Question

I am discovering random graph and I am trying to prove the following result. This is a follow-up on a previous question of mine

http://mathoverflow.net/questions/115748/whats-an-upper-bound-on-the-size-of-the-largest-biclique-in-random-bipartite-gra

Let G(X∪Y,p) be a random bipartite graph where the set of vertices is X∪Y, X and Y both have cardinality n and p is the proba of adding an edge between each node in X and each node in Y. p∈(0,1) is independent of $n$. A set $E_1∪E_2$, $E_1⊂X$ and $E_2⊂Y$ is a biclique if for each node $x∈X$ and each node $y∈Y$, there is an edge between $x$ and $y$.

Let $E=E_1∪E_2$ be a biclique satisfying ${Expectation}(∣E_1∣) \leq {Expectation}(∣E_2∣)$. The conjecture is that

for all α>0, Pr{∣E_1∣ is greater than α×n}→0 as n→∞.

Could any of you help me on this?

Thanks a lot!

If instead of requiring ${Expectation}(∣E_1∣) \leq {Expectation}(∣E_2∣)$, I were to focus on balanced biclique, i.e., require $∣E_1∣ = ∣E_2∣$, then the result is already known. Clearly, the result would also hold if I were to assume that $∣E_1∣ \leq ∣E_2∣$ (given that from such a biclique, I could "extract" a balanced biclique of size $∣E_1∣$).

Oliver

Answer 1

Thanks for the clarification. This set-up has enough flexibility that we can recover the star counterexample from before.

Almost every random graph has a vertex in each class of degree about $pn$. So we can almost always choose $E$ to be a large star, and we can ensure the centre of the star is in each class about equally often. Then the expectations of the class sizes are almost equal, so with a bit of tweaking we satisfy the condition on the choice of $E$; but $E_1$ will be too large with probability almost $1/2$.