Tutorials › Stochastic Calculus › Deriving the Brownian Self-Integral from Quadratic Variation

Home  /  Tutorials  /  Computing Stochastic Integrals

Computing Stochastic Integrals · Tutorial 131 of 600

Deriving the Brownian Self-Integral from Quadratic Variation

See how Brownian quadratic variation converts left-point sums into the exact self-integral identity, with the convergence error tracked explicitly.

Intermediate 10 min read

What You'll Learn

  • Define the Brownian self-integral as an Ito integral with a square-integrable adapted integrand.
  • Derive its identity from a telescoping sum of squared Brownian values.
  • Prove mean-square convergence of the quadratic-variation remainder.
  • Estimate the approximation error for specific partitions.
  • Distinguish the Ito self-integral from the corresponding Stratonovich integral.
  • Identify why applying the ordinary chain rule directly gives the wrong Ito answer.

From Left-Point Sums to the Brownian Self-Integral

The Ito integral of Brownian motion against itself is a basic example in which the choice of integration convention matters. Its value is not obtained by applying the ordinary chain rule to the square of a Brownian path: Brownian paths have nonzero quadratic variation. Instead, start with the left-point sums that define the Ito integral and use an exact algebraic identity on each partition interval. The remaining term is the quadratic-variation sum, whose convergence also gives a precise error estimate.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) adapted to \((\mathcal{F}_t)\), with increments independent of the past. Fix \(T>0\). A deterministic partition 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 \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\).

Definition: An Ito integrand on \([0,T]\) is taken here from \(\mathcal{L}^2\): the progressively measurable processes \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). For such an integrand, \(\int_0^T H_s\,dB_s\) is the Ito integral, defined as the limit in \(L^2(\mathbb{P})\) of adapted step-process integrals. The Brownian self-integral is \(\int_0^T B_s\,dB_s\).

This definition applies to \(H_s=B_s\). Brownian motion is continuous and adapted, hence progressively measurable, and \(\mathbb{E}[B_s^2]=s\). Therefore

$$ \mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=\frac{T^2}{2}<\infty. $$

For a partition \(\Pi\), let \(\eta(s)=t_{i-1}\) when \(t_{i-1}\le s<t_i\). The adapted left-point step process \(B_{\eta(s)}\) is also square-integrable. Its Ito integral is the sum \(\sum_i B_{t_{i-1}}\Delta B_i\). The central question is how these sums approach \(\int_0^T B_s\,dB_s\), and how to identify their limit.

The Exact Partition Identity

Proposition: For every partition of \([0,T]\), the left-point sum \(S_\Pi=\sum_{i=1}^n B_{t_{i-1}}\Delta B_i\) satisfies $$ S_\Pi=\frac12\left(B_T^2-Q_\Pi\right),\qquad Q_\Pi=\sum_{i=1}^n(\Delta B_i)^2. $$

Proof. For each interval, expand \(B_{t_i}=B_{t_{i-1}}+\Delta B_i\) and square both sides. This gives \(B_{t_i}^2-B_{t_{i-1}}^2=2B_{t_{i-1}}\Delta B_i+(\Delta B_i)^2\). Sum over \(i=1,\ldots,n\). The left side telescopes to \(B_T^2-B_0^2=B_T^2\), because \(B_0=0\). Rearranging the resulting equality gives the asserted identity. \(\square\)

The identity is exact for every partition; no limiting argument has been used. Its second term is the quadratic-variation sum. We next quantify its convergence to \(T\), so that the remainder in the left-point identity is controlled.

Theorem: For every deterministic partition of \([0,T]\), $$ \mathbb{E}[Q_\Pi]=T,\qquad \mathbb{E}[(Q_\Pi-T)^2]=2\sum_{i=1}^n(\Delta t_i)^2\le 2T|\Pi|. $$ Consequently \(Q_\Pi\to T\) in \(L^2(\mathbb{P})\) as \(|\Pi|\to0\).

Proof. The increments \(\Delta B_i\) are independent and each has distribution \(N(0,\Delta t_i)\). For a centered normal variable of variance \(v\), its second moment is \(v\) and its fourth moment is \(3v^2\). Thus \(\mathbb{E}[(\Delta B_i)^2]=\Delta t_i\), and \(\operatorname{Var}((\Delta B_i)^2)=3(\Delta t_i)^2-(\Delta t_i)^2=2(\Delta t_i)^2\). Summing the second moments gives \(\mathbb{E}[Q_\Pi]=\sum_i\Delta t_i=T\). Independence of the increments implies independence of their squares, so variances add: \(\operatorname{Var}(Q_\Pi)=2\sum_i(\Delta t_i)^2\). Since \(\Delta t_i\le|\Pi|\), we have \(\sum_i(\Delta t_i)^2\le|\Pi|\sum_i\Delta t_i=T|\Pi|\). Finally, the mean of \(Q_\Pi\) is \(T\), so \(\mathbb{E}[(Q_\Pi-T)^2]=\operatorname{Var}(Q_\Pi)\). The bound tends to zero with the mesh, proving the \(L^2\) convergence. \(\square\)

