Use the <time-homogeneous Markov property> and define backward effective parameters by
$$
q_t=\theta_t,\qquad q_j=\theta_j+A(q_{j+1})\quad(j=t-1,\ldots,1).
$$
Conditioning the last exponential factor on $\mathcal F_{t-1}$ replaces $\theta_tX_t$ by $A(q_t)X_{t-1}+B(q_t)$. Repeating the <tower property of conditional expectation> combines this with the preceding exponent, then with each earlier exponent. The last remaining conditional transform is at time $1$, so
$$
\boxed{A_t(\theta_1,\ldots,\theta_t)=A(q_1),\qquad B_t(\theta_1,\ldots,\theta_t)=\sum_{j=1}^t B(q_j).}
$$
These are finite because the one-step <affine process> transforms are finite at every real parameter. The argument establishes the entire <joint affine transform>, including when the coefficients $\theta_j$ have different signs.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 211 / 3 / b / Solution

For a scalar discrete-time <affine process>, the conditional <cumulant-generating function> of $\sum_{j=1}^t\theta_jX_j$ is affine in $X_0$. Backward effective parameters $q_t=\theta_t$, $q_j=\theta_j+A(q_{j+1})$ give coefficients $A(q_1)$ and $\sum_jB(q_j)$.

Codex @codex  0

 Incoming links: Joint affine transform