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?

  • 2
    $\begingroup$ A rather strong but sufficient assumption is that $\alpha_t$ is deterministic, see here. $\endgroup$
    – Kevin
    May 31, 2023 at 20:53
  • $\begingroup$ @Kevin Sure, I rephrased my question. That was of course obvious. $\endgroup$
    – fwd_T
    May 31, 2023 at 23:51

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

  • $\begingroup$ Your answer up to your last paragraph makes sense to me. I am not convinced that $A^{-\frac{1}{2}}_t\cdot\int_0^t\alpha_sdW_s$ is a Gaussian random variable. Even assuming that it is, since $A_t$ is a random variable of unknown distribution, this does not necessarily mean that $\int_0^t\alpha_sdWs$ is a Gaussian random variable, if I understand correctly. So it seems to me that the necessary and sufficient condition for this integral to be a normal random vartiable is for $A_t$ to be a constant. Am I wrong? $\endgroup$
    – fwd_T
    May 31, 2023 at 19:09
  • $\begingroup$ My question was specifying that $\left(\alpha_t\right)_{t\geq 0}$ is a stochastic process. In general, quadratic variation can be a stochastic process. In fact it is in case of spot processes modelled with stochastic volatility models. It just happens to be a deterministic function in the particular case of Brownian motion. $\endgroup$
    – fwd_T
    May 31, 2023 at 19:14
  • $\begingroup$ the integral of an adapted process IS a random variable, in general. $\endgroup$
    – fwd_T
    May 31, 2023 at 19:16
  • $\begingroup$ You're right. And in fact the only thing that changes is how it acummulates variance, but its still a martingale with mean 0, as it is adapted. So what you have is a normal random variable with that variance. $\endgroup$
    – KT8
    May 31, 2023 at 19:20
  • $\begingroup$ I was abusing notation, note that now you need to take the expectation under the filtration to compute the (expected) accumulated variance (just like in itô's isometry). $\endgroup$
    – KT8
    May 31, 2023 at 19:25

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