When a Brownian Stochastic Integral Has a Normal Distribution
Summary
The document asks when an integral of an adapted process against Brownian motion is normally distributed. It invokes Lévy’s characterization, which identifies a continuous martingale with the right quadratic variation as Brownian motion, and applies Itô’s isometry to relate the integral’s second moment to the integrated squared process. These ideas help distinguish martingale properties from distributional assumptions.
The answer’s conclusion needs a key qualification: an adapted integrand generally produces a martingale whose quadratic variation is random, and a random time change or variance does not by itself yield a normal variable. A deterministic quadratic variation is a sufficient condition for a Gaussian law, with variance equal to that quadratic variation. The argument also conflates expected quadratic variation with the pathwise quadratic variation required by Lévy’s theorem. Thus it is useful as a starting point, but its stated normality conclusion is not established for general adapted integrands.
Key ideas
- An adapted Brownian integrand gives a continuous local martingale when the relevant integrability conditions hold.
- The stochastic integral’s quadratic variation is the pathwise integral of the squared integrand.
- Itô’s isometry gives the expected squared integral, not its full probability distribution.
- A deterministic quadratic variation supports a Gaussian conclusion through martingale characterization.
- A random variance does not generally preserve normality.
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# Necessary conditions to ensure that stochastic integral is a normal variable
# Necessary conditions to ensure that stochastic integral is a normal variable
Let $\left(W_t\right)_{t\geq 0}$ be a Brownian motion with respect to filtration $\mathbb{F}=\left(\mathcal{F}_t\right)_{t\geq 0}$. Let $\left(\alpha_t\right)_{t\geq 0}$ be an $\mathbb{F}$-adapted stochastic process. What are necessary conditions to ensure that the stochastic integral $\int_0^t\alpha_sdW_s$ is a normal random variable?
## Answer by KT8 (score -1)
https://quant.stackexchange.com/a/75724
I'm going to be following Shreve's Stochastic Calculus for Finance II.
First, Lévy's theorem states that if $M(t)$ is a martingale relative to a filtration $\mathcal{F}(t)$, $M(0) = 0$ and accumulates quadratic variance as $[M, M] (t)=t$ for all $t\geq 0$, then $M(t)$ is a Brownian motion, and we agree that the increments of a Brownian motion are distributed according to a normal random variable.
We then need to prove that (i) $M(t)$ is a martingale, (ii) $M(0) = 0$.
(i) this can be proved as
$$ \begin{align} \mathbb{E} \big[ I(t) \vert \mathcal{F}(z) \big] & = \mathbb{E} \Bigg[ \int_0^t \alpha(s) dW_s \Bigg \vert \mathcal{F}(z) \Bigg] = \mathbb{E} \Bigg[ \int_z^t \alpha(s) dW_s + \int_0^z \alpha(s) dW_s \Bigg \vert \mathcal{F}(z)\Bigg] = \\ & = \mathbb{E} \Bigg[ \int_z^t \alpha(s) dW_s \Bigg \vert \mathcal{F}(z)\Bigg] + I(z), \end{align} $$ where you can see that the first integral vanishes by discretizing it and taking the expectation under the filtration $\mathcal{F}(z)$ is 0.
(ii) this follows straight forward from the integral itself.
Finally, the integral (let's call it $M(t)$ here) accumulates quadratic variance according to
$$\mathbb{E} \Big[ I^2 (t) \Big] = \mathbb{E} \Bigg[ \int_0^t \alpha^2(s) ds \Bigg] =: t A(t)$$
which follows from Itô's isometry.
Therefore, what you have is that the rescaled version of your integral $A(t)^{-1/2} \int_0^t \alpha(s) dW_s $ is a Brownian motion. You can see that the only difference with respect to the normal random variable at this stage is the variance. As the scaling factor does not introduce any source of randomness, we can claim that the integral is a normal random variable with mean $0$ and variance $\int_0^t \alpha^2(s) ds$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.