The conditional Gaussian mixture is a nonnegative supermartingale

For fixed $u\in {\mathbb{R}}^d$, define $M_t(u)=\exp \text{\hspace{0.17em}⁣}(u^{\mathsf{T}}{S}_t-\frac{1}{2}{u}^{\mathsf{T}}{G}_{t}u)$, $M_0(u)=1$, and $R_t(u)=e^{Q_t{\eta }_t-Q_t^2\mathrm{/}2}$ where $Q_t=u^{\mathsf{T}}{A}_t$. The model makes $Q_t$ ${\mathcal{F}}_t$-measurable and, by [F3], bounds $\mathrm{\mid }{Q}_t\mathrm{\mid }\le L\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Thus [F2] gives $\mathbb{E}[R_t(u)\mid {\mathcal{F}}_t]\le 1$ and [F6] gives $\mathbb{E}[R_t(u)]\le 1$. Also $M_t(u)=M_{t-1}(u)R_t(u)$, and $M_t(u)$ is ${\mathcal{H}}_t={\mathcal{F}}_{t+1}$-measurable.

For fixed $t$, each coordinate of $S_t$ and $G_t$ is a finite sum and product of variables observed by time $t$. On $(\mathrm{\Omega }\times {\mathbb{R}}^d,{\mathcal{H}}_t\otimes \mathcal{B}({\mathbb{R}}^d))$, the projections from each factor are measurable, so each coordinate of $S_t(\omega )$, $G_t(\omega )$, and $u$ is measurable. The Borel sigma-algebra of the finite-dimensional Euclidean product is generated by coordinate projections, since rational open boxes form a countable basis; hence $(\omega ,u)\mapsto (S_t(\omega ),G_t(\omega ),u)$ is measurable. The maps $(S,G,u)\mapsto u^{\mathsf{T}}S-\frac{1}{2}{u}^{\mathsf{T}}Gu$ and $x\mapsto e^x$ are continuous, so $(\omega ,u)\mapsto M_t(u,\omega )$ is product-measurable by [F11]. Tonelli's section-measurability clause [F10] therefore makes ${\bar{M}}_t=\int M_t(u)p_{\lambda }(u)du$ ${\mathcal{H}}_t$-measurable. The density $p_{\lambda }(u)=(\lambda \mathrm{/}(2\pi ){)}^{d\mathrm{/}2}{e}^{-\lambda \mathrm{\parallel }u{\mathrm{\parallel }}_2^2\mathrm{/}2}$ is nonnegative and integrates to $1$ by [F8] with $V=\lambda I_d$.

By the definitions of $S_t,G_t,V_t$ and the Gaussian density, ${\bar{M}}_t={\left(\frac{\lambda }{2\pi }\right)}^{d\mathrm{/}2}{\int }_{{\mathbb{R}}^d}\exp \text{\hspace{0.17em}⁣}(u^{\mathsf{T}}{S}_t-\frac{1}{2}{u}^{\mathsf{T}}{V}_{t}u)du\mathrm{.}$ Completing the square gives $u^{\mathsf{T}}{S}_t-\frac{1}{2}{u}^{\mathsf{T}}{V}_{t}u=\frac{1}{2}{S}_t^{\mathsf{T}}{V}_t^{-1}{S}_t-\frac{1}{2}(u-V_t^{-1}{S}_t{)}^{\mathsf{T}}{V}_t(u-V_t^{-1}{S}_t)\mathrm{.}$

For fixed $u,t$ and every $B\in {\mathcal{F}}_t$, step 1.1 and the defining event integrals [F5] give $\mathbb{E}[{\mathbf{1}}_B{R}_t(u)]\le P(B)$. Hence for any nonnegative finite-valued ${\mathcal{F}}_t$-measurable $W={\sum }_j{c}_j{\mathbf{1}}_{B_j}$, summing these inequalities gives $\mathbb{E}[W{R}_t(u)]\le \mathbb{E}[W]$. For a bounded nonnegative ${\mathcal{F}}_t$-measurable $W$, the finite-valued dyadic approximants $W_r=2^{-r}\lfloor 2^{r}W\rfloor$ increase to $W$. Since $R_t(u)\ge 0$, monotone convergence [F7] on both sides gives the same inequality for $W$.

