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Let $(B_t)_{t\in [0,\infty)}$ denote the standard Brownian motion and $\mathcal F_t$ its filtration. Let $\Omega$ be the underlying probability space and $T>0$.

If $f\in \mathcal L^2[0,T]$, we may think of $f$ as a deterministic process $f\in \mathcal L^2([0,T]\times\Omega)$ with $f_t(\cdot)$ constant on $\Omega$. Then the Ito integral $W(f)$ of $f$ is Gaussian distributed, where $$ W(f) = \int_0^T f_t(\cdot) \ dB_t$$ From the definition of the integral it is clear why: If we take a sequence $(f^{(n)})_{n\in\mathbb N}$ of deterministic step processes (ie. step functions) converging to $f$ in $\mathcal L^2$, the partial integrals $W(f^{(n)})$ are sums of Gaussians, and therefore Gaussian (and convergence in $\mathcal L^2$ implies convergence in distribution, so the distribution of $W(f)$ is Gaussian since it is the $\mathcal L^2$ limit of $W(f^{(n)})$).

From one angle, the fact that $W(f)$ is Gaussian distributed relies on the fact that the changes in $f$ over time are independent of the changes in the Brownian. By contradistinction, $W(B_s) = \frac{1}{2}(B_T^2 - T)$ is not Gaussian distributed because the process $B_s$ is dependent on $B_s$.

What I'm wondering is whether the integral of predictable processes produces a Gaussian distributed variable. Why do I think it might? If $(a_t)_{t\in [0,\infty)}$ is a predictable process then, vaguely, the random changes in the Brownian in time interval $(t,t+\epsilon)$ are independent of what happened before time $t$, which determines $a_t$.

Owen
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2 Answers2

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No. Let $f_t = 1$ for all $t$ with probability $1/2$ and $2$ with probability $1/2.$ $f_t$ is previsible. Then $W_T$ is a 50-50 mixture of normal with variance $T$ and $4T$, which is not a normal.

To put it a little more generally, it's true that the integral of a deterministic function is Gaussian with a variance that depends on the function, but then integral of a random function that is independent of the Brownian motion will have to be a mixture of normals with different variances, which are not necessarily normal. And there are plenty of random previsible processes that are independent of the Brownian motion.

Also, consider the theorem that any Martingale can be written as the integral with respect to brownian motion of a previsible process.

spaceisdarkgreen
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To follow up on spaceisdarkgreen's answer:

  1. For any distribution on the real line there is an $\mathcal F_T$-measurable random variable with that distribution. (Even of the form $g(W_T)$ for an appropriate Borel function $g$.)

  2. For any $\mathcal F_T$-measurable random variable $Y$ there is a predictable process $H$ such that $Y=\int_0^T H_s\,dW_s$ a.s. This is a result of R.M. Dudley [Wiener functionals as Itô integrals; Ann. Probability 5 (1977) 140–141; http://www.jstor.org/stable/pdf/2242810.pdf].

John Dawkins
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