Expectation number of cycles in a Erdős–Rényi random directed graph $G(n,p)$

Solution 1:

Let $X$ be the total number of cycles, $X_k$ the number of cycles of length $k$ in $G$, and $Y_k$ the number of cycles of length $k$ in a random graph of size $k$.

Note that:

• ${\sf E}[X] = \sum_{k=2}^n {\sf E}[X_k]$.
• ${\sf E}[X_k] = {n \choose k} {\sf E}[Y_k]$.
• ${\sf E}[Y_k] = (k!/k)p^k = (k-1)!p^k$.

The last one is because a cycle of length $k$ is a permutation of numbers from $1$ to $k$ but the first vertex does not matter (a cycle does not have a first vertex). The probability of each edge being present is $p$, so the probability of $k$ edges being present is $p^k$.

We have:

\begin{align*} {\sf E}[X] &= \sum_{k=2}^n E[X_k] \\ &= \sum_{k=2}^n {n \choose k} E[Y_k] \\ &= \sum_{k=2}^n {n \choose k} (k-1)! p^k \\ &= \sum_{k=2}^n \frac{n!}{k!(n-k)!} (k-1)! p^k \\ &= \sum_{k=2}^n \frac{n!}{(n-k)!k} p^k \\ &= \sum_{k=2}^n \frac{n\ldots (n-k+1)}{k} p^k \\ \end{align*}

For a simple lower bound we can drop the first half of the sum: \begin{align*} &\geq \sum_{k=n/2}^{n} \frac{n\ldots (n-k+1)}{k} p^k \\ &\geq \sum_{k=n/2}^{n} \frac{(n/2)^{k}}{n} p^k \\ &\geq \frac{1}{n}\sum_{k=n/2}^{n} (np/2)^{k} \\ &= \frac{(\frac{np}{2})^{n+1}-(\frac{np}{2})^{\frac{n}{2}}}{\frac{np}{2}-1} \end{align*}