Adapted from [math stack exchange](https://math.stackexchange.com/questions/4694934/just-how-regular-are-the-sample-paths-of-1d-white-noise-smoothed-with-a-gaussian).


**Background**: the prototypical example of---and way to generate---smooth noise is by convolving a one-dimensional white noise process with a Gaussian kernel.


**My question**: beyond smoothness, does one-dimensional white noise convolved with a Gaussian kernel have sample paths that are analytic functions with probability 1? Since there is a lot of existing literature on the sample path regularity of Gaussian processes I am hopeful that this already has been investigated somewhere, however, I haven't found an answer so far. Any help, pointers, or references are appreciated!


**Definitions:**


* White noise: White noise $\omega\_t$ is a mean-zero stationary Gaussian process on $t\in \mathbb R$ with autocovariance function given by: $\mathbb E[\omega\_t \omega\_s]= \delta(t-s)$, where $\delta$ is the Dirac delta.
* Gaussian kernel: is the function $\Phi\_t = e^{-\beta t^2}$ for all $t \in \mathbb R$ and some parameter $\beta >0$.
* White noise convolved with a Gaussian kernel: the process $w\_t=\int\_{\mathbb R} \omega\_s \Phi\_{t-s} d s$


**Known so far:**


* Properties of $ w\_t$: this is a mean-zero stationary Gaussian process with smooth sample paths and autocovariance given by


$$\kappa(h):=\mathbb E[w\_t w\_{t+h}] =e^{-\frac\beta 2h^2}\sqrt{\frac{\pi}{2 \beta}}.$$


* For any $n \in\mathbb N$, the derivative process $w^{(n)}\_t:=\frac{d^n}{dt^n}w\_t$ is a mean zero stationary Gaussian process with autocovariance
$\mathbb E[w^{(n)}\_t w^{(n)}\_{t+h}]= (-1)^n\kappa^{(2n)}(h)$. The latter comes from the **relation**:


\begin{equation}
\label{eq: autocov n m}
 \mathbb E[w^{(n)}\_t w^{(m)}\_{t+h}]= (-1)^n\kappa^{(n+m)}(h).
\end{equation}


To prove the relation we first **claim**:
$
 \mathbb E[w\_t w^{(m)}\_{t+h}]= \kappa^{(m)}(h).
$
We **prove the claim** by induction on $m$. The case $m=0$ is the definition. We show the statement for $m\geq 1$ assuming it holds for $m-1$:
\begin{align\*}
\mathbb E[w\_t w^{(m)}\_{t+h}] &= \lim\_{s \to 0}\frac 1 s\mathbb E[w\_{t}(w^{(m-1)}\_{t+h+s}-w^{(m-1)}\_{t+h})]\\
&= \lim\_{s \to 0}\frac 1 s \left( \kappa^{(m-1)}(h+s)-\kappa^{(m-1)}(h)\right) \\
&= \kappa^{(m)}(h).
\end{align\*}


We now **prove the relation**.
Fix $m\geq 0$. We prove the relation by induction on $n$. The case $n=0$ is the claim. We show the statement for $n\geq 1$ assuming it holds for $n-1$:
\begin{align\*}
\mathbb E[w^{(n)}\_t w^{(m)}\_{t+h}]&= \lim\_{s \to 0}\frac 1 s\mathbb E[(w^{(n-1)}\_{t+s}-w^{(n-1)}\_t)w^{(m)}\_{t+h}]\\
&= \lim\_{s \to 0}\frac 1 s \left( \mathbb E[w^{(n-1)}\_{t+s} w^{(m)}\_{t+h}]- \mathbb E[w^{(n-1)}\_{t} w^{(m)}\_{t+h}]\right)\\ %intermediate step
&= (-1)^{n-1}\lim\_{s \to 0}\frac 1 s \left( \kappa^{(n-1+m)}(h-s)-\kappa^{(n-1+m)}(h)\right) \\
&=(-1)^{n}\lim\_{s \to 0}\frac 1 s \left( \kappa^{(n-1+m)}(h+s)-\kappa^{(n-1+m)}(h)\right) \\ %intermediate step
&=(-1)^{n}\kappa^{(n+m)}(h).
\end{align\*}



Just how regular are the sample paths of 1D white noise smoothed with a Gaussian kernel?
Yes, $t \mapsto w\_t$ extends to an entire function $\mathbb{C} \to \mathbb{C}$ with probability $1$.
Here is an explicit construction for $(w\_t)\_{t \in \mathbb{R}}$.
Let $(X\_n)\_{n \ge 0}$ be independent real standard normal random variables, and let
$$w\_t := (\frac{\pi}{2 \beta})^\frac{1}{4} e^{-\frac{\beta}{2} t^2} \sum\_{n \ge 0} \frac{X\_n}{\sqrt{n!}} (\sqrt{\beta} t)^n$$
for $t \in \mathbb{R}$.
A basic fact about normal distributions:
If $d \ge 1$, $v\_n \in \mathbb{R}^d$ for $n \ge 0$, and $\sum\_n \|v\_n\|^2 < \infty$, then $V = \sum\_n X\_n v\_n$ is a real $d$-dimensional normal random vector, with $E[V] = 0$ and $E[V V^T] = (\sum\_n (v\_n)\_p (v\_n)\_q)\_{p,q \in \{1,\dotsc, d\}}$.
So $(w\_t)\_{t \in \mathbb{R}}$ is joint normal, $E(w\_t) = 0$, and $E[w\_t w\_{t+h}] = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} (t^2 + (t+h)^2)} \sum\_{n \ge 0} \frac{1}{n!} (\sqrt{\beta} t)^n (\sqrt{\beta} (t+h))^n = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} h^2}$.
We can extend $w\_t$ to $t \in \mathbb{C}$, then $E[\overline{w\_t} w\_{t+h}] = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} (\overline{t}^2 + (t+h)^2) + \beta \overline{t} (t+h)} = \sqrt{\frac{\pi}{2 \beta}} e^{-\frac{\beta}{2} (h + 2 \operatorname{Im}(t) i)^2}$.
Since $\lim\_{n \to \infty} \frac{C^n}{\sqrt{n!}} = 0$ for every $C>0$, to show that $t \mapsto w\_t$ is analytic with probability $1$, it is enough to show that if $q \in (0,1)$, then $\lim\_{n \to \infty} q^n X\_n = 0$ with probability $1$ (this is easy to check).


One could even use this method to simulate $(w\_t)\_{t \in \mathbb{R}}$ or $(w\_t)\_{t \in \mathbb{C}}$ for not too big $t$'s, by choosing a big enough $N$ and using $w\_t \approx (\frac{\pi}{2 \beta})^\frac{1}{4} e^{-\frac{\beta}{2} t^2} \sum\_{n = 0}^{N-1} \frac{X\_n}{\sqrt{n!}} (\sqrt{\beta} t)^n$ (for bigger $t$'s the convergence is slower, so one needs a bigger $N$).
I am sure there are faster algorithms, especially for big $t$'s, but at least this is pretty simple.


