Frozen Brownian Values as Random Integrands

A Brownian Value Held Fixed on an Interval

An Ito integrand may be random, provided it is adapted in the required way and has finite expected squared integral. A useful example is a Brownian value \(B_a\) held fixed while the integration time runs from \(a\) to \(b\). Since \(B_a\) is already known at time \(a\), it is a legitimate left-endpoint value for this interval. The integral is a product: the frozen value \(B_a\) times the later Brownian increment \(B_b-B_a\). Although both factors are Gaussian, their product is generally not Gaussian.

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 deterministic times \(0\le a<b\), and write \(h=b-a\). Brownian increments after time \(a\) are independent of \(\mathcal{F}_a\), and \(B_b-B_a\) has distribution \(N(0,h)\).

This integrand is progressively measurable: it is zero before \(a\), and on any interval after \(a\), its nonzero value \(B_a\) is measurable with respect to the information available at the interval's start. Its energy is finite, since

When \(a=0\), \(B_a=B_0=0\), so the integrand and integral are zero. The nondegenerate distributional formulas below require \(a>0\) and \(h>0\).

Evaluating the Integral

Proof. On \((a,b]\), the process \(H_s\) is the single random coefficient \(B_a\), which is \(\mathcal{F}_a\)-measurable. By the definition of the Ito integral for an adapted step process, the integral over this interval is its coefficient multiplied by the Brownian increment over the interval. The integrand is zero on \([0,a]\), so that part contributes zero. Therefore \(\int_0^b H_s\,dB_s=B_a(B_b-B_a)\), as asserted. The equality is almost sure, as is the step-integral definition. \(\square\)

The step-integral identity also gives a direct interpretation of the randomness: \(B_a\) is the value available at the left endpoint, while \(B_b-B_a\) is the new increment accumulated after that endpoint. These two factors are independent. In particular, if \(a>0\), then \(B_a\sim N(0,a)\) and \(B_b-B_a\sim N(0,h)\), independently.

The Product Distribution and Its Moments

Suppose now that \(a>0\) and \(h>0\). Let \(U=B_a/\sqrt{a}\) and \(V=(B_b-B_a)/\sqrt{h}\). By Brownian increment independence, \(U\) and \(V\) are independent standard normal variables. The integral \(I_{a,b}\) therefore has the representation \(I_{a,b}=\sqrt{ah}\,UV\). This representation determines its distribution and moments.

Proof. For \(a,h>0\), write \(I_{a,b}=\sqrt{ah}\,UV\), where \(U,V\) are independent \(N(0,1)\) variables. Conditional on \(U\), the normal characteristic function of \(V\) gives \(\mathbb{E}[e^{iu\sqrt{ah}UV}\mid U]=e^{-ah\,u^2U^2/2}\). Integrating over \(U\sim N(0,1)\), or evaluating the Gaussian integral, yields \(\mathbb{E}[e^{iuI_{a,b}}]=(1+ah\,u^2)^{-1/2}\).

Independence and the centered normal moments give \(\mathbb{E}[I_{a,b}]=\sqrt{ah}\,\mathbb{E}[U]\mathbb{E}[V]=0\), and \(\mathbb{E}[I_{a,b}^2]=ah\,\mathbb{E}[U^2]\mathbb{E}[V^2]=ah\). Hence the variance is \(ah\). Also, \(\mathbb{E}[U^4]=\mathbb{E}[V^4]=3\), so \(\mathbb{E}[I_{a,b}^4]=(ah)^2\cdot3\cdot3=9a^2h^2\).

To obtain the density, first consider \(Z=UV\). The change of variables from \((U,V)\) to \((U,Z)\), with \(V=Z/U\), has Jacobian factor \(1/|U|\). Since each standard normal density is \((2\pi)^{-1/2}e^{-x^2/2}\), for \(z\ne0\) the density of \(Z\) is \[ f_Z(z)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \exp\left(-\frac{x^2+z^2/x^2}{2}\right)\frac{dx}{|x|} =\frac{1}{\pi}\int_0^\infty \exp\left(-\frac{r^2+z^2/r^2}{2}\right)\frac{dr}{r}. \] In the last equality, the integrand is even and \(r=|x|\). Set \(r=\sqrt{|z|}e^{v/2}\). Then \(dr/r=dv/2\) and \((r^2+z^2/r^2)/2=|z|\cosh v\). The integral over \(v\in\mathbb{R}\) is twice its integral over \(v\ge0\), so \(f_Z(z)=K_0(|z|)/\pi\). Scaling by \(\sqrt{ah}\) gives the stated density. Finally, if \(a=0\), then \(B_a=0\) and the integral is identically zero. \(\square\)

The product law is symmetric about zero, but it is not normal when \(a,h>0\). A centered normal variable of variance \(ah\) would have fourth moment \(3a^2h^2\), whereas this integral has fourth moment \(9a^2h^2\). The difference is a useful way to detect the effect of multiplying two independent random factors.

Martingale Check and the Degenerate Endpoint

The step-integral process on the interval is \(M_t=0\) for \(t\le a\) and \(M_t=B_a(B_t-B_a)\) for \(t>a\). It is a true martingale on \([0,b]\), not merely a local martingale. Indeed, for each \(t\le b\), it is square-integrable: for \(t>a\), independence gives \(\mathbb{E}[M_t^2]=a(t-a)<\infty\). If \(a\le s\le t\), then \(B_a\) is \(\mathcal{F}_s\)-measurable and \(\mathbb{E}[B_t-B_s\mid\mathcal{F}_s]=0\), so \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). If \(s<a<t\), the tower property and \(\mathbb{E}[B_t-B_a\mid\mathcal{F}_a]=0\) give \(\mathbb{E}[M_t\mid\mathcal{F}_s]=0=M_s\). When \(s,t\le a\), both process values are zero. Thus the conditional-expectation identity holds in every case.

A Common Pitfall: Conditional Normality Is Not Unconditional Normality

Given \(B_a\), the increment \(B_b-B_a\) is independent and normal, so the integral conditional on \(B_a\) is normal with mean \(0\) and variance \(hB_a^2\). That conditional variance is random. It is therefore incorrect to conclude that the unconditional integral is normal merely because its conditional law is normal.

Wrong calculation: take a centered normal variable with variance \(\mathbb{E}[hB_a^2]=ah\), then assign the integral the normal fourth moment \(3(ah)^2\). Correction: compute the fourth moment from the independent product: \[ \mathbb{E}\big[(B_a(B_b-B_a))^4\big] =\mathbb{E}[B_a^4]\mathbb{E}[(B_b-B_a)^4] =(3a^2)(3h^2)=9a^2h^2. \] For \(a,h>0\), this is not \(3(ah)^2\), so the normal conclusion is false.

The distinction from an integral with a changing Brownian integrand is also important. Here the coefficient is fixed at \(B_a\) throughout the interval, so the step-integral definition gives the product \(B_a(B_b-B_a)\). Replacing the frozen coefficient with \(B_s\) changes the integrand and hence changes the integral.