Let $\xi$ be a random variable valued in the space of Schwartz distributions $\mathcal{S}'(\mathbb{R}^d)$.


For any open set $R\subset\mathbb{R}^d$ let $\Sigma(R)$ be the $\sigma$-algebra generated by $\{\xi(f)\,|\,f\in\mathcal{S}(\mathbb{R}^d), \,\textrm{supp}\,f\subset R\}$.


For any $\lambda\in(0,\infty)$ the random variable $\xi\_\lambda$ valued in $\mathcal{S}'(\mathbb{R}^d)$ is defined by $\xi\_\lambda(f) = \xi(f\_\lambda)$ for all $f\in\mathcal{S}(\mathbb{R}^d)$, where $f\_\lambda(x) = \lambda^{d/2}\,f(\lambda x)$.


Assume that:


(1) the $\sigma$-algebras $\Sigma(R\_1)$ and $\Sigma(R\_2)$ are independent for any disjoint open sets $R\_1,R\_2\subset\mathbb{R}^d$.


(2) for any $\lambda\in(0,\infty)$ the law of $\xi\_\lambda$ coincides with the law of $\xi$.


Is it true that under the above assumptions $\xi$ must be the white noise?



Abstract characterization of white noise
These assumptions are not sufficient. Take $d=1$ and for simplicity let us work with the antiderivative of $\xi$. You are then asking if a process with independent increments and Brownian scaling is necessarily a Brownian motion. Any $\alpha$-stable Levy process $L$ has independent increments and scaling $(L\_t)\_{t\geq 0}\sim\lambda^{1/\alpha}(L\_{\lambda t})\_{t\geq0}$, so the process $X\_t=L\_{t^{\alpha/2}}$ is a counterexample.


My guess is that translation adding translation invariance and nontriviality (note that actually $\xi\equiv 0$ is also a counterexample) to the assumptions should be sufficient.


