Suppose $(X_t)_{t\geq 0}$ diffuses as,
$$ dX_t = \mu(X_t)\, dt + \sigma(X_t) \, dW_t $$
and,
$$ g(t,x)=\mathbb{E}[g(T,X_T)\vert\mathcal{F}_t] $$
By Feynman-Kac we have,
$$ \frac{\partial}{\partial t}g(t,x)=-\mathcal{L}g(t,x) $$
Where,
$$ \mathcal{L}=\mu(X_t)\frac{\partial}{\partial x}+\frac{1}{2}\sigma^2(X_t)\frac{\partial^2}{\partial x^2} $$
Why is it that we are allowed to approximate $X_t$ by,
$$ X_t = e^{\mathcal{L}\sqrt{t}Z}X_0 $$
Where $Z\sim N[0,1]$?
EDIT: For those interested, there was a mistake in the question. This is a notation I'm not used to, since typically one knows $X_t$ has no time derivative almost surely. First, let $g(t,X_t)=X_t$ so that,
$$ \frac{\partial}{\partial t}X_t = -\mathcal{L}X_t $$
Now choose differential operators $V_0$ and $V_1$ so that,
$$ \mathcal{L} = V_0+\frac{1}{2}V_1^2 $$
Then apparently cubature of degree 3 is the approximation,
$$ X_t = e^{V_0 t + V_1\sqrt{t}Z}X_0 $$