of $\gamma_{jk}$ which are not zero. Since $\|{\widetilde{A}}(t, \omega)\| = \| A(t,\omega) \| / \lambda(t,\omega) = \alpha /
\lambda(t,\omega)$, we need a lower bound of $\lambda(t,\omega)$ to get a finite ${\widetilde{\alpha}}$ as defined in . Assume $\lambda(t,\omega)
\geqslant \ell$, and then the estimation becomes $$\label{eq:variance}
{\mathbb{E}}_{\Xi} \braket{\Phi_{\Xi}(t)} \leqslant
  \mathrm{e}^{t \Lambda [1 + (\alpha / \ell)^2]} \braket{\Psi(0)}.$$ Using the inequality $$\Lambda [1 + (\alpha / \ell)^2] \geqslant
  \Lambda [1 + (\alpha / \Lambda)^2] \geqslant 2\alpha,$$ we get a minimum value for the right hand side of , which can be achieved when $\lambda(t,\omega) \equiv \ell = \Lambda = \alpha$. Such a choice also turns and into equalities, and we eventually have $$\label{eq:optimal}
{\mathbb{E}}_{\Xi} \braket{\Phi_{\Xi}(t)} = \mathrm{e}^{2\alpha t} \braket{\Psi(0)}.$$ In fact, we can also show in this case that the variance is minimized by choosing $\lambda(t, \omega) \equiv \alpha$. To simplify the notation, we define $$w_{\Xi}(t) = \exp \left(
  \int_0^t \int_{\Omega} \lambda(s,\omega)
    \,\mathrm{d}\mu_{\omega} \,\mathrm{d}s
\right) \prod_{m=1}^M \frac{\alpha}{\lambda(t_m,\omega_m)}.$$ Then $\braket{\Phi_{\Xi}(t)} = [w_{\Xi}(t)]^2 \braket{\Psi(0)}$. Applying to $w_{\Xi}(t)$ yields ${\mathbb{E}}_{\Xi} w_{\Xi}(t) =
\mathrm{e}^{\alpha t}$, and thus we get the following estimation of the lower bound: $$\begin{split}
{\mathbb{E}}_{\Xi} \braket{\Phi_{\Xi}(t)} &=
  \braket{\Psi(0)} {\mathbb{E}}_{\Xi} [w_{\Xi}(t)]^2 \\
& \geqslant \braket{\Psi(0)} [{\mathbb{E}}_{\Xi} w_{\Xi}(t)]^2 =
  \mathrm{e}^{2 \alpha t} \braket{\Psi(0)}.
\end{split}$$ From 