Fluctuations from the Semicircle LawLecture 1
Ioana Dumitriu
University of Washington
Women and Math, IAS 2014
May 20, 2014
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 1 / 30
1 Review
2 Fluctuations
3 Calculation of the Variance
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 2 / 30
Review
Random (Real) Facts
Distributions are generalized functions, better understoodthrough the effect they have on functionsProbability distributions define random variables viacharacteristic functions:
X ∼ F if ∀[a, b] ∈ R , P[X ∈ [a, b]] = F([a, b]) .
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 3 / 30
Review
Random (Real) Facts
A probability distribution F may be given by a (positive) function(the probability density function) f with
∫R f (x)dx = 1. We say
dµ(x) = f (x)dx.
In this case P[X ∈ [a, b]] =∫ b
a f (x)dx.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 4 / 30
Review
The Moment Method
A “nice” distribution is described by the collection of its moments,{E[Xk] , ∀k ∈ Z, k ≥ 0}. This leads to the “moment method”.
More on the relationship between the distribution and itsmoments in today’s Review Session.
Convergence of moments is weak convergence, i.e., convergencein distribution. Stronger: in probability and almost surely.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 5 / 30
Review
Wigner Matrices
n× n real (or complex, quaternion) matrices W;symmetric: W = WT (or Hermitian W = W∗, self-dual W = WD);entries are independent up to symmetry (wij = wji);entries are identically distributed up to symmetry (all wij withi < j are equidistributed, resp. all wii are equidistributed);
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 6 / 30
Review
Wigner Matrices
the distributions have moments of all orders; in particular, E(Z4)is the 4th moment;all variables are centered (expectation 0) and all variances are 1(could also consider variance σ2 on the diagonal).
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 7 / 30
Review
Wigner Matrices
... actually, we consider the normalized Wigner matrices Wn = 1√n W.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 8 / 30
Review
The Semicircle Law
n = 500; A = rand(n);A = (A + A′)/√
n;hist(eig(A))semicircle
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Semicircle law with n=500
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 9 / 30
Review
Convergence to the Semicircle
To understand convergence through experiments:Convergence in distribution: pick an n× n random Wignermatrix W, pick one of its eigenvalues at random. Repeat manytimes. Plot histogram.
Almost surely: pick a single matrix W, and plot a histogram of allits eigenvalues.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 10 / 30
Review
LLN and CLT
1−D: let x1, x2, . . . , xn, . . . be independent samples from a distributionwith mean µ and variance σ2.
Law of Large Numbers: 1n
n∑i=1
xi − µ→ 0 as n→∞.
Central Limit Theorem:
n∑i=1
xi − nµ√
nσ∼ N(0, 1) .
What about matrices?
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 11 / 30
Review
LLN for Matrices
The Semicircle Law is akin to a Law of Large Numbers.Showed:
1nE(tr(Wk)) =
1nE
(n∑
i=1
λki
)→{
0, k odd,Ck/2, k even.
On the right hand side are the moments of the semicircle distribution,with density s(x) = 1
2π
√4− x2.
Convergence of moments means that the expected distribution of arandom eigenvalue converges in distribution to the semicircle law.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 12 / 30
Review
LLN for Matrices
What this implies immediately is that for all reasonable functionsf : [−2, 2]→ R,
1nE
(n∑
i=1
f(λi)
)→∫ 2
−2f(x)s(x)dx .
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 13 / 30
Review
Exact expressions for the distribution
The actual expected distributions can be computed, for theGOE/GUE/GSE, for any n. The expressions are not very complicated.
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5Distribution of one random eigenvalue, Wigner case, n=1
Figure: Distribution of one random eigenvalue, n = 1
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 14 / 30
Review
Exact expressions for the distribution
The actual expected distributions can be computed, for theGOE/GUE/GSE, for any n. The expressions are not very complicated.
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Distribution of one random eigenvalue, Wigner case, n=2
Figure: Distribution of one random eigenvalue, n = 2
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 15 / 30
Review
Exact expressions for the distribution
The actual expected distributions can be computed, for theGOE/GUE/GSE, for any n. The expressions are not very complicated.
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Distribution of one random eigenvalue, Wigner case, n=6
Figure: Distribution of one random eigenvalue, n = 6
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 16 / 30
Review
Exact expressions for the distribution
The actual expected distributions can be computed, for theGOE/GUE/GSE, for any n. The expressions are not very complicated.
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5Distribution of one random eigenvalue, Wigner case, n=100
Figure: Distribution of one random eigenvalue, n = 100
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Fluctuations
Bumps = Fluctuations
The actual expected distributions can be computed, for theGOE/GUE/GSE, for any n. The expressions are not very complicated.
