# How could I obtain this approximation of the May-Wigner theorem?

I'm trying to understand the complete proof of the May-Wigner theorem. We have a real random $$n\times n$$ matrix $$B$$ with its non-zero elements $$B_{ij}$$ are chosen independiently from a fixed distribution of mean 0 and variance $$\alpha^2$$. The porcentage of non-zero elements in $$B$$ is $$C$$, $$0. We consider a directed graph $$G$$ with $$n$$ vertices and one directed edge from vertex j to vertex I for each non-zero entry $$B_{ij}$$. Moreover, $$B^N$$ is the product of $$N$$ times the matrix B ($$n<) and its elements $$B_{ij}^N$$ are sums of non-zero N-fold products of entries of $$B$$: $$B_{ij_1}B_{j_1j_2}...B_{j_{N-1}j}$$ where subcrits may be repeated. Those products may be interpreted as a path of length $$N$$, with possible overlaps, from vertex $$j$$ to vertex $$i$$ in the graph $$G$$. Let $$b$$ be one non-zero summand of $$B_{ij}^N$$ and $$nC>1$$. We shall call a second such summand $$b'$$ a matcking summand for $$b$$ if their product involves only even exponents when written as a product of powers of distinct entries of $$B$$ (and in this case we know that $$bb'$$ is non-zero, because all odd moments are zero of the distribution of the hypothesis). We can obtain matching summands for $$b$$ in this way: first, the path in $$G$$ associated with $$b$$ may contain several subpaths from a vertex $$s$$ to a vertex $$r$$. We allow but do not require $$r=s$$. If there are t such subpaths ($$t\le N$$). All distinct, then $$t\!$$ matching summands for $$b$$ may be obtained by permuting these subpaths (no repetitions). If some of these subpaths are repeated, then there are less than $$t\!$$ such matching summands (because of the combination with repetition formule). We obtain an upper bound on the expected number of such summands (of $$B_{ij}^2$$, this is the objective of this section) by first fixing $$r$$ and $$s$$ (there are $$n^2$$ such pairs of species) and then maximizing the product $$B$$ of the probability of obtaining $$t$$ such subpaths by $$t!$$. An assymptotic counting argument yields: $$B\sim(\frac{N}{tn})^{2t}t^t$$. I'm trying to understand this last conclusion, if $$B$$ refers to the probability of obtaining $$t$$ subpaths multiplied by $$t!$$, then the $$t^t$$ on the right would be explained by the Stirling's approximation: $$t\!\sim t^t$$ for large $$t$$. But I don't know how to conclude the rest. The proof continues by the maximization of this conclusion by $$t\sim\sqrt\log n$$...

• Paragraph breaks are your friend... – YuiTo Cheng Apr 2 '19 at 14:18