How to obtain the formal scaling for the following PDE? In the paper by Mourrat and Weber( GLOBAL WELL-POSEDNESS OF THE DYNAMIC $\Phi^4_3$ MODEL ON THE TORUS) one reads:

Equation (1.1) is given by:

I couldn't arrive at (1.3)
Attempt: 
note that $\partial_{\hat t} = \lambda^2 \partial_t $ and $\partial_{\hat{x_i}} = \partial_{x_i}$ yields $\hat \Delta = \lambda^2 \Delta$ so:
$$\partial_{\hat t} \hat X = \lambda^2 \lambda^{\frac{2-d}{2}} \partial_t X = \lambda^2 \lambda^{\frac{2-d}{2}} \big(\Delta X  - X^3 + mX + \xi\big) \\=
\lambda^2 \lambda^{\frac{2-d}{2}}\Delta X  - \lambda^2 \lambda^{\frac{2-d}{2}}X^3 + \lambda^2 \lambda^{\frac{2-d}{2}}mX + \lambda^2 \lambda^{\frac{2-d}{2}}\xi$$
now note that 
$$\lambda^2\lambda^{\frac{2-d}{2}}\Delta X = \hat \Delta \hat X $$
$$\lambda^2\lambda^{\frac{2-d}{2}}mX = \hat m \hat X $$
So we would like to see that 
$$\lambda^2 \lambda^{\frac{2-d}{2}}X^3 = \hat X ^ 3 $$
$$ \lambda^2 \lambda^{\frac{2-d}{2}}\xi = \hat \xi $$
However $X^3 = \lambda^{-3\frac{2-d}{2}}\hat X^3$ so 
$$\lambda^2 \lambda^{\frac{2-d}{2}}X^3 =\lambda^2 \lambda^{\frac{2-d}{2}}\lambda^{-3\frac{2-d}{2}}\hat X^3 = \lambda^2 \lambda^{d-2}\hat X ^ 3  = \lambda ^d \hat X ^ 3$$
and $$  \lambda^2 \lambda^{\frac{2-d}{2}}\xi =  \lambda^2 \lambda^{\frac{2-d}{2}}\lambda^{-\frac{2 + d}{2}}\hat \xi  = \lambda ^ {2 - d}  \hat \xi$$
So I can' t arrive at the desired result. I am missing something here? 
 A: Here's my attempt,


*

*$\partial_tX$:


$$
\partial_tX = \frac{\partial X}{\partial t} = \frac{\lambda^2}{\lambda^{1 - d/2}}\frac{\partial \hat{X}}{\partial \hat{t}} = \lambda^{1+d/2}\partial_{\hat{t}}\hat{X}
$$


*$\Delta X$:


$$
\Delta X = \frac{\lambda^2}{\lambda^{1-d/2}}\frac{\partial \hat{X}}{\partial \mathbf{x}^2} = \lambda^{1+d/2}\hat{\Delta}\hat{X}
$$


*$X^3$:


$$
X^3 = \lambda^{3d/2 - 3}\hat{X}^3
$$


*$mX$:


$$
mX = \frac{\lambda^{d/2-1}}{\lambda^2} \hat{m}\hat{X} = \lambda^{d/2 - 3}\hat{m}\hat{X}
$$


*$\xi$:


$$
\xi = \lambda^{-1 - d/2}\hat{\xi}
$$
Putting all together we get
\begin{eqnarray}
\lambda^{1+d/2}\partial_{\hat{t}}\hat{X} &=& \lambda^{1+d/2}\hat{\Delta}\hat{X} - \lambda^{3d/2-3}\hat{X}^3 + \lambda^{d/2 - 3}\hat{m}\hat{X} + \lambda^{-1 - d/2}\hat{\xi} \\
\Rightarrow \qquad \partial_{\hat{t}}\hat{X} &=& \hat{\Delta}\hat{X} - \lambda^{d-4}\hat{X}^3 + \lambda^{-4} \hat{m}\hat{X} + \lambda^{-2-d}\hat{\xi}
\end{eqnarray}
Which is obviously not their Eq. (1.3). Unless there is something wrong with Eq. (1.3), I believe this should be the scaling:
$$
\hat{t} = \lambda^2t, \quad \hat{x} = \lambda x, \quad \hat{\xi} = \lambda^{(d-10)/4}\xi, \quad \hat{X} = \lambda^{(d-2)/4} \quad\mbox{and}\quad \hat{m} = \lambda^2m
$$
This last set is obtained just by simply assuming $\hat{t} = \lambda^{\alpha}t$, $\hat{x} = \lambda^{\beta}x$, $\cdots$ and then finding the numbers $(\alpha,\cdots)$ that match their Eq. (1.3)