Identifying the Limit and Its Error

The proposition and theorem show that the partition sums have the candidate limit \(\tfrac12(B_T^2-T)\). To identify that limit with the Ito integral, we also verify that the same left-point sums converge to the integral by its defining approximation. The Ito isometry states that for \(G\in\mathcal{L}^2\), \(\mathbb{E}[(\int_0^T G_s\,dB_s)^2]=\mathbb{E}\int_0^T G_s^2\,ds\). Applied to \(G_s=B_s-B_{\eta(s)}\), it gives

$$ \mathbb{E}\left[\left(S_\Pi-\int_0^T B_s\,dB_s\right)^2\right] =\mathbb{E}\int_0^T(B_s-B_{\eta(s)})^2\,ds =\frac12\sum_{i=1}^n(\Delta t_i)^2 \le\frac12T|\Pi|. $$

For the middle equality, on each interval \(s\in[t_{i-1},t_i]\), the increment \(B_s-B_{t_{i-1}}\) has mean zero and second moment \(s-t_{i-1}\); integrating that second moment gives \(\int_{t_{i-1}}^{t_i}(s-t_{i-1})\,ds=\tfrac12(\Delta t_i)^2\). The right side tends to zero, so \(S_\Pi\to\int_0^T B_s\,dB_s\) in \(L^2\). From the exact partition identity, \(S_\Pi=\tfrac12(B_T^2-Q_\Pi)\), and the theorem gives \(S_\Pi\to\tfrac12(B_T^2-T)\) in \(L^2\) as well. Limits in \(L^2\) are unique up to almost-sure equality. Therefore:

Theorem: For every \(T>0\), the Brownian self-integral satisfies $$ \int_0^T B_s\,dB_s=\frac12(B_T^2-T)\quad\text{almost surely}. $$ Moreover, for each deterministic partition, its left-point approximation has mean-square error \(\tfrac12\sum_i(\Delta t_i)^2\) relative to the Ito integral.

Proof. The preceding Ito-isometry calculation proves convergence of \(S_\Pi\) to the Ito integral in \(L^2\). The exact partition identity and the quadratic-variation theorem prove convergence of the same sums to \(\tfrac12(B_T^2-T)\) in \(L^2\). Since the \(L^2\) limit is unique, the two random variables agree almost surely. The displayed mean-square error is exactly the quantity computed using the Ito isometry. \(\square\)

Worked Partition Calculations

Worked Example: Two Equal Intervals on \([0,1]\)

Take the partition \(0<\tfrac12<1\). Set \(U=B_{1/2}-B_0\) and \(V=B_1-B_{1/2}\). Then \(B_{1/2}=U\), \(B_1=U+V\), and the left-point sum is \(S_\Pi=B_0(B_{1/2}-B_0)+B_{1/2}(B_1-B_{1/2})=0\cdot U+UV=UV\). The quadratic-variation sum is \(Q_\Pi=U^2+V^2\). Applying the exact identity gives

$$ \frac12(B_1^2-Q_\Pi) =\frac12\left((U+V)^2-(U^2+V^2)\right) =\frac12(2UV)=UV=S_\Pi. $$

Thus the identity agrees term by term for this partition. The limiting integral is \(\tfrac12(B_1^2-1)\), since \(T=1\); the correction is \(1/2\), not zero. As an expectation check, \(\mathbb{E}[B_1^2]=1\), so the limiting expression has mean zero. Also \(\mathbb{E}[UV]=0\), since \(U\) and \(V\) are independent and centered.

Worked Example: Error on a Two-Interval Partition of \([0,2]\)

Use \(0<1<2\), so both time increments equal \(1\). The quadratic-variation sum is \(Q_\Pi=(B_1-B_0)^2+(B_2-B_1)^2\). Substituting \(\Delta t_1=\Delta t_2=1\) into the variance formula gives \(\mathbb{E}[Q_\Pi]=1+1=2\) and \(\mathbb{E}[(Q_\Pi-2)^2]=2(1^2+1^2)=4\). The partition identity says \(S_\Pi=\tfrac12(B_2^2-Q_\Pi)\), while the integral is \(\tfrac12(B_2^2-2)\). Subtracting gives the exact error \(S_\Pi-\int_0^2B_s\,dB_s=-\tfrac12(Q_\Pi-2)\). Therefore its mean-square error is \(\tfrac14\cdot4=1\).

Check: the Ito-isometry error formula gives \(\tfrac12(1^2+1^2)=1\), the same value. Its mean is zero because \(\mathbb{E}[Q_\Pi-2]=0\), consistent with the mean-square computation.

