P(t) := (x_{1}t + \cdots +x_{n}t^{n})^{r} $ and calculate its coefficient at $t^{n} $. By the polynomial formula this coefficient coincides with the left side of . On the other side, by direct gathering the components in the product $ (x_{1}t + \cdots +x_{n}t^{n})^{r} $ at $ t^{n} $, we get the right side of the formula .

Thus equation  is proved. Then rewrite the Faa di Bruno’s formula:

$$\frac{d^{n}f(F(z))}{dz^{n}} = \sum_{r=1}^{n} \frac{f^{(r)}(F(z))}{r!} \cdot \sum_{ \stackrel{i_{1},\ldots,i_{r}>0}{i_{1}+\cdots+i_{r}=n}}
\frac{n!}{i_{1}!\cdots i_{n}!} F^{(i_{1})}(z) \cdots  F^{(i_{r})}(z).$$

Let $ F(z) = F_{1}(z; t, x, y) $. Then $$\begin{gathered}
\delta_{0}(x)\partial_{z}^{n}f(F_{1}(z; t, x, y)) =  \sum_{r=1}^{n} \frac{f^{(r)}(F_{1}(z; t, x, y)) }{r!}\\
\times \sum_{ \stackrel{i_{1},\ldots,i_{r}>0}{i_{1}+\cdots+i_{r}=n}}
\frac{n!}{i_{1}!\cdots i_{n}!} \partial_{z}^{i_{1}} F_{1}(z; t,x,y) \cdots  \partial_{z}^{i_{r}}F_{1}(z, t, x, y).\end{gathered}$$ Using the expression of the moments in terms of generating functions, by the passage to the limit with respect to $z\rightarrow 0$ we get the required equation .

\[L4\] For $ n> 0$ and $ t\geq 0 $ the higher moments of subpopulations: $\eta_{x,t}$ and $\eta_{x,t}(y)$, at $y\in \mathbb{Z}^{d}$, whose initial position was at point $x$, satisfy the following integral equations: $$\label{E1.10}
M_{n}(t, x) = M_{1}(t, x) + \delta_{0}(x)\int_{0}^{t}M_{1}(t-q, x, 0)g_{n}(M_{1}(q,0),\ldots, M_{n-1}(q, 0))dq,$$ $$\begin{gathered}
\label{E1.11}
M_{n}(t, x, y) = M_{1}(t, x, y)\\ + \delta_{0}(x)\int_{0}^{t}M_{1}(t-q, 