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


[what's an upper bound on the size of the largest biclique in random bipartite graph?](https://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



expected size of unbalanced biclique in random bipartite graph
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 $1/2$.


