Proof of Pre-Stopping-Time Envelope and Restricted Moment Bounds for the State Fluctuation Process

Throughout, "the extended lemma" is the extended good-set stopping-time lemma, "the tracking lemma" is the pathwise tracking lemma, "the covariation lemma" is the stopped covariation lemma for the martingale part, and "the restricted-moments lemma" is the restricted-moments lemma. All notation is that of the statement.

Preliminary: the trajectory satisfies the flow hypotheses of the extended lemma. The map $S$ has continuous components and values in $\Delta^l$ with $S_0=x_0$, and for all $t,\gamma$, by clause 2 of the trajectory-pair definition, $S^\gamma_t=S^\gamma_0+\int_0^tb^\gamma(S_s,A_s)\,ds$ with a Riemann integral of a continuous integrand; by claim 3 of the integral toolkit this Riemann integral equals the Lebesgue integral $\int_{[0,t]}b^\gamma(S_s,A_s)\,ds$. Hence, with $S^*=S$, the hypotheses of claims 4 and 5 of the extended lemma hold, and in particular (claim 4 there) $S=S(x_0,\zeta_A)$. Since $\varepsilon_Y>0$ and $|x_0-S^*_0|=0<\varepsilon_Y$, the stopped deviation bound of claim 3 of the extended lemma is available:

Claim 1. Fix $\omega\in\Omega_0$ and $t\in[0,T]$, and put $u=\min(t,\tau^*(\omega))$. By clause (vii)(a) of the existence and uniqueness theorem, $\Sigma_u(\omega)\in\Delta^l$ and $\Sigma_0(\omega)\in\Delta^l$. By the triangle inequality for the Euclidean norm,

since $S_u(x_0,\hat{\alpha}(\omega))=\Phi_u(\omega)$ and $Y_u(\omega)=|\Phi_u(\omega)-S^*_u|$ with $S^*=S$. The first summand is at most $|M_u(\omega)|+\Lambda_be^{\Lambda_bu}\mathcal{M}_u(\omega)\le|M_u(\omega)|+\Lambda_be^{\Lambda_bT}\mathcal{M}_u(\omega)$ by claim 2 of the tracking lemma (whose $S^\omega$ is the flow $S(\Sigma_0(\omega),\hat{\alpha}(\omega))$) and the fact that the exponential function is nondecreasing. The second summand is at most $e^{\Lambda_bT}\,|\Sigma_0(\omega)-x_0|$ by claim 4 of the flow stability lemma, applied with base pair $(x_0,\hat{\alpha}(\omega))$ and perturbed pair $(\Sigma_0(\omega),\hat{\alpha}(\omega))$: the perturbed control equals the base control, so every $g^\gamma_r$ of claim 3 there vanishes and $G=0$ is admissible. Since $S_0=x_0$, $|\Sigma_0(\omega)-x_0|=N^{-1/2}|\mathfrak{s}_0(\omega)|$. The third summand is at most $\varepsilon_Y$ by (P1). Altogether

Next, $\mathfrak{b}_s(\omega)\le\overline{\mathfrak{b}}(\omega)$ for every $s\in[0,T]$ and $\omega\in\Omega_0$: by the supremum lemma, $\overline{M}^\gamma(\omega)=\sup_{r\in[0,T]}|M^\gamma_r(\omega)|$ for $\omega\in\Omega_0$, so $|M^\gamma_s(\omega)|\le\overline{M}^\gamma(\omega)$ for each $\gamma$, whence $|M_s(\omega)|^2=\sum_\gamma(M^\gamma_s(\omega))^2\le\sum_\gamma(\overline{M}^\gamma(\omega))^2=\overline{M}(\omega)^2$ and $|M_s(\omega)|\le\overline{M}(\omega)$, the nonnegative square root being nondecreasing; and $\mathcal{M}_s(\omega)\le\mathcal{M}_T(\omega)=I(\omega)$, the paths of $\mathcal{M}$ being nondecreasing by claim 1 of the tracking lemma and $I=\mathcal{M}_T$ on $\Omega_0$ by part (b) of the restricted-moments lemma. In particular, for $t<\tau^*(\omega)$ one has $\min(t,\tau^*(\omega))=t$, so $\mathbf{1}_{\{t<\tau^*\}}(\omega)|\mathfrak{s}_t(\omega)|\le\sqrt{N}\,\mathfrak{b}_t(\omega)\le\sqrt{N}\,\overline{\mathfrak{b}}(\omega)$; for $t\ge\tau^*(\omega)$ the left side vanishes and the bound is trivial, $\overline{\mathfrak{b}}$ being nonnegative.

