/2})^2\). Its expectation is \(\tfrac12\cdot\tfrac12=\tfrac14\), and its variance is \((\tfrac12)^2\cdot2(\tfrac12)^2=\tfrac18\). Substituting \(n=2\) into the general expressions gives \((2-1)/(2\cdot2)=\tfrac14\) and \((2-1)(4-1)/(3\cdot2^3)=\tfrac18\), respectively. These finite-partition values need not equal the limit; convergence concerns increasingly fine partitions.
Worked Example: A Brownian Weight \(G_t=B_t\)
On \([0,1]\), \(G_t=B_t\) is adapted and continuous. It satisfies the theorem's integrability condition: the standard Brownian maximal inequality gives \(\mathbb{E}[\sup_{s\le1}|B_s|^2]\le4\mathbb{E}[B_1^2]=4\). Thus $$ \sum_iB_{t_{i-1}}(\Delta B_i)^2\longrightarrow\int_0^1B_s\,ds \quad\text{in }L^2. $$ For the partition \(0<\tfrac12<1\), the first weight is \(B_0=0\), so the sum is \(B_{1/2}(B_1-B_{1/2})^2\). Its expectation is zero because the two factors are independent and \(B_{1/2}\) is centered.
As a check, its second moment is $$ \mathbb{E}[B_{1/2}^2]\mathbb{E}[(B_1-B_{1/2})^4] =\frac12\cdot3\left(\frac12\right)^2=\frac38. $$ The limiting time integral also has mean zero. Its variance is $$ \mathbb{E}\left[\left(\int_0^1B_s\,ds\right)^2\right] =\int_0^1\int_0^1\mathbb{E}[B_sB_t]\,ds\,dt =\int_0^1\int_0^1\min(s,t)\,ds\,dt =2\int_0^1\int_0^t s\,ds\,dt =\frac13. $$ The finite sum's second moment \(\tfrac38\) is not yet the limiting value; the theorem says the sums converge in \(L^2\) as the mesh decreases, not that one coarse sum already has the limit's moments.
A Common Mistake: Replacing Each Square by Its Mean
It is incorrect to replace every \((\Delta B_i)^2\) in a finite sum by \(\Delta t_i\) as if the two were equal. For example, on the two-interval partition in the preceding deterministic-weight example, this mistaken substitution gives \(\tfrac12\cdot\tfrac12=\tfrac14\). The actual sum is \(\tfrac12(B_1-B_{1/2})^2\), a random variable with variance \(\tfrac18\), not the constant \(\tfrac14\).
The correction is to use the conditional expectation and control the centered terms across the whole partition. Each squared increment has mean \(\Delta t_i\), but a single squared increment fluctuates. The proof shows that after multiplying by the adapted weights and summing, the total centered fluctuation tends to zero in \(L^2\). Likewise, first-order sums converge to Itô integrals because their step-integrand approximations converge in the norm specified by the Itô isometry.
Check Your Understanding
Use the stated hypotheses when applying either convergence result; for conditional expectations, remember that the Brownian increment is independent of the information available at the left endpoint.
- For a continuous adapted \(H\) with \(\mathbb{E}[\sup_{s\le T}|H_s|^2]<\infty\), what quantity does \(\sum_iH_{t_{i-1}}\Delta B_i\) approach, and in what mode of convergence?
- Why do the centered terms \(G_{t_{i-1}}((\Delta B_i)^2-\Delta t_i)\) at distinct intervals have zero pairwise expectation?
- For \(G_t=t\) on \([0,1]\), calculate the expectation and variance of the weighted squared-increment sum on four equal intervals.
- Explain why \((\Delta B_i)^2\) cannot be replaced by \(\Delta t_i\) in each term of a finite partition sum.
- What integrability and path regularity assumptions in the second-order theorem ensure that its Riemann-sum part converges in \(L^2\)?