Disjoint union of random graphs again a random graph?

Question

Let $G_{n,p}, n\in \mathbb{N}, p\in(0,1)$ be the binomial random graph, i.e. a graph on $n$ vertices where an edge is in $G_{n,p}$ with probability $p$. Also, let $q\in (0,1)$.

Can one regard $G_{n,q}$ as a disjoint union of many $G_{n,p}$?

Answer 1

If one is willing to drop the disjointness requirement, then the answer is yes, for appropriate values of $q$. For example, if $q = 2p - p^2$, then $G_{n,q}$ will have the distribution of the union of two random graphs with distribution $G_{n,p}$. (Since each edge $e$ has probability $p$ of being in either of the two graphs with distribution $G_{n,p}$, the probability that $e$ is an at least one of them is $2p - p^2$.)

In general, if $G_1$ has the distribution of $G_{n,p_1}$, $G_2$ has the distribution of $G_{n,p_2}$, and $q = p_1 + p_2 - p_1 p_2$, then a random graph with distribution $G_{n,q}$ will have the distribution of the union of $G_1$ and $G_2$.

The fact that it is possible to couple $G_{n,p}$ and $G_{n,q}$ in this way is often useful. For example, this type of argument can be used to show that if $\mathcal{P}$ is an increasing property of graphs, then the probability that a random graph with distribution $G_{n,p}$ is in $\mathcal{P}$ is increasing in $p$.

Answer 2

Note: I read your question as having chosen a single $p$ and $q$ ($p < q$) in advance. If this is not what you intended, then my answer may not be relevant.

If there is a large difference between $p$ and $q$, then the random graph $G_{n,q}$ possesses properties that disjoint unions of $G_{n,p}$ do not.

As an example, suppose $nq$ tends to a constant greater than $1$ while $np$ is less than $1$. The random graph $G_{n,q}$ almost surely has a giant connected component. Each component of $G_{n,p}$, however, is almost surely of at most logarithmic size, so their disjoint union cannot possibly have a giant component.

(See Wikipedia for more on the evolution of random graphs.)