Claim 6. Fix $\omega\in\Omega_0$ and $t\in[0,T]$. The displayed triangle inequality of Claim 1 and the two estimates following it were derived for an arbitrary point of $[0,T]$, the stopping time entering only through the substitution of the barrier $\varepsilon_Y$ for the third summand by (P1); carrying them out at the point $t$ itself rather than at $u=\min(t,\tau^*(\omega))$ therefore gives

By the second part of Claim 1, $|M_t(\omega)|\le\overline{M}(\omega)$ and $\mathcal{M}_t(\omega)\le I(\omega)$, so the sum of the first three summands is at most $\overline{\mathfrak{b}}(\omega)-\varepsilon_Y=Q(\omega)$. Multiplying by $\sqrt{N}$ gives $|\mathfrak{s}_t(\omega)|\le\sqrt{N}(Y_t(\omega)+Q(\omega))$, which is Claim 6.

Claim 2. Fix $\gamma$ and consider the process $X^\gamma=(\mathbf{1}_{\Omega_0}M^\gamma_t)_{t\in[0,T]}$. By claim 2 of the covariation lemma, $X^\gamma$ is progressively measurable with respect to $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$ — hence adapted, each $X^\gamma_t$ being $\mathcal{F}^{\mathrm{sys}}_t$-measurable by claim 1 of the progressive measurability toolkit — it is bounded in absolute value everywhere by the constant of that claim, whose value $1+2(l-1)BT$ is at most $K_M=2+2(l-1)BT$ (the two lemmas use the same letter for different constants), so each $X^\gamma_t$ is square-integrable (a bounded random variable is square-integrable by linearity and monotonicity of the integral), and for every $\omega\in\Omega_0$ its path agrees with $t\mapsto M^\gamma_t(\omega)$, which is right-continuous at every $t\in[0,T)$ in the sense required by the supremum lemma, again by claim 2 of the covariation lemma. The constant function $\tau\equiv T$ is a stopping time of $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$ by claim 1 of the stopping-time toolkit, so claim 4 of the covariation lemma gives, for all $0\le r\le t\le T$ and every $D\in\mathcal{F}^{\mathrm{sys}}_r$,

so by the averaged-form characterization in the martingale definition, $X^\gamma$ is a square-integrable martingale with respect to $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$, with time index restricted to $[0,T]$. All hypotheses of the fourth-moment maximal inequality thus hold for $X^\gamma$ with the event $\Omega_0$, and its conclusion reads $\mathbb{E}[(\overline{M}^\gamma)^4]\le4\,\mathbb{E}[(X^\gamma_T)^4]$. Pointwise $(X^\gamma_T)^4=\mathbf{1}_{\Omega_0}(M^\gamma_T)^4\le|M_T|^4$, since $(M^\gamma_T)^2\le\sum_\delta(M^\delta_T)^2=|M_T|^2$; so by part (b) of the restricted-moments lemma,

For reals $x_1,\dots,x_l$ one has $(\sum_\gamma x_\gamma)^2\le l\sum_\gamma x_\gamma^2$ (sum the inequalities $2x_\gamma x_\delta\le x_\gamma^2+x_\delta^2$ over all pairs); applying this with $x_\gamma=(\overline{M}^\gamma)^2$ gives $\overline{M}^4=\bigl(\sum_\gamma(\overline{M}^\gamma)^2\bigr)^2\le l\sum_\gamma(\overline{M}^\gamma)^4$, whence $\mathbb{E}[\overline{M}^4]\le4\,l^2\,c_M\kappa_T\,N^{-2}$. Each $\overline{M}^\gamma$ is a random variable with $0\le\overline{M}^\gamma\le K_M$ (the supremum lemma), so $\overline{M}$ is a random variable with $0\le\overline{M}\le\sqrt{l}\,K_M$, and $I$ is a random variable with $\mathbb{E}[I^4]\le T^4c_M\kappa_TN^{-2}$ by part (b) of the restricted-moments lemma; $|\mathfrak{s}_0|$ is a random variable (the Euclidean norm of a tuple of random variables, by measurability of continuous functions of measurable maps), so $Q$ and $\overline{\mathfrak{b}}=\varepsilon_Y+Q$ are random variables.

For nonnegative reals, $(a+b+c)^2\le3(a^2+b^2+c^2)$ (the case $l=3$ of the display above), so $(a+b+c)^4\le9(a^2+b^2+c^2)^2\le27(a^4+b^4+c^4)$. Hence

using $\mathbb{E}[|\mathfrak{s}_0|^4]\le\kappa_0$ and $\kappa_0\ge1$ to absorb the first two terms. Next, $(a+b)^2\le2(a^2+b^2)$ and $(a+b)^4\le4(a^2+b^2)^2\le8(a^4+b^4)$ give

and $\mathbb{E}[Q^2]=\mathbb{E}[Q^2\cdot1]\le\bigl(\mathbb{E}[Q^4]\bigr)^{1/2}$ by the Cauchy-Schwarz inequality for the mean-square norm, so $\mathbb{E}[\overline{\mathfrak{b}}^2]\le2\varepsilon_Y^2+2c_Q^{1/2}\kappa_0^{1/2}N^{-1}$.

