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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.

saz
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jpv
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  • 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. – zhoraster Oct 09 '15 at 07:13
  • @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 $T>0$. – saz Oct 09 '15 at 07:38
  • @saz, I assume that everything is well-defined. Of course, to this end we need some integrability. – zhoraster Oct 09 '15 at 08:44
  • @zhoraster I see. – saz Oct 09 '15 at 08:48
  • @saz Could you name a source to prove the sufficiency of $sup_{t\leq T}\mathbb{E}(|X_t|)<\infty$ ? – Leoncino Dec 07 '20 at 14:57

1 Answers1

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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.

saz
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  • saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something. – jpv Oct 09 '15 at 17:35
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    @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. – saz Oct 09 '15 at 17:38
  • saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so? – jpv Oct 09 '15 at 18:06
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    @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). – saz Oct 09 '15 at 18:40
  • saz: I am assuming that $Y_t(\omega) = \int_0^t X_s(\omega) ds$ exists for all $\omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity. – jpv Oct 10 '15 at 03:42
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    @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). – saz Oct 10 '15 at 04:30
  • I still have some troubles with the proof. For simplicity assume $f \geq 0$, then the simple functions which approximate $f$ are $f_n (t):= \sum_{k=0}^{n2^n-1}(k/2^n)\chi_{[k/2^n \leq f(t) <(k+1)/2^n)}$. Then, $\int_0^T f_n(s)ds = \sum_{k=0}^{n2^n-1}(k/2^n)\lambda[k/2^n \leq f <(k+1)/2^n)$ and it is not directly related to the Riemann sums. – jpv Oct 10 '15 at 05:41
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    @jpv See my edited answer. (And please note that the integral $\int_0^t f(s) , ds$ exists if and only if $\int_0^t |f(s)| , ds < \infty$.) – saz Oct 11 '15 at 06:24
  • Thanks. I completely understand the part where the sums are Gaussian and hence their a.s. limit is Gaussian. I am unable to get my head around this: "If $f : [0,T] \to \mathbb{R}$ is Lebesgue measurable and $\int_0^T|f(s)|ds < \infty$, then $f$ is Reimann integrable and both the integrals coincide". If this is false, then we atleast need to show that the Lebesgue integral can be approximated by Riemann sums. Lebesgue integral is defined using partitioning the range of the function so I am unable to see a direct connection. It may be that I am missing some very important knowledge. – jpv Oct 11 '15 at 07:23
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    @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). – saz Oct 11 '15 at 07:53
  • I will go with this extra assumption for now. Thanks for your effort and time. – jpv Oct 11 '15 at 08:10
  • I assume the correct integral is $\int_0^t f(s) , ds = \lim_{n \to \infty} \frac{t}{n} \sum_{j=0}^{n-1} f \left( t \frac{j}{n} \right)$. – Daniel Camarena Perez Dec 08 '18 at 19:15
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    @DanielCamarenaPerez Right, there were some $t$'s missing; thank you. – saz Dec 08 '18 at 19:45