and $\psi$ satisfy the *semi-flow property*, i.e. for all $0 < s \le r \le t$ and $u \in \cU$, $$\begin{aligned}
    \label{eq:flow}
    \begin{array}{r@{}lrl}
    \phi_s(t,u) &= \phi_r(t,u) + \phi_s(r,\psi_r(t,u)), &\qquad \phi_t(t,u) &= 0\\
    \psi_s(t,u) &= \psi_s(r,\psi_r(t,u)), &\qquad \psi_t(t,u) &= u.
    \end{array}\end{aligned}$$

To show the first property, recall that by Equation we have $$\label{eq:affine_repeat}
E\big[ e^{\scal{u}{X_{t}}} |\cF_{s} \big] = \exp\big( \phi_s(t,u) + \scal{\psi_s(t,u)}{X_s} \big)$$ for all $u \in \cU$ and $0 \le s \le t$. Since $\scal{\Re u}{X_{t}} \le 0$, a.s., the left hand side is bounded by one in absolute value. Thus, also $$\Re \phi_s(t,u) + \scal{\Re \psi_s(t,u)}{X_s} \le 0,\  a.s.$$ and consequently $$\Re \phi_s(t,u) + \scal{\Re \psi_s(t,u)}{x} \le 0, \quad \text{for all $x \in \supp(X_s)$.}$$ Taking arbitrary convex combinations of these inequalities and using that $\conv(\supp(X_s)) = D$ by Condition \[def:full\_support\], we obtain that the inequality must in fact hold for all $x \in D$. Since $D$ is a cone this implies that $\Re \phi_s(t,u) \le 0$ and $\psi_s(t,u) \in \cU$, proving (i).

To show the semi-flow equations we apply iterated expectations to the left hand side of , yielding $$\begin{aligned}
E\big[ E\big[ e^{\scal{u}{X_t}} |\cF_{r} \big] |\cF_{s} \big] &= E\big[ \exp\big( \phi_r(t,u) + \scal{\psi_r(t,u)}{X_r} \big) |\cF_{s} \big] = \\
&= 