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Are there any stochastic processes $(X_t)_{t \in \mathbb{R}^d}$ such that

  1. almost surely paths are continuous but nowhere differentiable and
  2. sampling of $n$ points $X_{t_n}$ on a path can be done in $O(n)$ time?

Most sampling techniques I have in mind require at least $O(n \log n)$ time.

On the the other hand, all linear-time samplers I know, create processes such that 1) fails; like Perlin-noise, White noise, Pink noise, etc.

Is it even theoretically possible? So this might be a very fundamental question.

darksun
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  • This is a very interesting question also in relation to noise generation, shading. If you restrict your sampling points to a dense subset of $\mathbb{R}$ then the answer is yes. You could for instance use the standard Brownian motion and (following the proof of the Wiener theorem) take the dyadic rationals as sampling points. See https://www.math.uni-frankfurt.de/~neiningr/MP.pdf page 10. – JHT May 24 '21 at 17:49

2 Answers2

1

Let $(X_i)_{i\in \mathbb{N}_0}$ be a sequence of iid standard Normal distributed numbers. In dimension one define $$B_t^H = X_0 t + \sum_{k=0}^\infty X_{2^k+\lfloor 2^k t\rfloor}\frac{|2^k t-\lfloor 2^k t + \tfrac{1}{2}\rfloor|}{(2^k)^H}, \ t\in [0,1].$$ For $H=\tfrac{1}{2}$ this is the standard Brownian motion and its paths are continuous but nowhere differentiable (almost surely).

When $t$ is a dyadic rational ($t \in \{\tfrac{i}{2^m} : i,m \in \mathbb{N}_0\}$) the above series has only $m$ summands.

That means, sampling at these positions can be done in linear time. As this is a dense subset and numbers in a computer are represented anyway by binaries this should do the trick.

In dimension two take $(X_{i,j})_{i,j\in \mathbb{N}_0}$ and define $$B_{s,t}^H = X_{0,0} \tfrac{s+t}{2} + \sum_{k=0}^\infty \frac{(\tilde X +\dot X) a(2^k s,2^k t)+(\tilde X -\dot X)b(2^k s,2^k t)}{(2^k)^H}$$ for $ \ (s,t)\in [0,1]^2$, where $$\tilde X = X_{2^k+\lfloor 2^k s\rfloor,2^k+\lfloor 2^k t\rfloor},\\ \dot X = X_{2^k+(\lfloor 2^k s\rfloor+ c(2^k s,2^k t)) \text{mod}\, 2^k,2^k+ (\lfloor 2^k t\rfloor+ c(2^k t,2^k s)) \text{mod}\, 2^k},\\ c(x,y)=\tfrac{\text{sign}(\{x\}+\{y\}-1)+\text{sign}(\{x\}-\{y\})}{2},\\ a(x,y)=1 -||\{x\}+\{y\}-1|-|\{x\}-\{y\}||,\\ b(x,y)=1- |\{x\}+\{y\}-1|-|\{x\}-\{y\}| $$ and $\{\cdot\}$ denotes the fractional part.

JHT
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1

I believe the Ornstein-Uhlenbeck process has these properties. This is based on another answer on this site about simulating the OU process I read a while back, which unfortunately I cannot find at the moment.

If $dX_t = -\theta X_t dt + \sigma dW_t$, with $X_0=0$, then $X$ is a Gaussian process with covariance function $\Gamma(s,t)=\frac{\sigma^2}{2\theta}e^{-\theta|t-s|}$. To sample $X_{t_{i+1}}$ from $X_{t_i}$, we will use the fact that the Ornstein-Uhlenbeck process has a stationary distribution, $X_t \sim N(0,\frac{\sigma^2}{2\theta})$ for all $t$. Let $(Z_i)$ be a sequence of i.i.d. $N(0,\frac{\sigma^2}{2\theta})$ random variables, define $\rho_i := \frac{\sigma^2}{2\theta}e^{-\theta(t_{i+1}-t_i)}$, and set $X_{t_{i+1}} := \rho_i X_{t_i} + \sqrt{1-\rho_i^2}Z_i$. We can compute $(X_{t_i},X_{t_{i+1}})$ has the correct joint distribution, and because of the structure of $\Gamma$ we can show that $(X_{t_1},X_{t_2},...,X_{t_n})$ also has the correct joint distribution. Hence this gives a way to sample $n$ points from a path of $X$. Since the $X_{t_i}$ are generated through a simple linear recurrence, sampling all $n$ points takes only $O(n)$ time.

user6247850
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