involves $$\begin{aligned}
{\mathrm{d}}\tilde{W}_k(t) = dW_k(t)+ \rho_k (a^i_kV_k(t)^{1/2}+b^i_kV_k(t)^{-1/2}) {\mathrm{d}}t,\end{aligned}$$ where under classical assumptions $\tilde{W}_k$ should be a Wiener process under the putative risk neutral measure $\mathbb{Q}^i$. Under this measure the process $V_k$ would then solve the SDE $$\begin{aligned}
 \label{varprocess2}
{\mathrm{d}}V_k(t) &= \kappa_k  ( \theta_k - V_k(t)) {\mathrm{d}}t - \rho_k \sigma_k (a^i_kV_k(t)+b^i_k){\mathrm{d}}t+ \sigma_k V_k(t)^{\frac{1}{2}} {\mathrm{d}}\tilde{W}_k(t)\nonumber \\
&= (\kappa_k  \theta_k- \rho_k \sigma_k b^i_k ){\mathrm{d}}t -\kappa_k (1+\frac{\rho_k \sigma_k a^i_k}{\kappa_k})V_k(t){\mathrm{d}}t+ \sigma_k V_k(t)^{\frac{1}{2}} 
{\mathrm{d}}\tilde{W}_k(t). 
$$ Under $\mathbb{P}$, the process $V_k$ does not reach 0 if the Feller condition is satisfied, i.e. $$2\kappa_k \theta_k \geq \sigma_k^2,
$$ while under the putative risk neutral measure the process $V_k$ would not reach 0 if the corresponding Feller condition would be satisfied, that is $$2\kappa_k \theta_k \geq\sigma_k^2+2\rho_k \sigma_k b^i_k.
$$ Therefore, the process $V_k$ would have a different behavior at 0 under the two measures, provided that $$\sigma_k^2\leq 2\kappa_k \theta_k <\sigma_k^2+2\rho_k \sigma_k b^i_k.
\label{eq:strictLocalMartingaleTest}$$ In this case the putative risk neutral measure would not be an equivalent probability measure and classical risk neutral pricing would not be well-founded.\
In order to get an intuition of what is the typical path behavior when dealing with true and strict local martingales, we simulate some paths of the Radon-Nikodym derivative for the putative 