What is the integral of a Gaussian white noise

Question

Let us define $w(t)$ be a Gaussian white noise process, in the sense that

• $w(t) \sim N(0, 1)$ for all $t \in [0, \infty)$.
• $w(t_1)$ and $w(t_2)$ are independent for every $t_1, t_2 \in [0, \infty)$.

So $w(t)$ is in some sense a finite-energy white noise process.

I would like to ask, can the integral

$$ \int_0^T w(t) dt $$

be well-defined? Does $\lim_{|\Pi| \to 0} \sum_{i=1}w(t_i) (t_{i+1} - t_i)$ or $\lim_{|\Pi| \to 0} \sum_{i=1}w(t_{i+1}) (t_{i+1} - t_i)$ converge to something useful?

Answer 1

Not exactly what you are asking for but if you use the framework developed by Takeyuki Hida, then the White noise process $[0,\infty)\ni t\mapsto w(t)$ can be seen as a $(S)^*$-valued process (i.e. a Hida-distribution-valued family of "stochastic distributions" indexed by time).

In that case the integral $$\int_0^T w(t)dt$$ can be defined rigorously as a Pettis integral in $(S)^*$ and using the fact that formally $w(t)=\frac{d}{dt} B(t)$ where $B$ is a Brownian motion you have that $$\int_0^T w(t)dt=B(T).$$

You can find more discussions about White noise analysis in the following books 1, 2.

Answer 2

To long for a comment

The Wikipedia article you found mentions under the section Continuous-time white noise that "However, a precise definition of these concepts is not trivial, because some quantities that are finite sums in the finite discrete case must be replaced by integrals that may not converge. Indeed, the set of all possible instances of a signal $w$ is no longer a finite-dimensional space $\mathbb {R} ^{n}$, but an infinite-dimensional function space. Moreover, by any definition a white noise signal $w$ would have to be essentially discontinuous at every point; therefore even the simplest operations on $w$, like integration over a finite interval, require advanced mathematical machinery."

Nonetheless, at first glance, I find it an interesting question if one could reconcile at least intuitively $W(t)=\int_0^tw(s)\,ds\sim N(0,t)\quad $ with $\quad w(s)\,ds\sim N(0,ds^2)$.