On some estimates for Erd\"os-R\`enyi random graph

TL;DR

This paper derives asymptotic formulas for the factorial moments of the number of components in Erdős-Rényi graphs and provides bounds for the probabilities of connectivity and isolated vertices.

Contribution

It introduces new asymptotic relations for factorial moments and bounds for key graph properties in Erdős-Rényi models using generating functions.

Findings

Asymptotic relation for factorial moments of component counts

Bounds for the probability of graph connectivity

Bounds for the probability of isolated vertices

Abstract

We consider a number $\nu_n$ of components in a random graph $G(n,p)$ with $n$ vertices, where the probability of an edge is equal to $p$. By operating with special generating functions we shows the next asymptotic relation for factorial moments of $\nu_n$: $$ \mathsf{E}(\nu_n-1)^{\underline s} = (1+o(1))\left( \frac 1p \sum\limits_{k=1}^\infty\frac{k^{k-2}}{k!}(npq^n)^k\right)^s + o(1) $$ as $n$ tends to $\infty$ and $q=1-p$. And the following inequations hold: $$ 1-2nq^{n-1} \le p_n\le\frac{1}{nq^n}, $$ $$ 1-\frac{1}{nq^n}\le pi_n\le nq^{n-1}, $$ where $p_n$ is the probability that $G(n,p)$ is connected and $pi_n$ is the probability that $G(n,p)$ has an isolated vertex.

Equations

$$\mathsf{E}({\nu }_n-1{)}^{\underline{s}}=(1+o(1)){\left(\frac{1}{p}\sum _{k=1}^{\infty }\frac{k^{k-2}}{k!}(np{q}^n{)}^k\right)}^s+o(1)$$

$$1-2n{q}^{n-1}⩽p_n⩽\frac{1}{n{q}^n},$$

$$1-\frac{1}{n{q}^n}⩽p{i}_n⩽n{q}^{n-1},$$

$$\mathsf{P}\{G(n,p)=g\}=(p/q{)}^{E(g)}{q}^{n(n-1)/2},$$

$$\mathsf{P}\{{\nu }_n=s\}=\sum _{k=0}^{\infty }{A}_{n,k,s}(p/q{)}^k{q}^{n(n-1)/2}$$

$$p_n=\sum _{k=0}^{\infty }{B}_{n,k}(p/q{)}^k{q}^{n(n-1)/2}.$$

$$R=R(x,q)=\sum _{n=0}^{\infty }\frac{x^n}{n!q^{n(n-1)/2}}{r}_n(q),$$

$$\hat{R}=\sum _{n=0}^{\infty }\frac{x^n}{n!q^{n(n-1)/2}}\frac{d{r}_n(q)}{dq},$$

$$A$$

$$B$$

$$E$$

$$M_k$$

$${\mathcal{M}}_k$$

$$A_{n,k,s}=\sum _{\genfrac{}{}{0ex}{}{n_1+\cdots +n_s=n}{k_1+\cdots +k_s=k}}\frac{n!}{s!}\frac{B_{n_1,k_1}\cdots B_{n_s,k_s}}{n_1!\cdots n_s!},$$

$${\bar{A}}_{n,k,s}=\sum _{\genfrac{}{}{0ex}{}{n_1+\cdots +n_s=n}{k_1+\cdots +k_s=k}}\frac{n!B_{n_1,k_1}\cdots B_{n_s,k_s}}{n_1!\cdots n_s!}$$

$$A(x,y)=\sum _{n,k}\frac{A_{n,k}}{n!}{x}^n{y}^k,B(x,y)=\sum _{n,k}\frac{B_{n,k}}{n!}{x}^n{y}^k.$$

$$A(x,y)=e^{B(x,y)}$$

$$\frac{A_{n,k,s}}{n!}{x}^n{y}^k=\frac{1}{s!}\sum _{\genfrac{}{}{0ex}{}{n_1+\cdots +n_s=n}{k_1+\cdots +k_s=k}}\frac{B_{n_1,k_1}{x}^{n_1}{y}^{k_1}\cdots B_{n_s,k_s}{x}^{n_s}{y}^{k_s}}{n_1!\cdots n_s!}=[x^n{y}^k]B(x,y{)}^s.$$

