Threshold for apperance of quite short paths/cycles in random graphs

Question

We say that a graph $G$ is distributed with $\mathcal{G}_{n,p}$ if it is a graph on $n$ vertices, and for which each of the ${n\choose 2}$ possible edges is chosen independently of the other edges and with probability $p$.

A monotone property $P$ of a graph is a set of graphs (on $n$ vertices) that is closed from above (that is, if $G\in P$ and $G\subseteq H$ then $H\in P$.

A function $f(n)$ is said to be a threshold for a property $P$ if for any $p(n)=\omega(f(n))$, $G\sim\mathcal{G}_{n,p}$ has $P$ asymptotically almost surely (a.a.s.), and for any $p(n)=o(f(n))$, $G\sim\mathcal{G}_{n,p}$ does not have $P$ a.a.s.

For example, if $P$ is "has a triangle as a subgraph", then $P$ is clearly monotone, and $f(n)=n^{-1}$ is a threshold for $P$. $f(n)=\frac{\ln{n}+\ln\ln{n}}{n}$ is a threshold for the Hamiltonicity property (in a stronger sense).

My question is this: what are the thresholds for the properties of having "quite short" paths or cycles? By "quite short" I mean of length $\Theta(n^\varepsilon)$ for some $0<\varepsilon<1$, or of length $\Theta(\ln{n})$.

Answer 1

For example you should be able to show that $f(n) = n^{-3/2}$ is the threshold for the property of having a path of length $2$, by a birthday paradox argument. This argument or something similar probably also holds for paths of any fixed finite length, but I'm not sure how to show it.

Answer 2

The threshold function is $f(n) = 1/n$ for each of your cases. In fact, Łuczak proved that if $v_1$ denotes the number of vertices of degree 1 in $G(n,p)$, then for $n p \to \infty$, w.h.p. $G(n,p)$ contains cycles of all lengths $r$, $3 \leq r \leq n - (1+\epsilon)v_1$. In particular, since $v_1$ is concentrated around its mean for this range of $p$, we have that: for any fixed $\delta \in (0,1)$, if $np \to \infty$, then $G(n,p)$ contains cycles of all lengths $r$, $3 \leq r \leq \delta n$.