Induct on $t$ to prove integrability and the supermartingale inequality. The base $M_0(u)=1$ is integrable. If $M_{t-1}(u)$ is integrable, put $W_k=\min \{M_{t-1}(u),k\}$, a bounded nonnegative ${\mathcal{F}}_t$-measurable variable. Step 2.1 gives $\mathbb{E}[W_k{R}_t(u)]\le \mathbb{E}[W_k]$. Since $W_k{R}_t(u)\uparrow M_t(u)$, [F7] gives $\mathbb{E}[M_t(u)]\le \mathbb{E}[M_{t-1}(u)]\le 1$. Replacing $W_k$ by ${\mathbf{1}}_B{W}_k$ for any $B\in {\mathcal{F}}_t$ and applying [F7] again gives $\mathbb{E}[{\mathbf{1}}_B{M}_t(u)]\le \mathbb{E}[{\mathbf{1}}_B{M}_{t-1}(u)]$. Let $Y_t$ be a conditional-expectation version of $M_t(u)$ given ${\mathcal{F}}_t$, which exists by [A1] and [F5]. For each $k\ge 1$, $B_k:=\{Y_t\ge M_{t-1}(u)+1\mathrm{/}k\}\in {\mathcal{F}}_t$. On $B_k$, the event-integral property in [F5] and the preceding inequality imply ${\int }_{B_k}{Y}_{t}dP={\int }_{B_k}{M}_t(u)dP\le {\int }_{B_k}{M}_{t-1}(u)dP,$ whereas the definition of $B_k$ makes the left side at least the right side plus $P(B_k)\mathrm{/}k$. Thus $P(B_k)=0$ for every $k$; since $\{Y_t>M_{t-1}(u)\}={\bigcup }_{k\ge 1}{B}_k$, we have $Y_t\le M_{t-1}(u)$ almost surely. This is the supermartingale conditional-expectation inequality.

For any $B\in {\mathcal{H}}_{t-1}={\mathcal{F}}_t$, Tonelli [F10] and step 3.1 give $\mathbb{E}[{\mathbf{1}}_B{\bar{M}}_t]=\int \mathbb{E}[{\mathbf{1}}_B{M}_t(u)]p_{\lambda }(u)du\le \int \mathbb{E}[{\mathbf{1}}_B{M}_{t-1}(u)]p_{\lambda }(u)du=\mathbb{E}[{\mathbf{1}}_B{\bar{M}}_{t-1}]\mathrm{.}$ Taking $B=\mathrm{\Omega }$ shows ${\bar{M}}_t$ is integrable. Let $Y_t$ be a conditional-expectation version of ${\bar{M}}_t$ given ${\mathcal{H}}_{t-1}$; it exists by [A1] and [F5]. For each $k\ge 1$, $B_k:=\{Y_t\ge {\bar{M}}_{t-1}+1\mathrm{/}k\}\in {\mathcal{H}}_{t-1}$. The event-integral property [F5] and the displayed event inequality imply ${\int }_{B_k}{Y}_{t}dP=\mathbb{E}[{\mathbf{1}}_{B_k}{\bar{M}}_t]\le \mathbb{E}[{\mathbf{1}}_{B_k}{\bar{M}}_{t-1}],$ while the definition of $B_k$ makes the left side at least the right side plus $P(B_k)\mathrm{/}k$. Thus every $B_k$ is null and $Y_t\le {\bar{M}}_{t-1}$ almost surely. Together with adaptedness from step 1.2 and ${\bar{M}}_0=1$, this is the integrable supermartingale property.

The translation $w=u-V_t^{-1}{S}_t$ is a $C^1$ diffeomorphism with Jacobian determinant $1$, so [F9] and [F8] evaluate the last integral as $(2\pi {)}^{d\mathrm{/}2}\mathrm{/}\sqrt{\det V_t}$. Since $\det V_0={\lambda }^d$ by [F12], this gives ${\bar{M}}_t=\sqrt{\frac{\det V_0}{\det V_t}}\exp \text{\hspace{0.17em}⁣}\left(\frac{1}{2}{S}_t^{\mathsf{T}}{V}_t^{-1}{S}_t\right)=Z_t\mathrm{.}$ Thus the process in the Statement is the supermartingale established in step 4.1.