$$\sum _{n,k}\frac{A_{n,k,s}}{n!}{x}^n{y}^k=\frac{1}{s!}B(x,y{)}^s$$

$$A(x,y)=\sum _{s=0}^{\infty }\frac{B(x,y{)}^s}{s!}=e^{B(x,y)}.$$

$$A_{n,k}=\left(\genfrac{}{}{0ex}{}{n(n-1)/2}{k}\right),$$

$$\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{n(n-1)/2}{k}\right)y^k=\sum _{k=0}^{n(n-1)/2}\left(\genfrac{}{}{0ex}{}{n(n-1)/2}{k}\right)y^k=(1+y{)}^{n(n-1)/2},$$

$$A(x,y)=\sum _{n=0}^{\infty }(1+y{)}^{n(n-1)/2}\frac{x^n}{n!}.$$

$$B(x,y)=\ln \sum _{n=0}^{\infty }(1+y{)}^{n(n-1)/2}\frac{x^n}{n!}.$$

$$\sum _k{A}_{n,k}{\left(\frac{p}{q}\right)}^k{q}^{n(n-1)/2}=1,\sum _k{B}_{n,k}{\left(\frac{p}{q}\right)}^k{q}^{n(n-1)/2}=p_n,$$

$$A\left(x,\frac{p}{q}\right)=\sum _{n,k}\frac{A_{n,k}}{n!}{x}^n(p/q{)}^k=\sum _n\frac{x^n}{q^{n(n-1)/2}n!}=A,$$

$$B\left(x,\frac{p}{q}\right)=\sum _{n,k}\frac{B_{n,k}}{n!}{x}^n(p/q{)}^k=\sum _n\frac{p_n{x}^n}{q^{n(n-1)/2}n!}=B.$$

$$A=e^B.$$

$$\sum _{n=0}^{\infty }\frac{\mathsf{P}\{{\nu }_n=s\}}{q^{n(n-1)/2}}\frac{x^n}{n!}=\sum _{n,k}\frac{A_{n,k,s}}{n!}{x}^n(p/q{)}^k,$$

$$\sum _{n=0}^{\infty }\frac{x^n}{q^{n(n-1)/2}n!}\mathsf{P}\{{\nu }_n=s\}=\frac{1}{s!}B(x,p/q{)}^s=\frac{1}{s!}{B}^s,$$

$$R=\sum _{n=0}^{\infty }\frac{r_n{x}^n}{n!q^{n(n-1)/2}},T=\sum _{n=0}^{\infty }\frac{t_n{x}^n}{n!q^{n(n-1)/2}},RT=\sum _{n=0}^{\infty }\frac{z_n{x}^n}{n!q^{n(n-1)/2}}.$$

$$z_n=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)q^{k(n-k)}{r}_k{t}_{n-k}.$$

$$p_n=1-\sum _{k=1}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)q^{k(n-k)}{p}_{n-k}.$$

Full text

[TABLE]

as $n$ tends to $\infty$ and $q=1-p$. And the following inequations hold:

[TABLE]

[TABLE]

where $p_{n}$ is the probability that $G(n,p)$ is connected and $pi_{n}$ is the probability that $G(n,p)$ has an isolated vertex.

1 Notations

Let $G_{n}$ be a set of undirected graphs with $n$ labeled vertices. For any graph $g\in G_{n}$ let $C(g)$ be a number of connected components in the graph $g$ and $E(g)$ be a number of edges in thr graph $g$. Besides we denote by $F_{s,n}$ the number of all forests in $G_{n}$, that contains exactly $s$ trees. We also suppose that components in $G_{n}$ are not ordered.

Further, let $A_{n,k,s}$ be a number of graphs in $G_{n}$, which contains $n$ vertices, $k$ edges and $s$ components, $A_{n,k}$ be a number of graphs, which contains $n$ vertices and $k$ edges, and $B_{n,k}$ — a number of connected graphs with $n$ vertices and $k$ edges. For definiteness we suppose that $A_{0,k}=A_{0,k,s}=A_{n,k,0}=0$ in all cases, except $n=k=s=0$, where we set by definition $A_{0,0}=A_{0,0,0}=1$. Besides, let $B_{0,k}=0$ for all $k$. It’s clear that $A_{n,k}=\sum_{s}A_{n,k,s}$, where index $s$ runs on all integer non-nagative numbers.

Let us consider the random graph $G(n,p)$, which contains $n$ labeled vertices, where each of $\binom{n}{2}$ edges is present with the probability $p$ independently of other edges. Each concrete realization of random graph $G(n,p)$ is a graph from $G_{n}$. This model of random graphs was firstly described by Erdös and Rényi in [1, 2] and then has been well studied by Béla Bollobás [3], Valentin Kolchin [4] and other authors.

