Past exam of the random matrix theory course of the University of Oxford / 2025 / 1 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

W^{T} = W

, so the spectral theorem for real symmetric matrices gives real eigenvalues and an orthonormal basis of eigenvectors. Moreover, for every real vector

v

,

v^{T} W v = \frac{1}{n} v^{T} X X^{T} v = \frac{1}{n} ∥ X^{T} v ∥^{2} \geq 0.

Thus

W

is a positive semidefinite matrix. For any nonzero eigenvector

v

with eigenvalue

λ

,

λ ∥ v ∥^{2} = v^{T} W v \geq 0, λ \in [0, \infty) .

This matrix is a scaled example of the Wishart distribution.

ii

Solution

Let

g_{i} = (X_{i 1}, \dots, X_{in})

be row

i

of

X

, viewed as a real vector. These are independent standard Gaussian random vectors, and

(X X^{T})_{ij} = g_{i} \cdot g_{j}

. The trace expression for the empirical spectral measure gives

m_{k} (n) = n^{- k - 1} \sum_{i_{1}, \dots, i_{k}} (g_{i_{1}} \cdot g_{i_{2}}) \dots (g_{i_{k}} \cdot g_{i_{1}}) .

We need the norm moments

E ∥ g_{i} ∥^{2} = n, E ∥ g_{i} ∥^{4} = n (n + 2), E ∥ g_{i} ∥^{6} = n (n + 2) (n + 4) .

For completeness,

S = ∥ g_{i} ∥^{2}

has the chi-squared distribution with

n

degrees of freedom. By independence and the Gaussian integral, its moment-generating function is

E e^{s S} = (1 - 2 s)^{- n /2}

for

s < 1/2

. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,

E [m_{1} (n)] = n^{- 2} i \sum E ∥ g_{i} ∥^{2} = 1.

For the second moment, if

i \neq = j

, conditional expectation given

g_{i}

gives

E [(g_{i} \cdot g_{j})^{2} ∣ g_{i}] = ∥ g_{i} ∥^{2}

, hence an unconditional value

n

. Separating equal and unequal row indices yields

E [m_{2} (n)] = n^{- 3} {n n (n + 2) + n (n - 1) n} = 2 + \frac{1}{n} .

For the third moment, the summand is

(g_{i} \cdot g_{j}) (g_{j} \cdot g_{k}) (g_{k} \cdot g_{i})

. There are three disjoint cases.

If $i, j, k$ are all different, average over $g_{k}$ first. Its covariance matrix is the identity, so the conditional expectation of the last two factors is $g_{j} \cdot g_{i}$ . The summand's expectation is therefore $E (g_{i} \cdot g_{j})^{2} = n$ . There are $n (n - 1) (n - 2)$ such triples.
If exactly two indices agree, cyclic symmetry reduces the calculation to $i = j \neq = k$ . The product is $∥ g_{i} ∥^{2} (g_{i} \cdot g_{k})^{2}$ , whose conditional expectation given $g_{i}$ is $∥ g_{i} ∥^{4}$ . Its expectation is $n (n + 2)$ , and there are $3 n (n - 1)$ such triples.
If all three indices agree, the product is $∥ g_{i} ∥^{6}$ , with expectation $n (n + 2) (n + 4)$ . There are $n$ choices.

Consequently

E [m_{3} (n)] = n^{- 4} {n^{2} (n - 1) (n - 2) + 3 n^{2} (n - 1) (n + 2) + n^{2} (n + 2) (n + 4)} = 5 + \frac{6}{n} + \frac{4}{n ^{2}} .

The three limiting expected values are therefore

n \to \infty lim E [m_{k} (n)] = ⎩ ⎨ ⎧ 1, 2, 5, k = 1, k = 2, k = 3.

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