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5Distribution of one random eigenvalue, Wigner case, n=1,2,6,100
n=1
n=2
n=6
n=100
Figure: Distribution of one random eigenvalue, n = 1, 2, 6, 100
How can we compute the fluctuation?Compute moments more carefully.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 18 / 30
Fluctuations
Bumps = Fluctuations
The actual expected distributions can be computed, for theGOE/GUE/GSE, for any n. The expressions are not very complicated.
−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5Distribution of one random eigenvalue, Wigner case, n=1,2,6,100
n=1
n=2
n=6
n=100
Figure: Distribution of one random eigenvalue, n = 1, 2, 6, 100
How can we compute the fluctuation?Compute moments more carefully.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 19 / 30
Fluctuations
Format of CLT
Recall that in 1−D,
n∑i=1
Xi − nµ√
nσ=
n∑i=1
Xi − E(
n∑i=1
Xi
)√
nσ→ N(0, 1)
• For Wigner matrices we will have something similar:
n∑i=1
f (λi)− E(
n∑i=1
f (λi)
)σf
→ N(0, 1) ,
provided that f is “smooth enough”.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 20 / 30
Fluctuations
Format of CLT
Recall that in 1−D,
n∑i=1
Xi − nµ√
nσ=
n∑i=1
Xi − E(
n∑i=1
Xi
)√
nσ→ N(0, 1)
For Wigner matrices we will have something similar:
n∑i=1
f (λi)− E(
n∑i=1
f (λi)
)σf
→ N(0, 1) ,
provided that f is “smooth enough”.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 21 / 30
Fluctuations
Format of CLT
Start with the simplest smooth functions, f (x) = xk; must then showthat for a Wigner real matrix Wn
Xn,k := tr(Wkn)− E(tr(Wk
n)) ,
has the property that
Xn,k√Var(Xn,k)
→ N(0, 1) ,
where N(0, 1) is the standard normal variable, and convergence is indistribution.
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Calculation of the Variance
Variance
Need to show that all moments of Xn,k/√
Var(Xn,k) converge to thoseof the standard normal variable.
The first step is to calculate the variance
Var(Xn,k) = E((
tr(Wkn))2)−(E(tr(Wk
n)))2
.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 23 / 30
Calculation of the Variance
Expansion of the trace power
Writetr(Wk
n) =∑I∈I
wI ,
whereI := {I = (i1, i2, . . . , ik), | 1 ≤ i1, i2, . . . , ik ≤ n} ,
that is, ordered k-tuples; we also use the notation
wI = wi1i2wi2i3 . . .wiki1 .
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 24 / 30
Calculation of the Variance
Expansion of the trace power
Then
Var(Xn,k) = E((
tr(Wkn))2)−(E(tr(Wk
n)))2
=∑I,J∈I
E(wIwJ
)− E(wI)E(wJ) .
To each I ∈ I there corresponds a graph GI, with vertex labels{i1, . . . , ik}, with v vertices and e edges, having an edge betweenvertices ij and il if they occur consecutively in I (loops are ok).
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 25 / 30
Calculation of the Variance
Graphs and labels
Consider the graph which is the union of the two graphscorresponding to I and J (for a given I, J).
In the walks corresponding to I and J, edges may be repeated;loops are also possible.Total # of edges in the walks, with multiplicities, = 2k.Enough to consider the case when the graph is connected;otherwise wI, wJ independent and
E(wIwJ)− E(wI)E(wJ) = 0 .
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Calculation of the Variance
Graphs and labels
If any edge has multiplicity 1 in the union of the walks,E(wIwJ) = 0 = E(wI)E(wJ).... therefore only need to consider walks where all edges arerepeated.
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Calculation of the Variance
Graphs and labels
SoTotal # of edges in the union of walks with multiplicities = 2k,every edge repeated.The graph is connected.So total # of actual edges e ≤ k and e ≥ v− 1, v ≤ k + 1.
Given i1, i2, . . . , ik, j1, j2, . . . , jk, the total number of such graphs isindependent of n.Asymptotics are given by those graphs for which v is as large aspossible.
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Calculation of the Variance
First Attempt: v = k + 1
No such terms are relevant.
Ioana Dumitriu (UW) Fluctuations from the Semicircle Law Lecture 1 May 20, 2014 29 / 30
Calculation of the Variance
First Attempt: v = k + 1
Must have v = k + 1, e = v− 1; so join graph is a tree on whicheach edge is walked on twice.Hence the two closed walks that form it are trees on which eachedge is walked on twice.No edge overlap, so wI is independent from wJ.Term contributes 0 to covariance.
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