On different sources I found different parabolic scalings for space time white noise that I believe are in contradicton one with the other.
Let $\xi(t,x)$ be space-time white noise on $\mathbb{R}\times\mathbb{R}^d$. I apply a scaling $t\to t\epsilon^{-\alpha}$, $x\to x\epsilon^{-\beta}$ and I want to find $\gamma$ such that the new noise $$\epsilon^{\gamma}\xi(t\epsilon^{-\alpha},x\epsilon^{-\beta})$$ has the same distribution as $\xi(t,x)$.
Which is the right $\gamma$? How can I compute it?
A covariance calculation suggests $$ \gamma=-\alpha\frac{d}{2}-\frac{\beta}{2} $$ but I found some sources which say that $\dot W(t,dx)dt$ has the same distribution as $$\epsilon^{\frac{d}{2}+1}\dot W(t\epsilon^{-2},dx\epsilon^{-1})dt$$ where $W$ is a cylindrical Wiener process on $L^2(\mathbb{R}^d)$, and this would correspond to saying that $$ \gamma=\alpha\frac{d}{2}+\frac{\beta}{2} \ .$$ Which is the right $\gamma?$
Am I missing something?
One year later, I suppose the issue has been solved. Clearly your computations are correct, but I personally found it surprising that at large scales white noise "decays" instead of exploding, which is what brought me to this question. Here's a simple and obvious argument as to why your result is not only correct, but also reasonable. We just consider white noise on the real line, then we can say: $$\xi (t) = \partial_t B(t)$$ for a Brownian motion $B$. Now we use the scaling invariance of the Brownian motion, namely $$B_{\epsilon}(t) : =\sqrt{\epsilon} B(t \epsilon^{-1}) \stackrel{d}{=}B(t).$$ Then by taking the time derivative we find: $$ \xi_{\epsilon}(t) = \partial_t B_{\epsilon}(t) = \frac{1}{\sqrt{\epsilon}} \partial_t B(t\epsilon^{-1}) \stackrel{d}{=} \partial_t B(t) = \xi(t). $$