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A Gaussian process $z(s)$ can be established by convolving a gaussian white noise process $x(s)$ with a smoothing kernel $k(s)$ http://ftp.stat.duke.edu/WorkingPapers/01-03.pdf

$$z(s)=\int_{S}^{} \! k(u-s) x(u).du \ \ \text{where } s\in R $$

As a result, the covariance between any two points $s$ and $s'$ can be expressed as follows

$$ c(s, s')=\int_{S}^{} \! k(u-s) k(u-s').du= \int_{S}^{} \! k(u-(s-s')) k(u).du$$

I need help understanding two issues :

1) white noise process is discontinuous and thus Riemann integration cannot be used in the first equation

2) How is the covariance equation derived

3) How did $s'$ jump to the first kernel

raK1
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  • The input process should not necessarily be Gaussian. – msm Sep 02 '16 at 00:46
  • @msm Yup i understand that, but in the case that $x(s)$ is a gaussian white noise, how are these equations derived ? thanks – raK1 Sep 02 '16 at 00:48
  • a white noise is a random i.i.d. zero mean sequence $W_n$. The "continuous" analog doesn't exists (the limit when the sample rate $\to \infty$) but there is the Wiener process $B(t)$ , whose "derivative" is such a white noise with infinite sample rate : $B(t) = \lim_{N \to \infty} \frac{\sum_{n =1}^{\lfloor N t \rfloor} W_n}{\sqrt{N}}$ (the $\frac{1}{\sqrt{N}}$ is necessary for $\mathbb{E}[ |B(t+a)-B(t)|^2] = a\mathbb{E}[|W_n|^2]$). $\quad$ So what I mean is that you can convoluate $B(t)$ with $k'(t)$ and you should get the covariance you'd like – reuns Sep 02 '16 at 01:04
  • @user1952009. Thank you. My question is related to the gaussian white noise $x(u)$. In the first page of the paper linked, an illustration of convoltution with white noise is provided. I dont understand how this integral is managed and how they derived the covariance – raK1 Sep 02 '16 at 01:07
  • read https://en.wikipedia.org/wiki/White_noise#Continuous-time_white_noise and think to the continuous white noise of your paper and the $w(t)$ of wiki as the derivative of my $B(t)$ (that isn't defined, but it isn't a problem since $B \ast k'(t)$ is) https://en.wikipedia.org/wiki/Wiener_process – reuns Sep 02 '16 at 01:11
  • @user1952009 please can you please elaborate more on how the covariance is then derived. – raK1 Sep 02 '16 at 17:50
  • Possible duplicate: https://math.stackexchange.com/q/1912591/532409 "Convolute White noise" – Quillo Feb 25 '23 at 16:10
  • https://math.stackexchange.com/q/134011/445105 Scalar product with white noise is closely related to convolutions – Felix Benning Jun 19 '23 at 10:23

1 Answers1

1

White noise can only be defined in the sense of distributions or as a measure. A good definition can be found in Adler and Taylor (2007, Sec. 1.4.3), see also this SE answer.

To calculate second moments you want to use stochastic integration Adler and Taylor (2007, sec. 5.2) (also see below) for deterministic functions $f,g$ $$ \mathbb{E}[W(f)W(g)] \overset{\text{def.}}=\mathbb{E}\Bigl[\Bigl(\int f(x)W(dx)\Bigr)\Bigl(\int g(x)W(dx)\Bigr) \Bigr] =\int f(x)g(x) dx, \tag{1} $$ which can be viewed as a special case of the Itô Isometry.

Convolution

We can consider convolutions as a special case $$ (f*W)(t) = \int f(t-s)W(ds) = W(f(t-\cdot)) $$ then the covariance function (expectation is zero) is given by $$ C(t,s) = \mathbb{E}[ (f*W)(t)(f*W)(s)] = \int f(t-x)f(s-x)dx $$

Stochastic Integration

The trick to prove (1), is to show that the mapping $$ W:\begin{cases} L^2(\mathbb{R}^n, \mathcal{B}, \nu) &\to L^2(\Omega, \mathcal{A}, \mathbb{P})\\ f &\mapsto W(f) := \int f(t) W(dt) \end{cases} $$ preserves the scalar product. We first consider simple functions $f=\sum_{i=1}^n a_i \mathbf{1}_{A_i}$ for disjoint $A_i$, then $$ W(f)=\int f(t) W(dt) \overset{\text{def.}}= \sum_{i=1}^n a_i W(A_i) $$

Comment: in particular the expectation is zero and variance given by $\sum_{i=1}^n a_i \nu(A_i)$ considering the definition of $W$.

To calculate the scalar product between $f$ and $g=\sum_{i=1}^n b_i \mathbf{1}_{B_i}$ we assume without loss of generality $A_i=B_i$ (consider all of their interesections). Then $$\begin{aligned} \langle W(f), W(g) \rangle_{L^2(\mathbb{P})} &= \mathbb{E}\Bigl[\sum_{i=1}^n a_i W(A_i) \sum_{j=1}^n a_j W(A_j)\Bigr]\\ &= \sum_{i=1}^n a_i b_i \mathbb{E}[W(A_i)^2]\\ &= \int f(t) g(t) \nu(dt)\\ &= \langle f, g\rangle_{L^2(\nu)} \end{aligned}$$ Since the simple function are dense in $L^2$ and the scalar product is continuous we can deduce that $W$ is an isometry (where $W(f)$ for general $f$ is defined as the limit of $W(f_n)$ for simple functions $f_n$ approximating $f$). (1) then directly follows from the respective definitions of the scalar product (i.e. the second and penultimate term).

Felix Benning
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