It is easy to see that the parobability distribution of such random graph is defined as follows:

[TABLE]

where $g\in G_{n}$ and $q=1-p$.

Let denote by $\nu_{n}$ the number of connected components of $G(n,p)$, i. e. $\nu_{n}=C(G(n,p))$, and let $p_{n}$ be the probability that random graph $G(n,p)$ is connected, thus $p_{n}=\mathsf{P}\{\nu_{n}=1\}$. It’s clear that

[TABLE]

and

[TABLE]

Froom the above agreements it follows that $p_{0}=0$ and $p_{1}=1$.

Below we’ll need the special generated function, which we define as follows: for a sequence of functions $\{r_{n}(q)\}$ we put

[TABLE]

where we often will skip arguments $x$ and $q$, except such cases when we will use special values of them. Below in this text we will call such functions as SG-functions (SG = special generated).

It is easy to see that SG-functions are formal power series which are not converges at all. But most of all usual operations with SG-functions (such as adding, production, differentiation and integration on both arguments) does not lead to conflicts when counting coefficients before $x^{n}$.

Let denote

[TABLE]

i.e. the operator $\widehat{}$ denotes SG-function for the sequence of derivatives of $r_{n}(q)$.

Let also:

[TABLE]

where $z^{\underline{k}}=z(z-1)\dots(z-k+1)$ denotes the factorial power $k\geqslant 0$. Therefore, $A$ is a SG-function of $\{1\}$, $B$ is a SG-function of probabilies that graph is connected, $E$ is a SG-function of expectations of components quantity, $M_{k}$ is a SG-function of $k$-th factorial moments of $\nu_{n}$, and $\mathcal{M}_{k}$ is a SG-function of $k$-th factorial moments of $(\nu_{n}-1)$. It’s easy to see that $A$ converges only if $q=1$. Below we’ll see that all of theese series are converges in the same conditions.

2 Basical Relations

Lemma 1**.**

If the relation $n=k=s=0$ does not holds, then

[TABLE]

where the summation is over all integer non-negative $n_{i}$, $k_{i}$.

Proof.

Let consider the set of graphs $\bar{G}_{n}$ with $n$ vertices, where the components are ordered. It is clear that the number $\bar{A}_{n,k,s}$ of such graphs with $n$ vertices, $k$ edges and $s$ components is equal to $s!A_{n,k,s}$.

By the other side, any graph from $\bar{G}_{n}$ with $n$ vertices and $s$ components we can make by getting some ordered partition of the set of $n$ vertices with non-empty parts, which has the volumes $n_{1},\dots,n_{s}$. The number of such partitions is equal to $n!/(n_{1}!\cdots n_{s}!)$. For every set of vertices, included in connected components, we can find the number of connected graphs with $n_{i}$ vertices and $k_{i}$ edges. It is equal to $B_{n_{i},k_{i}}$. By choosing $k_{i}$ in such a way that $k_{1}+\dots+k_{s}=k$, and summing over all partitions of $n$ vertices, we get the equation:

[TABLE]

From this we get (2) for positive $n,n_{i},s$ and non-negative $k$. Extention of this relation for zero values of $n,n_{i}$ and $s$ follows from the previous agreements. ∎

Now we consider the next generated functions, which are exponential by parameter $x$:

[TABLE]

The summation is over integer non-negative $n,k$.

Lemma 2**.**

[TABLE]

Proof.

By multyplying the relation (2) by $x^{n}y^{k}/n!$ we get:

[TABLE]

The last notation denotes a coefficient before $x^{n}y^{k}$ in the series $B(x,y)^{s}$. Now, by summing over integer non-negative $n,k$ for $s>0$ we get the following:

[TABLE]

Note, that by virtue of the agreements this equation stays also true for $s=0$. Finally, by summing over integer non-negative $s$ we get:

[TABLE]

∎

From the relation (3) we can obtain any exact expressions for probabilities of random graph $G(n,p)$. First of all, it is clear that:

[TABLE]

where we suppose that $\binom{m}{k}=0$ for $k>m$. It is easy to see that

[TABLE]

hence,

[TABLE]

From this and from (3) it follows that

[TABLE]

One can see that $B(x,y)$ is the generated function for a sequence [5] where the nulled element is equal to zero.

By putting $y=p/q$ and from the obvious equtions

[TABLE]

we get that for previously defined series $A$ and $B$ the next relations are true:

[TABLE]

Thus, we have

Lemma 3**.**

[TABLE]

This proved equation is the base fact, which we will use anythere below without a special link.

From (1) it follows that:

[TABLE]

and by (4), where we put $y=p/q$, we get following:

[TABLE]

i.e. the formal series $B^{s}/s!$ is SG-function of probabilities $\mathsf{P}\{\nu_{n}=s\}$ for a fixed number $s$ of connected components.

Let us consider two SG-functions and their product:

[TABLE]

One can easily proof the following

Lemma 4** (Convolution Formula).**

For $n\geqslant 0$:

[TABLE]

Further we will use this formula without a special link to it. The next recursion formula for probabilities $p_{n}$ is an anlogue of a recursion formula for a number of connected graphs, that was obtained in [6].

Lemma 5**.**

For any $n\geqslant 1$

[TABLE]

Proof.

By differentiating the relation $A=e^{B}$ by the parameter $x$ we get:

[TABLE]

hence, from the convolution formula it follows that

[TABLE]

Since $p_{0}=0$ and $\frac{k}{n}\binom{n}{k}=\binom{n-1}{k-1}$ follows

[TABLE]

and by replacing $k$ by $n-k$ we get the statement of Lemma. ∎

By analogue we can get a recursive formula for probabilities $\mathsf{P}\{\nu_{n}=s\}$.

Lemma 6**.**

[TABLE]

for $n\geqslant s>1$.

Proof.

Let us denote

[TABLE]

then by (7) we get:

[TABLE]

then by differentiating by $x$ it follows that:

[TABLE]

hence,

[TABLE]

From this and according to $\mathsf{P}\{\nu_{k}=s-1\}=0$ as $k<s-1$ we get Lemma statement. ∎

Lemma 7**.**

The following relations hold:

[TABLE]

[TABLE]

Proof.

If we put $x/q$ instead of $x$ in the definition of series $A$, we get that $A^{\prime}=A(x/q)=e^{B(x/q)}$. On the other side, $A^{\prime}=B^{\prime}e^{B}$. Therefore,

[TABLE]

hence,

[TABLE]

Now we take the corresponding coefficients before $x^{n}$ in theese series and get the relation (11). The ineqution (12) follows from (11) if we left in this summa only the summand with $s=1$. ∎

3 Several Equations

Lemma 8**.**

For $s\geqslant 0$

[TABLE]

and in particulary, $E=AB$.

Proof.

By definition,

[TABLE]

hence by (7) we get:

[TABLE]

∎

Now we consider the connection between moments of $\nu_{n}$ and $\nu_{n}-1$.

Lemma 9**.**

For $s\geqslant 1$

[TABLE]

[TABLE]

Proof.

The first equation is follows from

[TABLE]

and the second one not hard to proof by induction with the obvious start eqation $\mathcal{M}_{0}=A=M_{0}$. ∎

Lemma 10**.**

For $s\geqslant 1$

[TABLE]

Now we ready to use the operator $\widehat{}$ for SG-functions of moments. First of all, we get:

Lemma 11** (Derivative Relashionship Formula).**

If $R$ is a SG-function, then:

[TABLE]

Here and below the single quote without a parameter notation denotes the derivative by $x$, and the derivative by $q$ is marked by index $q$.

The following equations hold.

Lemma 12**.**

[TABLE]

Proof.

By the convolution formula and from $\widehat{A}=0$ we get:

[TABLE]

From here it follows that:

[TABLE]

Hence by Derivative Relashionship Formula we get that:

[TABLE]

so we have the first equation of statement.

To get the equations for $\mathcal{M}_{s}$ it is sufficient to use Lemmas 9, 10 and previous relation. ∎

4 Several Inequations

Let denote by $\gg$ that the inequation $\geqslant$ holds for all coefficient before $x^{n}$ in the considering series. For example, the notation $\sum a_{n}x^{x}\gg\sum b_{n}x^{n}$ means that for all $n$ the inequation $a_{n}\geqslant b_{n}$ holds. It is easy to verify that:

if $X\gg Y$ and $Z\gg 0$, then $XZ\gg YZ$;

if $X\gg Y$ and $V\gg W$, then $X+V\gg Y+W$.

Lemma 13**.**

For $n>0$

[TABLE]

Proof.

Left inequation follows from:

[TABLE]

with help of convolution formula and because of $B^{\prime}\gg 1$. Right inequation follows from:

[TABLE]

∎

Lemma 14**.**

For all $n\geqslant 1$ and $s\geqslant 1$ the following inequations hold:

[TABLE]

Proof.

Left inequation.

[TABLE]

hence, by the convolution formula we get:

[TABLE]

where the last summation we can estimate by the summand as $k=n-s$, and therefore we have:

[TABLE]

here we get the left equation of Lemma statement.

Right inequation. Following relations one can get from the results that were proved above.

[TABLE]

[TABLE]

[TABLE]

Since $n-1\geqslant 0$, $n/2-1\geqslant 0$ for $n\geqslant 2$, $n/2-1\geqslant-1/2$ for $n=1$ it follows that

[TABLE]

and from the equations (13) we get the next inequation:

[TABLE]

Now we get coefficients before $x^{n}$:

[TABLE]

Dividing by n we get:

[TABLE]

It is esy to see that

[TABLE]

where we use derivative of product. Therefore from this and (14) we get

[TABLE]

Hence, dividing by $\mathsf{E}(\nu_{n}-1)^{\underline{s}}+q^{n-1}\mathsf{E}(\nu_{n-1})^{\underline{s}}$ we find the inequation

[TABLE]

Note, that the function $f(t)=(s+t)/(1+t)$ not increases as $t$ increases, if $s\geqslant 1$ and $t>0$. From Lemma 13 it follows that:

[TABLE]

Therefore from this and (16) we get:

[TABLE]

Let $q\leqslant q_{1}\leqslant 1$, and let $\mathsf{E}_{1}=\mathsf{E}|_{q=q_{1}}$. By integrating (17) on the interval $[q;q_{1}]$ we get follows:

[TABLE]

Then we put both sides of this inequation into the argument of function $e^{x}$, and get:

[TABLE]

or

[TABLE]

Then we set $q_{1}=1$ and finally find that

[TABLE]

because as $q_{1}=1$ we have $\mathsf{E}_{1}(\nu_{n}-1)^{\underline{s}}=(n-1)^{\underline{s}}$.

This proofs the right inequation of Lemma. ∎

Note, that if we put $s=n-1$, then we have an equation

[TABLE]

where the right hand side is equal to half of the just proved estimation.

5 Asymptotic behavior of $\nu_{n}$

In this section we consider an asymptotics of moments $\mathsf{E}(\nu_{n}-1)^{\underline{s}}$ as $s$ is fixed and positive. We will study a behavior of moments in the following zones of parameters:

$p\to 0$, $n=\mathrm{const}$; 2. 2.

$q^{n}\to e^{-\alpha}$, where fixed $\alpha\geqslant 0$ and $n\to\infty$; 3. 3.

$q^{n}\to 0$ as $n\to\infty$

1. 3.1

$nq^{n}\to\infty$, 2. 3.2

$nq^{n}\to\alpha>0$, 3. 3.3

$nq^{n}\to 0$ (in this case $p$ can be a positive constant $<1$).

5.1 Asymptotics for $n=\mathrm{const}$

It is easy to prove the following Lemma, besause the minimal graph with $n$ vertices and $s$ components is a forest with $s$ trees.

Lemma 15**.**

If $p\to 0$ and $n=\mathrm{const}$, then for any $s\leqslant n$ the following equation holds: $\mathsf{P}\{\nu_{n}=s\}=F_{s,n}p^{n-s}+O(p^{n-s+1})$. In particular, $p_{n}=n^{n-2}p^{n-1}+O(p^{n})$.

Hence we have the following

Theorem 1**.**

If $p\to 0$ and $n=\mathrm{const}$, then for any $s\leqslant n$:

[TABLE]

5.2 Asymptotics for $q^{n}\to e^{-\alpha}$

Let

[TABLE]

This series converges as $|x|\leqslant e^{-1}$ and this is a generated function for sequence of numbers of labelled trees [4].

Theorem 2**.**

If $q^{n}\to e^{-\alpha}$ as $n\to\infty$, where $\alpha\geqslant 0$, then for $s\geqslant 0$

[TABLE]

In particular, for $\alpha=0$ we have the relation $\mathsf{E}(\nu_{n}-1)^{\underline{s}}\sim n^{s}$.

Proof.

We will use a mathematical induction on the parameter $s$. It is clear that the statement of Theorem holds for $s=0$. Let us suppose that it holds for $s-1$ and will show it for $s\geqslant 1$.

From the relation $x\mathcal{M}^{\prime}_{s}=xB^{\prime}M_{s}=xB^{\prime}BM_{s-1}=Bx\mathcal{M}^{\prime}_{s-1}$ (see Lemma 10) and from the convolution formula we get:

[TABLE]

where

[TABLE]

As $k$ is fixed, one can get next relations: $\binom{n-1}{k}q^{k(n-k)}\sim(nq^{n})^{k}/k!$, $p_{k}\sim k^{k-2}p^{k-1}$ (Lemma 15). And from the induction hypothesis we get: $\mathsf{E}(\nu_{n-k})^{\underline{s-1}}\sim\left(\frac{n}{\alpha}\beta(\alpha e^{-\alpha}\right)^{s-1}$. Hence:

[TABLE]

where we use the asymptotics $np\to\alpha$ and $npq^{n}\to\alpha e^{-\alpha}$, which is follows from Theorem conditions.

So, it is easy to see that $S_{1}/\left(\frac{n}{\alpha}\beta(\alpha e^{-\alpha})\right)^{s}$ as closed to 1 as $k_{0}$ is bigger, because of the convergence of the series $\beta(x)$ for $x=\alpha e^{-\alpha}$.

Let we estimate $S_{2}$. It is clear that $\mathsf{E}(\nu_{n-k}-1)^{\underline{s-1}}\leqslant n^{s-1}$. From this and from the equation (9) we get

[TABLE]

Therefore,

[TABLE]

i.e. the ratio $S_{2}/\left(\frac{n}{\alpha}\beta(\alpha e^{-\alpha})\right)^{s}$ tends to 0 as $k_{0}\to\infty$.

Thus, $\mathsf{E}(\nu_{n}-1)^{\underline{s}}=\left(\frac{n}{\alpha}\beta(\alpha e^{-\alpha})\right)^{s}(1+o(1))$.

In the case of $\alpha=0$ the proof of Theorem is similary, but instead of $\beta(\alpha e^{-\alpha})/\alpha$ we should write 1 at all places. ∎

5.3 Asymptotics for $q^{n}\to 0$

Theorem 3**.**

Let $q^{n}\to 0$ and $nq^{n}\geqslant C$ as $n\to\infty$, where fixed $C>0$, then

[TABLE]

Proof.

We will use an induction by $s$. It is clear that the statement of Theorem holds for $s=0$. Let us suppose that it holds for $s-1$ and will show it for $s\geqslant 1$.

The following relations hold:

[TABLE]

where we use the fact, that $(1-q^{n})\leqslant n(1-q)=np$. Therefore we get that

[TABLE]

or

[TABLE]

From the equation (16) it follows that

[TABLE]

By the induction hypothesis and from Lemma 14 we get that

[TABLE]

where the positive constants $C_{k}$, generally speaking, are depends on the parameter $s$. By putting this inequation into (18) we get:

[TABLE]

Let $q_{1}=\varepsilon^{1/(n-1)}$, where $\varepsilon$ is an arbitrary small positive number, hence $q_{1}^{n-1}=\varepsilon$ and $q<q_{1}$ (it follows from $q^{n}\to 0$). Besides let denote $\mathsf{E}_{1}=\mathsf{E}|_{q=q_{1}}$ as it was above.

Now, we integrate the inequation (19) on the interval $[q;q_{1}]$ and get that

[TABLE]

or

[TABLE]

because $q_{1}^{n-1}=\varepsilon$ and $n(q_{1}-q+q^{2}/2-q_{1}^{2}/2)=n(q_{1}-q)(1-q/2-q_{1}/2)=n(p-p_{1})(p/2+p_{1}/2)\leqslant np^{2}\to 0$. The last expression is follows from $np^{2}\cdot nq^{n}=(np)^{2}e^{n\ln q}\to 0$ and from the conditions of Theorem.

By Theorem 2 we get that

[TABLE]

where $\alpha=-\ln\varepsilon$.

