3
$\begingroup$

Given a random zero-mean gaussian random variable $X(t)$ with parameter $t$, such that $E [X(t) X(t^\prime)] = \sigma^2 (t) \delta_{tt^\prime}$, is it possible to produce a single gaussian random variable that is the weighted sum of all $X(t)$ by a non-random function $f(t)$? In other words, is $Z=\int_0^T f(t) X(t) dt$ well defined, and if so, is $Z$ a gaussian normal random variable, and what is its variance?

My naive attempt is to discretize the integral \begin{equation} Z = \int_0^T f(t) X(t) dt \approx \sum_{n=0}^{N-1} f(n \delta T) X(n \delta T) \delta T \end{equation} where $\delta T = T/N$. Now, if we take the simplest limit where $\sigma(t) = \sigma_0$ and $f(t) = 1$, this reduces to \begin{eqnarray} Z & = & \frac{T}{N}\sum X_n \end{eqnarray} But this is a gaussian random variable with $\sigma = \sqrt{N} \delta T \sigma_0 = T \sigma_0 / \sqrt{N}$, which has $\sigma \rightarrow 0$ as $N \rightarrow \infty$.

If I were to take a guess, I'd assume $Z$ would be gaussian and have variance $\sigma^2 = \int_0^T f^2(t) \sigma^2(t) dt$, but I have been unable to show this.

$\endgroup$
3
  • $\begingroup$ Are you familiar with Ito integrals? Also, your $Z$ should converge to 0 by the central limit theorem. $\endgroup$
    – Kirill
    Jul 18, 2013 at 17:03
  • $\begingroup$ I am not. Perhaps some context would be helpful. I am trying to solve a differential equation with spatial and temporal variables where I have added a stochastic noise term. I want to reduce the spatial dimensions by assuming a spatial profile along one of the spatial direction, and then averaging over that coordinate. This results in the integral listed in the original question. I don't really need a formally correct proof, just a way to deal with this spatial averaging. $\endgroup$ Jul 18, 2013 at 17:20
  • $\begingroup$ Also, you're correct that it converges to 0. Fixed. $\endgroup$ Jul 18, 2013 at 18:59

1 Answer 1

1
$\begingroup$

Kudos for your logic in studying this problem (except that your $\sigma$ goes to $0$ when $N\to\infty$). Indeed, as you realized, the construction of the object $Z$ runs into serious problems, which are of two kinds mainly.

First, an i.i.d. process $(X(t))_{t\in\mathbb R_+}$ is a wild beast, for example, which probability space are we going to use? Second, the way to define the integral $\int\limits_0^TX(t)\mathrm dt$ is not obvious at all. The Riemann way is doomed since every upper Darboux sum is $+\infty$ and every lower Darboux sum is $-\infty$, and the Lebesgue way runs into measurability issues.

Note that stochastic integrals are based on quite different processes, for instance on a Brownian motion $(W_t)_{t\in\mathbb R_+}$, which is far from being i.i.d. To begin with, its paths being almost surely continuous, this process is entirely determined by $(W_t)_{t\in\mathbb Q_+}$, that is, a countable collection of random variables--and now one can begin to work...

$\endgroup$
3
  • $\begingroup$ Sorry, I don't think I understood your post. See my above comment. I'm not a mathematician, and I don't need a formal proof, but just a qualitatively correct way to handling such an integral. If it helps, I think I can formulate the problem such that $\sigma(t)$ is constant. $\endgroup$ Jul 18, 2013 at 17:31
  • 1
    $\begingroup$ Take-home message: the integral you are trying to define does not exist and cannot exist. To expand a little bit: reasonable limiting procedures yield $Z=0$ with full probability (as @Kirill indicated and as you confirmed yourself in a comment). $\endgroup$
    – Did
    Jul 18, 2013 at 18:22
  • $\begingroup$ Okay, thanks. So would it be possible to decompose in terms of basis functions like $X(t) = \sum_n e^{2 \pi i n t/T} c_n$, where $c_n$ is a random variable with $E[c_n^* c_m] \sim \delta_{nm}$? (I know I'm being sloppy about real vs. complex RVs.) If so, that would be sufficient. But it seems like this would require the original integral to converge to a finite value. $\endgroup$ Jul 18, 2013 at 18:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .