Wigner Theorem for Random Matrices
Julien Riposo, Ph.D, CQF
Mathematician, Researcher - Quant (Wilmott Award)
We give the main steps for a proof of the theorem.
For any natural integer n, we define the nth Catalan number given by
Let q be a natural integer.
Result 1: The number of Dyck paths of length 2q is C_q.
Result 2: We have the series expansion:
2. Words
Let n be a natural integer. The set {1,2,...,n} can be seen as an alphabet. Any sequence s_1, ..., s_q of letters in {1,2,...,n} is a word. Let M be the set of words from the alphabet {1,2,...,n}. Two words m and m' are equivalent if there is a bijection between their letters (thus 12321 and 42524 are equivalent), and we write m~m'.
the binary relation ~ is an equivalence relation, that is it is reflexive (m~m), symmetrical (m~m' if and only if m'~m) and transitive (if m~m' and m'~m'', then m~m''). There exists a partition of M made of equivalence classes, each of these having a representant (arbitrary).
Let M_q,p be the set of such representants of the words of length q+1 having p distinct letters and such that each pair of letters is consecutively repeated at least twice, no matter the order.
Result 3: the cardinal (number of elements) of M_2q,q+1 is C_q.
3. Wigner Theorem
Let M_n be a n x n symmetric random matrix such that its elements are Z_i,jI/sqrt(n), and the Z_i,j are all independent and identically distributed random variables with finite moments (see below Wigner Theorem for more precisions). We let λ_n,i be the ith eigenvalue (increasing order) of M_n.
Let
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Consider \epsilon>0. With the Bienaymé-Tchebychev inequality, we have:
It is tedious to prove that the right-hand-side tends to 0, so that the sequence (γ_n,2q)_n in fact tends to the number C_q (as (E(γ_n,2q))_n does).
Using the Markov and Bienaymé-Tchebychev inequalities, we can prove that
for any positive integer k and B>0.
Now, let f be a continuous and bounded function on R. According to the Wierstrass approximation theorem, there exists a polynomial P such that (uniform norm):
We finally have:
Three points:
and bearing with this in mind, the second term tends to 0.
Hence, we deduce the theorem:
For all the theoretical justifications, find them in the formidable book:
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