Besides that,

[TABLE]

because $q_{1}^{n-1}=\varepsilon$ and again from Theorem 2. From this and from (20) it follows that

[TABLE]

Therefore, by choosing an arbitrary small $\varepsilon>0$ and using the relationship $\beta(x)\sim x$ as $x\to 0$ we get the relation:

[TABLE]

From Lemma 14 we have $\mathsf{E}(\nu_{n}-1)^{\underline{s}}\geqslant(n-1)^{\underline{s}}q^{s(n-1)}=(nq^{n})^{s}(1+o(1))$. Now we see that Theorem follows from theese both equations. ∎

Let $nq^{n}\to\alpha$, where $\alpha$ is a positive constant. From Theorem 3 we see that $\mathsf{E}(\nu_{n}-1)^{\underline{s}}\to\alpha^{s}$.

It is known that in this case random variable $(\nu_{n}-1)$ tends to Poisson distribution with the parameter $\alpha$.

Thus we have

Theorem 4**.**

If $nq^{n}\to\alpha$ as $n\to\infty$ and $\alpha$ is a fixed positive constant, then for any fixed integer $k\geqslant 1$:

[TABLE]

From Lemma 14 it follows that if $nq^{n}\to 0$, then $\mathsf{E}(\nu_{n}-1)\asymp nq^{n-1}$. So we can conclude that $\nu_{n}$ tends to 1. Below we’ll show an estimation of $p_{n}$ in this case.

6 Several Consequences

Generally, we can conclude that in all zones of parameters $p$ and $n$

[TABLE]

and

[TABLE]

It is easy to verify, because if $np\to\infty$ or $np\to 0$, then it follows that $npq^{n}\to 0$ and $\beta(npq^{n})/p\sim nq^{n}$.

Now we can estimate the probability $p_{n}$ that graph $G(n,p)$ is connected.

[TABLE]

and from Lemma 14 we get:

[TABLE]

If we put $p=\frac{c\ln n}{n}$ and $c>1$, then we have $nq^{n-1}=n\exp\{-c\ln n+O(\ln^{2}n)/n\}=n^{1-c}(1+O(\ln^{2}n)/n)$. Therefore we finally get:

[TABLE]

If $nq^{n}\to\alpha$ (for example, $p=(\ln n+c+o(1))/n$, where $\alpha=e^{-c}$), then from Theorem 4 we get that:

[TABLE]

To estimate $p_{n}$ as $nq^{n}\to\infty$ we now consider the isolating probability. Let $pi_{n}$ be a probability that $G(n,p)$ has an isolated vertex. Let $A_{i}$ be an event that $i$-th vertex is isolated, then from the Inclusion–exclusion principle we get:

[TABLE]

It is easy to see that $\mathsf{P}\{A_{1}\dots A_{k}\}=q^{k(k-1)/2}q^{k(n-k)}$, so

[TABLE]

According to convolution formula we can find that SG-function $PI$ of $\{pi_{n}\}$ is equal to $RT+A$, where $R$ and $T$ are SG-functions of the corresponding sequences $\{r_{n}\}$ and $\{t_{n}\}$, which are defined as follows: $r_{n}=(-1)^{n-1}q^{n(n-1)/2}$ and $t_{n}=1$.

Hence we have

[TABLE]

Thus $PI=A-e^{-x}A=A(1-e^{-x})$.

Since $(1-e^{-x})\leqslant x$ it follows that $PI\ll Ax$, and from the convolution formula we obtain

[TABLE]

It is easy to see that $PI^{\prime}=A^{\prime}(1-e^{-x})+Ae^{-x}=PI\cdot B^{\prime}+A-PI\gg PI\cdot B^{\prime}$, because $A-PI\gg 0$, and from the convolution formula we get:

[TABLE]

or

[TABLE]

So, if $nq^{n}\to\alpha>0$, then $pi_{n}\geqslant\alpha e^{-\alpha}+o(1)$.

And also we have

[TABLE]

Since $PI=A(1-e^{-x})$ it follows that $PIe^{x}=Ae^{x}-A$ and, therefore, $(PI-A)(e^{x}-1)=-PI$. From the relation $e^{x}-1>x$ we get that $PI\gg(A-PI)x$, therefore from the convolution formula we find that $pi_{n}\geqslant(1-pi_{n})nq^{n-1}$, then $(1-pi_{n})\leqslant 1/(nq^{n-1})$ and we get finally

[TABLE]

Now we can combine all obtained results (22), (26), (25), (24) and (27) in the following

Theorem 5**.**

For all $n\geqslant 1$

[TABLE]

And if $nq^{n}\geqslant C>0$ as $n\to\infty$, then we can substitute $nq^{n}$ by $\mathsf{E}(\nu_{n}-1)(1+o(1))$ in theese relations.