and $ z \geqslant 0$. Besides, by the theorem about differentiability of the solutions of differential equations with respect to a parameter the functions $ F_{1}(z; t, \cdot) $ and $ F_{1}(z; t, \cdot, y) $ are continuously differentiable with respect to $z$ for $ z>0 $ as many times as the function $f(u)$, $ u \in [0, 1] $.

Let us now turn to the study of the first moments. To derive the differential equations for the first moments we will use the generating functions and obtained earlier Lemma \[L1\] with Remark \[Rem1\]. For it, we express the moments in terms of the Laplace generating functions: $$M_{1}(t, x) = - \lim_{z\to 0+}\partial_{z}F_{1}(z; t, x),\quad
M_{1}(t, x, y) = - \lim_{z\to 0+}\partial_{z}F_{1}(z; t, x, y).$$

\[L2\] The moments $M_{1}(t, x)$ and $M_{1}(0, x, y)$ satisfy the following backward Kolmogorov differential equations: $$\begin{aligned}
\label{E1.6}
\frac{\partial{M_{1}(t, x)} }{\partial{t}} &= (\mathscr{H}M_{1}(t, \cdot))(x),&&
M_{1}(0, x) \equiv 1,\quad x\in \mathbb{Z}^d,~t\geq 0,\\
\label{E1.7}
\frac{\partial{M_{1}(t, x, y)} }{\partial{t}} &= (\mathscr{H}M_{1}(t, \cdot, y))(x),&&
M_{1}(0, x, y) = \delta_{x}(y),\quad t\geq 0\end{aligned}$$

Let us derive equation ; equation can be derived similarly. By differentiating the equation $$\partial_{t}F_{1}(z; t, x, y) = (\mathscr{A}F_{1}(z; t, \cdot, y))(x) + \delta_{0}(x)f(F_{1}(z; t, x, y)),$$ 