Claim 3. Let $D'\in\mathcal{F}$ and $s,t\in[0,T]$. Measurability. By claim 2 of the covariation lemma, each $\mathbf{1}_{\Omega_0}\Sigma^\gamma$ is progressively measurable with respect to $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$, and $\tau^*$ is a stopping time of that filtration (claim 3 of the extended lemma), so by claim 4 of the stopping-time toolkit the sampled function $\omega\mapsto\mathbf{1}_{\Omega_0}(\omega)\Sigma^\gamma_{\min(t,\tau^*(\omega))}(\omega)$ is a random variable; the function $\omega\mapsto S^\gamma_{\min(t,\tau^*(\omega))}$ is a random variable as the composition of the measurable $\omega\mapsto\min(t,\tau^*(\omega))$ (for real $q$, $\{\min(t,\tau^*)\le q\}$ is $\Omega$ or $\{\tau^*\le q\}\in\mathcal{F}$) with the continuous $S^\gamma$, by measurability of continuous functions of measurable maps; hence $\mathbf{1}_{\Omega_0}\mathfrak{s}^\gamma_{\min(t,\tau^*)}=\sqrt{N}(\mathbf{1}_{\Omega_0}\Sigma^\gamma_{\min(t,\tau^*)}-\mathbf{1}_{\Omega_0}S^\gamma_{\min(t,\tau^*)})$ and $\mathbf{1}_{\Omega_0}|\mathfrak{s}_{\min(t,\tau^*)}|^2$ are random variables by the same composition lemma. Likewise $\{s<\tau^*\}\in\mathcal{G}_s\subseteq\mathcal{F}$ by claim 3 of the extended lemma, and $\mathbf{1}_{\Omega_0}\mathbf{1}_{\{s<\tau^*\}}|\mathfrak{s}_s|^2$ is a random variable ($\mathbf{1}_{\Omega_0}\mathfrak{s}^\gamma_s=\sqrt{N}(\mathbf{1}_{\Omega_0}\Sigma^\gamma_s-\mathbf{1}_{\Omega_0}S^\gamma_s)$ being a random variable, since $\mathbf{1}_{\Omega_0}\Sigma^\gamma_s$ is $\mathcal{F}^{\mathrm{sys}}_s$-measurable by progressive measurability and claim 1 of the progressive measurability toolkit, and $S^\gamma_s$ is a constant). Bounds. By claim 1, at every point of $\Omega_0$ one has $|\mathfrak{s}_{\min(t,\tau^*)}|^2\le N\overline{\mathfrak{b}}^2$ and $\mathbf{1}_{\{s<\tau^*\}}|\mathfrak{s}_s|^2\le N\overline{\mathfrak{b}}^2$; off $\Omega_0$ the left-hand integrands below vanish. Hence, by monotonicity of the expectation,

Moreover $\overline{\mathfrak{b}}^2\le2\varepsilon_Y^2+2Q^2$ pointwise, so

the middle step by Cauchy-Schwarz applied to $\mathbf{1}_{D'}$ and $Q^2$ together with $\mathbb{E}[\mathbf{1}_{D'}^2]=P(D')$. Multiplying by $N$ gives the asserted bounds. The fourth-moment bound follows the same way from $|\mathfrak{s}_{\min(t,\tau^*)}|^4\le N^2\overline{\mathfrak{b}}^4$ on $\Omega_0$ and claim 2: $\mathbb{E}[\mathbf{1}_{\Omega_0}|\mathfrak{s}_{\min(t,\tau^*)}|^4]\le N^2\mathbb{E}[\overline{\mathfrak{b}}^4]\le8N^2\varepsilon_Y^4+8c_Q\kappa_0$.

Claim 4. Since $\overline{\mathfrak{b}}=\varepsilon_Y+Q$ with $Q\ge0$, $\{\overline{\mathfrak{b}}\ge\varepsilon_Y+\theta\}=\{Q\ge\theta\}=\{Q^4\ge\theta^4\}$, the fourth power being nondecreasing on $[0,\infty)$. By Markov's inequality applied to the nonnegative random variable $Q^4$ at level $\theta^4>0$ and by claim 2,

Claim 5. Let $\omega\in\Omega_0$ and $t\in[0,T]$. By claim 2 of the realized-control lemma, $\hat{\alpha}(s,\omega)=\alpha_s(\omega)$ for every $s\in[0,T]$, so at every $s$,

The two integrands agree at every point of $[0,T]$, so their Lebesgue integrals over $[0,t]$ agree, and by linearity of the integral the constant factor $1/N$ may be taken outside: