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Let $(\Omega,\Sigma,P)$ be a probability space and $X: [0,\infty) \times \Omega \to \mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $\sum_i a_i X_{t_i}$ are Gaussian random variables). If the process is continuous, it seems to be clear that the process $Y_t (\omega) = \int_0^t X_s(\omega) ds$ is a Gaussian process.

Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.

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  • $\begingroup$ Yes, the measurability in the sense of $([0,\infty)\times \Omega, \mathcal{B}([0,\infty))\otimes \mathcal F) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is enough. $\endgroup$
    – zhoraster
    Commented Oct 9, 2015 at 7:13
  • $\begingroup$ @zhoraster Why do you think so? For example, if $Z \sim N(0,1)$, then $X_t := \frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $\int_0^t X_s \, ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $\sup_{t \leq T} \mathbb{E}(|X_t|)<\infty$ for $T>0$. $\endgroup$
    – saz
    Commented Oct 9, 2015 at 7:38
  • $\begingroup$ @saz, I assume that everything is well-defined. Of course, to this end we need some integrability. $\endgroup$
    – zhoraster
    Commented Oct 9, 2015 at 8:44
  • $\begingroup$ @zhoraster I see. $\endgroup$
    – saz
    Commented Oct 9, 2015 at 8:48
  • $\begingroup$ @saz Could you name a source to prove the sufficiency of $sup_{t\leq T}\mathbb{E}(|X_t|)<\infty$ ? $\endgroup$
    – Leoncino
    Commented Dec 7, 2020 at 14:57

1 Answer 1

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Question 1: Is $Y_t(\omega)$ well-defined?

Answer: No, in general, $Y_t(\omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$\int_0^t |X_s(\omega)| \, ds <\infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $\sup_{t \leq T} \mathbb{E}(|X_t|)<\infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z \sim N(0,1)$; then $X$ is Gaussian, but the integral $\int_0^t X_t \, ds$ does not exist.)

Question 2: Is $\omega \mapsto Y_t(\omega)$ a random variable for fixed $t \geq 0$?

Answer: If the process $X: (0,\infty) \times \Omega \to \mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t \geq 0$. Otherwise, measurability of $Y_t$ might fail.

Question 3: Is $(Y_t)_{t \geq 0}$ Gaussian?

Answer: If $t \mapsto X_t(\omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] \to \mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)

Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] \to \mathbb{R}$ it is known that the (Riemann) integral

$$\int_0^t f(s) \, ds$$

can be approximated by Riemann sums

$$\sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$

where $0=t_0 < \ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j \in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t \frac{j}{n}$, we find

$$\int_0^t f(s) \, ds = \lim_{n \to \infty} \frac{t}{n} \sum_{j=0}^{n-1} f \left( t \frac{j}{n} \right).$$

Applying this in order (stochastic) setting, we get

$$\int_0^t X_s \, ds = \lim_{n \to \infty} \frac{t}{n} \sum_{j=0}^{n-1} X_{t j/n};$$

and $\frac{t}{n} \sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.

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    $\begingroup$ @jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit. $\endgroup$
    – saz
    Commented Oct 9, 2015 at 17:38
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    $\begingroup$ @jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter). $\endgroup$
    – saz
    Commented Oct 9, 2015 at 18:40
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    $\begingroup$ @jpv Just write down how the Lebesgue integral of a step function of the form $$f(s) = 1_{[a,b]}(s),$$ or more generally, $$ f(s) = \sum_{j=1}^n c_j 1_{[a_j,b_j]}(s)$$ looks like - then you will see that this gives Riemann sums. So, by approximating the integrand by step functions, we get Riemann sums. Moreover, note that the Riemann integral and the Lebesgue integral coincide in this particular case. (Finaly, mind that $\int_0^t f(s) \, ds$ is even differentiable (in $t$), if $f$ is continuous, but we do no not need this property here). $\endgroup$
    – saz
    Commented Oct 10, 2015 at 4:30
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    $\begingroup$ @jpv See my edited answer. (And please note that the integral $\int_0^t f(s) \, ds$ exists if and only if $\int_0^t |f(s)| \, ds < \infty$.) $\endgroup$
    – saz
    Commented Oct 11, 2015 at 6:24
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    $\begingroup$ @jpv Ah, sorry, I did a stupid mistake there. You are right, we have to assume that $t \mapsto X_t(\omega)$ is Riemann integrable; otherwise we are lost (at least using the argumentation I sketched above). $\endgroup$
    – saz
    Commented Oct 11, 2015 at 7:53

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