Worked Example: A Unequal Partition of \([0,1]\)

Take \(0<\tfrac14<1\), with time increments \(\tfrac14\) and \(\tfrac34\). The quadratic-variation variance is \[ 2\left(\left(\frac14\right)^2+\left(\frac34\right)^2\right) =2\left(\frac1{16}+\frac9{16}\right)=\frac54. \] The left-point sum differs from the integral by \(-\tfrac12(Q_\Pi-1)\). Its mean-square error is therefore \(\tfrac14\cdot\tfrac54=\tfrac5{16}\). The Ito-isometry formula independently gives \(\tfrac12(\tfrac1{16}+\tfrac9{16})=\tfrac12\cdot\tfrac{10}{16}=\tfrac5{16}\). The two calculations agree. The larger interval contributes \(\tfrac12\cdot\tfrac9{16}=\tfrac9{32}\) of the total error, while the shorter contributes \(\tfrac12\cdot\tfrac1{16}=\tfrac1{32}\); their sum is \(\tfrac{10}{32}=\tfrac5{16}\).

Ito and Stratonovich: The Correction Matters

The Ito integral uses left endpoints. A Stratonovich integral, written \(\int_0^T B_s\circ dB_s\), instead uses midpoint values in the symmetric-sum definition. For each interval, the midpoint value is \(\tfrac12(B_{t_{i-1}}+B_{t_i})\), so its sum satisfies

$$ \sum_i\frac{B_{t_{i-1}}+B_{t_i}}2\,\Delta B_i =\frac12\sum_i(B_{t_i}^2-B_{t_{i-1}}^2) =\frac12B_T^2. $$

Thus the symmetric sums give \(\int_0^T B_s\circ dB_s=\tfrac12B_T^2\), whereas the Ito integral is \(\tfrac12(B_T^2-T)\). Their difference is \(T/2\), the one-dimensional correction \(\tfrac12[B,B]_T\), since Brownian quadratic variation is \([B]_T=T\). In differential shorthand, \(dB_t\,dB_t=dt\); this quadratic term is precisely what the ordinary chain rule omits. The rules \(dB_t\,dt=0\) and \(dt\,dt=0\) express that mixed and time-only second-order terms vanish in Ito calculations.

A related ordinary-calculus identity is valid for a \(C^1\) path \(x:[0,T]\to\mathbb{R}\): defining \(\int_0^T x\,dx:=\int_0^T x(t)x'(t)\,dt\), the ordinary chain rule gives \(x(T)^2-x(0)^2=2\int_0^T x\,dx\). The regularity condition matters; differentiability alone does not guarantee absolute continuity or this endpoint formula. Brownian paths are not \(C^1\), and their quadratic variation cannot be discarded.

Common Mistake: Dropping the Quadratic-Variation Term

A wrong calculation is to write \(\int_0^T B_s\,dB_s=\tfrac12B_T^2\) by treating \(d(B_t^2)\) as \(2B_t\,dB_t\). The partition identity shows the missing term explicitly: \(B_T^2=2S_\Pi+Q_\Pi\). Since \(Q_\Pi\to T\), not zero, the left-point sums converge to \(\tfrac12(B_T^2-T)\), not \(\tfrac12B_T^2\). For instance, the proposed answer has expectation \(\tfrac12\mathbb{E}[B_T^2]=T/2\), whereas the Ito integral has expectation zero: for each left-point sum, \(\mathbb{E}[B_{t_{i-1}}\Delta B_i]=0\) by independence of the increment from \(\mathcal{F}_{t_{i-1}}\) and its zero mean, and the sums converge in \(L^2\), hence in \(L^1\).

The exact remainder is \(Q_\Pi-T\). Its mean is zero, and its mean square is \(2\sum_i(\Delta t_i)^2\le2T|\Pi|\). Consequently, the integral approximation error is exactly \(-\tfrac12(Q_\Pi-T)\), with mean square at most \(T|\Pi|/2\). This both explains why the correction survives in the limit and quantifies how quickly the left sums converge.

Check Your Understanding

Use the exact telescoping identity and the stated mean-square estimates in each question.

  1. For the partition \(0<\tfrac13<1\), write the left-point sum using the two Brownian increments and verify the partition identity algebraically.
  2. For a partition of \([0,3]\) with three equal intervals, calculate \(\mathbb{E}[Q_\Pi]\), \(\operatorname{Var}(Q_\Pi)\), and the mean-square error of the left-point approximation.
  3. Explain why the estimate \(2T|\Pi|\) proves convergence in \(L^2\) but does not by itself assert convergence for every sample path.
  4. Compute the difference between the Ito and Stratonovich self-integrals on \([0,T]\), and identify the quadratic-variation term responsible for it.
  5. For \(x\in C^1([0,T])\), state the ordinary endpoint identity for \(\int_0^T x(t)x'(t)\,dt\), and explain why differentiability alone is not the stated sufficient hypothesis.

Previous

Next

Back to Tutorials