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There is a family of random graphs $G(n, p)$ with $n$ nodes (due to Gilbert). Each possible edge is independently inserted into $G(n, p)$ with probability $p$. Let $X_k$ be the number of cliques of size $k$ in $G(n, p)$.

I know that $\mathbb{E}(X_k)=\tbinom{n}{k}\cdot p^{\tbinom{k}{2}}$, but how do I prove it?

How to show that $\mathbb{E}(X_{\log_2n})\ge1$ for $n\to\infty$? And how to show that $\mathbb{E}(X_{c\cdot\log_2n}) \to 0$ for $n\to\infty$ and a fixed, arbitrary constant $c>1$?

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So basically there are three questions involved.

I know that $E(X_k)=\tbinom{n}{k}\cdot p^{\tbinom{k}{2}}$, but how do I prove it?

You use the linearity of expectation and some smart re-writing. First of all, note that $$ X_k = \sum_{T \subseteq V, \, |T|=k} \mathbb{1}[T \text{ is clique}].$$ Now, when taking the expectation of $X_k$, one can simply draw the sum out (due to linearity) and obtain $$ \mathrm{E}(X_k) = \sum_{T \subseteq V, \, |T|=k} \mathrm{E}(\mathbb{1}[T \text{ is clique}]) = \sum_{T \subseteq V, \, |T|=k} \mathrm{Pr}[T \text{ is clique}]$$ By drawing out the sum, we eliminated all possible dependencies between subsets of nodes. Hence, what is the probability that $T$ is a clique? Well, no matter what $T$ consists of, all edge probabilities are equal. Therefore, $\mathrm{Pr}[T \text{ is clique}] = p^{k \choose 2}$, since all edges in this subgraph must be present. And then, the inner term of the sum does not depend on $T$ anymore, leaving us with $\mathrm{E}(X_k) = p^{k \choose 2} \sum_{T \subseteq V, \, |T|=k} 1 = {n \choose k} \cdot p^{k \choose 2}$.

How to show that for $n\rightarrow\infty$: $E(X_{\log_2n})\ge1$

I am not entirely sure whether this is even correct. Applying a bound on the binomial coefficient, we obtain

$$E(X_{\log n}) = {n \choose \log n} \cdot p^{\log n \choose 2} \leq \left(\frac{n e p^\frac{(\log n)}{4}}{\log n}\right)^{\log n} = \left(\frac{ne \cdot n^{(\log p) / 4}}{\log n}\right)^{\log n}.$$ (Note that I roughly upper bounded $p^\frac{-1 + \log n}{2}$ by $p^\frac{\log n}{4}$.) However, now one could choose $p = 0.001$, and obtain that $\log_2 0.001 \approx -9.96$, which makes the whole term go to $0$ for large $n$. Are you maybe missing some assumptions on $p$?

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  • $\begingroup$ Is that right? $E(X_{\log n}) = {n \choose \log n} \cdot p^{\log n \choose 2} \leq \left(\frac{n e p^\frac{(\log n)}{4}}{\log n}\right)^{\log n}$. Does it not have to be $$E(X_{\log n}) = {n \choose \log_2 n} \cdot p^{\log_2 n \choose 2} \leq \left(\frac{n e}{\log_2 n}\right)^{\log n}\cdot p^\frac{(\log_2 (n)\cdot e)^2}{4}$$ but now i dont know how to go on $\endgroup$
    – user1374864
    May 28, 2012 at 20:25
  • $\begingroup$ I applied said bound only on ${n \choose \log n}$. For $p$, you can observe that ${\log n \choose 2} = (\log n) (\log n -1)/2$. Now since $p \leq 1$, we want to make its exponent smaller (convince yourself why). For reasonably big $n$, we have that $(\log n) (\log n -1)/2 > (\log n)^2 / 4$. Therefore the above computation should be correct... $\endgroup$
    – HdM
    May 28, 2012 at 20:47
  • $\begingroup$ What is with the third question? $\endgroup$
    – Queue
    May 28, 2012 at 22:45
  • $\begingroup$ It suffers the same problem as the second question. Sorry, I should have clarified that. $\endgroup$
    – HdM
    May 28, 2012 at 23:32

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