Necessary conditions to ensure that stochastic integral is a normal variable

Question

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 1

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 $\int_0^t \alpha^2(s) ds$.