Why the Interval Start Matters
A Brownian self-integral over an interval records the Brownian motion’s fluctuations relative to its value at the start of that interval. For \(0\le a<b\), the relevant integrand is \(B_s-B_a\), not \(B_s\). Subtracting \(B_a\) makes the integral depend only on the Brownian increment within \([a,b]\). That observation gives an exact formula and, through the normal distribution of Brownian increments, an exact distribution.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to \((\mathcal{F}_t)\). In particular, \(B_b-B_a\) is independent of \(\mathcal{F}_a\) and has distribution \(N(0,b-a)\). Ito integrals below are defined for progressively measurable integrands \(H_s\) in \(\mathcal{L}^2\): on every finite time interval \([0,T]\), this means \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\).
The Endpoint Formula
The key identity relates the integral to the squared increment and the length of the interval. Its proof uses the defining approximation of the Ito integral and the quadratic variation of Brownian motion. For partitions of \([a,b]\) with mesh tending to zero, the Brownian quadratic variation sums converge in probability to \(b-a\).
Proof. Fix \(0\le a<b\) and a sequence of partitions \(a=t_0<t_1<\cdots<t_n=b\) whose mesh tends to zero. Write \(\Delta B_i=B_{t_i}-B_{t_{i-1}}\). Expanding each squared increment gives \((B_{t_i}-B_a)^2-(B_{t_{i-1}}-B_a)^2=2(B_{t_{i-1}}-B_a)\Delta B_i+(\Delta B_i)^2\). Summing over \(i\) telescopes the left side to \((B_b-B_a)^2\). Thus \(\sum_i(B_{t_{i-1}}-B_a)\Delta B_i=\tfrac12((B_b-B_a)^2-\sum_i(\Delta B_i)^2)\). The left side converges in \(L^2(\mathbb{P})\) to the Ito integral \(J_{a,b}\), by the definition of that integral. The quadratic variation sums on the right converge in probability to \(b-a\). Therefore the right side converges in probability to \(\tfrac12((B_b-B_a)^2-(b-a))\). A sequence cannot have two different limits in probability: indeed, if it converges in probability to \(X\) and \(Y\), then for every \(\varepsilon>0\), \(\mathbb{P}(|X-Y|>\varepsilon)\le\mathbb{P}(|X-X_n|>\varepsilon/2)+\mathbb{P}(|X_n-Y|>\varepsilon/2)\to0\). Hence the two limits agree almost surely, proving the identity. \(\square\)
The correction by \(-(b-a)/2\) is essential. The square of a centered normal increment has positive mean, while the Ito integral here has mean zero. The correction makes those facts consistent.
Worked Example: A Half-Unit Interval
Find the distribution, mean, variance, and support of \(J_{2,5/2}\).
The interval length is \(h=5/2-2=1/2\), and \(B_{5/2}-B_2\) has distribution \(N(0,1/2)\). Write \(B_{5/2}-B_2=\sqrt{1/2}\,Z\), where \(Z\sim N(0,1)\). The endpoint formula gives \(J_{2,5/2}=\tfrac12((1/2)Z^2-1/2)=\tfrac14(Z^2-1)\). Thus the distribution is a scaled and shifted chi-square distribution with one degree of freedom. Since \(Z^2\ge0\), its support is \([-1/4,\infty)\).
Using \(\mathbb{E}[Z^2]=1\) and \(\mathbb{E}[Z^4]=3\), its mean is \(\tfrac14(1-1)=0\). Its variance is \(\tfrac1{16}\operatorname{Var}(Z^2)=\tfrac1{16}(3-1)=1/8\). Check: the general variance formula below gives \(h^2/2=(1/2)^2/2=1/8\), and the lower endpoint \(-h/2=-1/4\) agrees with the expression \(\tfrac14(Z^2-1)\).
Conditional Distribution and Moments
Proof. Brownian increment properties give \(B_b-B_a=\sqrt{h}\,Z\) in distribution, with \(Z\) standard normal and independent of \(\mathcal{F}_a\). The endpoint formula therefore gives the asserted conditional law. The standard normal moments \(\mathbb{E}[Z^2]=1\) and \(\mathbb{E}[Z^4]=3\) yield \(\mathbb{E}[\tfrac h2(Z^2-1)]=\tfrac h2(1-1)=0\) and \(\mathbb{E}[(\tfrac h2(Z^2-1))^2]=\tfrac{h^2}{4}(3-2+1)=h^2/2\). The second expression is the variance because the mean is zero. Finally \(Z^2\ge0\), and \(Z^2\) can be arbitrarily large, giving support \([-h/2,\infty)\). If \(a=b\), the integral is instead the degenerate random variable \(0\), so the positive-length distribution statement is not applied. \(\square\)
For \(h>0\), the density can also be written explicitly. Set \(Y=Z^2\), whose density on \(y>0\) is \(f_Y(y)=e^{-y/2}/\sqrt{2\pi y}\). Under the change of variable \(x=\tfrac h2(y-1)\), we have \(y=1+2x/h\) and \(dy/dx=2/h\). Therefore the density of \(J_{a,b}\) is \(f_J(x)=e^{-1/2-x/h}/\sqrt{\pi h(x+h/2)}\) for \(x>-h/2\), and it is zero for \(x\le-h/2\). There is no point mass at the lower endpoint, since a normal random variable has probability zero of being exactly zero.
Worked Example: A Conditional Tail Probability
For the interval \([1,3]\), find the conditional mean and variance of \(J_{1,3}\), and express \(\mathbb{P}(J_{1,3}>3\mid\mathcal{F}_1)\) in terms of the standard normal CDF \(\Phi\).
