\[ll4\] Denote as in Lemma \[ll1\], formula (\[rods\]) $$\rho_{d-k}(y_1,\dots ,y_{d-k};\mu_a^{(d-k)})=\frac{A(a)}{B(a)},\;k=0,\dots ,d-2.$$ Then we have for $y_1,\dots ,y_{d-k}$ with different orientations $$B(a)\geq \sum_{n=0}^\infty\left(\frac{aT}{d}\right)^n{\sum\dots\sum}_{n_1+\dots +n_d=n}\frac{1}{n_1!\dots n_d!}\times$$$$\times\exp\left(\nu_{d-k}(2b)^k\sum_{\{l_j\}_{j=1}^{d-k}\subset [d]}\prod_{i=1}^{d-k}n_{l_j}\right),$$

$$A(a)\leq \sum_{n=0}^\infty\left(\frac{aT}{d}\right)^n{\sum\dots\sum}_{n_1+\dots +n_d=n}\frac{1}{n_1!\dots n_d!}\times$$$$\times\exp\left(\nu_{d-k}b^k \left( \sum_{\{l_j\}_{j=1}^{d-k}\subset [d]}\prod_{i=1}^{d-k}n_{l_j}+\vartheta \right)\right),$$ where $\vartheta$ is the number of $(d-k)$-tuples (with different orientations) $$\{x_1,\dots ,x_{d-k}\}\subset\{u_1,\dots ,u_n,y_1,\dots ,y_{d-k}\}$$ such that there is at least one of $y_j$ among $x_1,\dots ,x_{d-k}.$

[**Proof:**]{} It follows from Lemma \[ll1\], inequality (\[boh\]) and integration w.r.t. (\[rdf\]). Here $n_j$ are interpreted as the number of $u_l,\;l=1,\dots n$ with orientation $e_j,\;j=1,\dots d.$ $\Box$

\[ll5\]Under the notation from Lemma \[ll4\] it holds $$B(a)\geq \exp(\frac{1}{d}aT(d-k-1)),\;k=0,\dots ,d-2.$$

[**Proof:**]{} Substituting $\beta=\frac{aT}{d}$ we have $$\begin{gathered}
\nonumber \sum_{n=0}^{\infty} \beta^{n} \sum_{n_{1}+ \ldots n_{d} = n} \frac{1}{n_{1}! \ldots n_{d}!} \exp(\nu_{d-k} (2b)^k \sum_{\{l_{j}\}_{j=1}^{d-k} \subset [d]} \prod_{j=1}^{d-k}n_{l_{j}}) =  \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d}}}{n_{1}! \ldots n_{d}!}  \exp(\nu_{d-k} (2b)^k \sum_{\{l_{j}\}_{j=1}^{d-k} \subset [d]} \prod_{j=1}^{d-k}n_{l_{j}}) = \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-1}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-1}}}{n_{1}! \ldots n_{d-1}!}\exp\left( \beta\exp(\nu_{d-k} (2b)^k \sum_{\{l_{j}\}_{j=0}^{d-k-1} \subset [d-1]} \prod_{j=1}^{d-k-1} n_{l_{j}}) \right) \times  \\
\nonumber \times \exp\left( \nu_{d-k} (2b)^k \sum_{\{l_{j}\}_{j=1}^{d-k} \subset [d-1]} \prod_{i=1}^{d-k}n_{l_{j}}\right) \geq  \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-1}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-1}}}{n_{1}! \ldots n_{d-1}!} \exp( \nu_{d-k} (2b)^k \sum_{\{l_{j}\}_{j=1}^{d-k} \subset [d-1]} \prod_{j=1}^{d-k}n_{l_{j}}) \geq \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-k}}}{n_{1}! \ldots n_{d-k}!} \exp( \nu_{d-k} (2b)^k \sum_{\{l_{j}\}_{j=1}^{d-k} 