Here \(h=3-1=2\). The conditional mean is \(0\), and the conditional variance is \(h^2/2=2^2/2=2\). For \(Z\sim N(0,1)\), the distribution formula gives \(J_{1,3}\) the conditional law of \(\tfrac22(Z^2-1)=Z^2-1\). Hence \(J_{1,3}>3\) is equivalent in this representation to \(Z^2-1>3\), or \(|Z|>2\). Its conditional probability is \(\mathbb{P}(J_{1,3}>3\mid\mathcal{F}_1)=\mathbb{P}(|Z|>2)=2(1-\Phi(2))\). The strict inequality is appropriate: replacing \(>\) by \(\ge\) leaves the probability unchanged because this distribution has no atoms at \(3\). Check: the support starts at \(-h/2=-1\), and the threshold \(3\) lies above that endpoint, so the tail event is possible; the two normal tails are equal by symmetry.
Self-Integrals on Disjoint Intervals
The same formula also identifies which parts of the Brownian path determine each interval integral. In particular, a self-integral on \([a,b]\) is a function only of \(B_b-B_a\). Brownian increments on non-overlapping intervals are independent, so the corresponding self-integrals are independent as well.
Proof. For each \(i\), the endpoint formula gives \(J_{a_i,b_i}=\tfrac12((B_{b_i}-B_{a_i})^2-(b_i-a_i))\) almost surely. Brownian increments over the intervals \([a_i,b_i]\) are independent, including when there is a gap between consecutive intervals. Applying a deterministic measurable function separately to each of a collection of independent random variables preserves independence: for Borel sets \(A_i\), the event that the \(i\)-th transformed variable lies in \(A_i\) is the event that its original increment lies in the inverse image of \(A_i\). Independence of the increments therefore factors the joint probability of these events. This proves the proposition. \(\square\)
Worked Example: Adding Two Equal-Length Self-Integrals
Let \(J_1=J_{0,1}\) and \(J_2=J_{1,2}\). Find the distribution, mean, and variance of \(J_1+J_2\), and compute \(\mathbb{P}(J_1+J_2>0)\).
Each interval has length \(1\), so \(J_i\) has the distribution of \(\tfrac12(Z_i^2-1)\), where \(Z_1,Z_2\) are independent standard normal random variables. Independence follows from the proposition. Therefore \(J_1+J_2=\tfrac12(Z_1^2+Z_2^2-2)\) in distribution. The sum \(Z_1^2+Z_2^2\) has a chi-square distribution with two degrees of freedom. Equivalently, \((Z_1^2+Z_2^2)/2\) has an exponential distribution with rate \(1\). Thus \(J_1+J_2\) has the distribution of \(E-1\), where \(E\) is exponential with rate \(1\), and its support is \([-1,\infty)\).
An exponential random variable with rate \(1\) has mean \(1\) and variance \(1\), so \(J_1+J_2\) has mean \(1-1=0\) and variance \(1\). Since \(J_1+J_2>0\) is equivalent to \(E>1\), its probability is \(e^{-1}\). Check: each summand has mean \(0\) and variance \(1^2/2=1/2\); independence gives mean \(0+0=0\) and variance \(1/2+1/2=1\), matching the shifted-exponential calculation.
A Common Mistake: Forgetting to Center the Integrand
The interval self-integral is not the same as \(\int_a^b B_s\,dB_s\). The latter uses the uncentered Brownian value and includes the starting value \(B_a\). Algebraically, \(B_s=(B_s-B_a)+B_a\), so the distinction is the term \(B_a(B_b-B_a)\).
Wrong calculation: applying the self-integral formula directly to \(\int_1^2 B_s\,dB_s\) would give \(\tfrac12((B_2-B_1)^2-1)\). This is the formula for \(J_{1,2}\), not for the uncentered integral.
Correction: the quadratic-variation identity for the uncentered integral is \(\int_1^2 B_s\,dB_s=\tfrac12(B_2^2-B_1^2-1)\). Subtracting the correct centered expression gives \(\tfrac12(B_2^2-B_1^2-1)-\tfrac12((B_2-B_1)^2-1)=B_1(B_2-B_1)\). For example, substituting \(B_1=2\) and \(B_2-B_1=1/2\), the centered self-integral is \(\tfrac12((1/2)^2-1)=-3/8\), whereas the uncentered expression is \(\tfrac12((5/2)^2-2^2-1)=5/8\). Their difference is \(5/8-(-3/8)=1\), equal to \(2\cdot(1/2)\). The numerical endpoint values illustrate the algebra; the identity itself holds almost surely.
The interval length controls the scale of the self-integral: its variance is \(h^2/2\), and its lower support endpoint is \(-h/2\). When intervals are disjoint, their self-integrals are independent, but their distributions need not be identical if the interval lengths differ. These facts make interval self-integrals useful building blocks for computations involving Brownian paths over several time steps.
Check Your Understanding
Use the endpoint formula and Brownian increment properties to answer the following.
- For \(J_{4,7}\), find the conditional mean, conditional variance, and lower endpoint of the support.
- Express \(\mathbb{P}(J_{0,2}\le 1\mid\mathcal{F}_0)\) in terms of \(\Phi\), taking care with the inequality after rewriting the event using a standard normal variable.
- For two disjoint intervals of lengths \(1\) and \(3\), compute the mean and variance of the sum of their self-integrals.
- Explain why the endpoint formula alone implies independence of self-integrals on disjoint intervals.
- Write the difference between \(\int_a^b B_s\,dB_s\) and \(J_{a,b}\) in terms of \(B_a\) and the Brownian increment over \([a,b]\).